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Response:

Calculated Shard: 164 (from laksa094)

2. Crawled Status Check

Query:

curl -X POST \
  'http://laksa164.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://www.khanacademy.org/science/physics/v/intro-to-springs-and-hooke-s-law\')), getAhrefsUnparsedNoserviceFromURL(\'https://www.khanacademy.org/science/physics/v/intro-to-springs-and-hooke-s-law\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/www.khanacademy.org\/science\/physics\/v\/intro-to-springs-and-hooke-s-law","crawl_time":1775948244,"first_indexed_time":1450324966,"http_code":200,"src_unparsed":"org,khanacademy!www,\/science\/physics\/v\/intro-to-springs-and-hooke-s-law s443","src_root_hash":"3750796715567959364","history_drop_reason":null,"meta_title":"Intro to springs and Hooke's law (video) | Khan Academy","meta_descriptions":["Discover the phenomena of springs and Hooke's Law. Explore how force applied to a spring results in compression or elongation, and how this relationship is linear. Uncover the concept of restorative force and how it counteracts applied force, keeping our spring in equilibrium."],"attrs_boilerpipe_text":"Video transcript\nLet's learn a little\nbit about springs. So let's say I have a spring. Let me draw the ground so that\nwe know what's going on with the spring. So let me see, this\nis the floor. That's the floor, and\nI have a spring. It's along the floor. I'll use a thicker one, just\nto show it's a spring. Let's say the spring looks\nsomething like this. Whoops, I'm still using\nthe line tool. So the spring looks like this. This is my spring, my amazingly\ndrawn spring. Let's say at this end it's\nattached to a wall. That's a wall. And so this is a spring when I\ndon't have any force acting on it, this is just the natural\nstate of the spring. And we could call this, where it\njust naturally rests, this tip of the spring. And let's say that when I were\nto apply a force of 5 Newtons into the spring, it looks\nsomething like this. Redraw everything. So when I apply a force of 5\nNewtons-- I'll draw the wall in magenta now. When I apply a force\nof 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every\nday or a sofa. So let's say it compresses\nto here. If this was the normal resting--\nso this is where the spring was when I applied no\nforce, but when I applied 5 Newtons in that direction, let's\nsay that this distance right here is 10 meters. And so a typical question that\nyou'll see, and we'll explain how to do it, is a spring\ncompresses or elongates when you apply a certain force\nby some distance. How much will it compress when\nyou apply a different force? So my question is how much will\nit compress when I apply a 10-Newton force? So your intuition that it'll\ncompress more is correct, but is it linear to how much\nI compress it? Is it a square of how\nmuch I compress it? How does it relate? I think you probably\ncould guess. It's actually worth\nan experiment. Or you could just keep\nwatching the video. So let's say I apply\na 10-Newton force. What will the spring\nlook like? Well, it'll be more\ncompressed. Drop my force to 10 Newtons. And if this was the natural\nplace where the spring would rest, what is this distance? Well, it turns out that\nit is linear. What do I mean by linear? Well, it means that the more\nthe force-- it's equally proportional to how much the\nspring will compress. And it actually works\nthe other way. If you applied 5 Newtons in this\ndirection, to the right, you would have gone 10 meters\nin this direction. So it goes whether you're\nelongating the spring or compressing the spring within\nsome reasonable tolerance. We've all had this experience. If you compress something too\nmuch or you stretch it too much, it doesn't really go back\nto where it was before. But within some reasonable\ntolerance, it's proportional. So what does that mean? That means that the restoring\nforce of the spring is minus some number, times the\ndisplacement of the spring. So what does this mean? So in this example right here,\nwhat was the displacement of the spring? Well, if we take positive x to\nthe right and negative x to the left, the displacement\nof the spring was what? The displacement, in this\nexample right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the\nrestorative force is going to be equal to minus K times\nhow much it's distorted times minus 10. So the minuses cancel out,\nso it equals 10K. What's the restorative force\nin this example? Well, you might say, it's 5\nNewtons, just because that's the only force I've drawn here,\nand you would be to some degree correct. And actually, since we're doing\npositive and negative, and this 5 Newton is to the\nleft, so to the negative x-direction, actually, I should\ncall this minus 5 Newtons and I should call this\nminus 10 Newtons, because obviously, these are vectors and\nwe're going to the left. I picked the convention that\nto the left means negative. So what's the restorative\nforce? Well, in this example-- and we\nassume that K is a positive number for our purposes. In this example, the restorative\nforce is a positive number. So what is the restorative\nforce? So that's actually the force,\nthe counteracting force, of the spring. That's what this formula\ngives us. So if this spring is stationary\nwhen I apply this 5-Newton force, that means that\nthere must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring\nwould keep compressing. And if the force was more than 5\nNewtons, the spring would go back this way. So the fact that I know that\nwhen I apply a 5-Newton force to the left, or a negative\n5-Newton force, the spring is no longer moving, it means that\nthere must be-- or no longer accelerating, actually,\nit means that there must be an equal and opposite force to\nthe right, and that's the restorative force. Another way to think about it is\nif I were to let-- well, I won't go in there now. So in this case, the restorative\nforce is 5 Newtons, so we can\nsolve for K. We could say 5 is\nequal to 10K. Divide both sides by 10. You get K is equal to 1\/2. So now we can use that\ninformation to figure out what is the displacement\nwhen I apply a negative 10-Newton force? When I push the spring\nin with 10 Newtons in the leftward direction? So first of all, what's the\nrestorative force here? Well, if the spring is no longer\naccelerating in either direction, or the tip of\nthe spring is no longer accelerating in either\ndirection, we know that the restorative force must be\ncounterbalancing this force that I'm compressing\nwith, right? The force that the spring wants\nto expand back with is 10 Newtons, positive\n10 Newtons, right? And we know the spring constant,\nthis K for this spring, for this material,\nwhatever it might be, is 1\/2. So we know the restorative force\nis equal to 1\/2 times the distance, right? And the formula is\nminus K, right? And then, what is\nthe restorative force in this example? Well I said it's 10 Newtons, so\nwe know that 10 Newtons is equal to minus 1\/2x. And so what is x? Well, multiply both sides\nby minus 1\/2, and you get minus 20. I'm sorry, multiply both sides\nby minus 2, you get minus 20 is equal to x. So x goes to the\nleft 20 units. So that's all that\nit's telling us. And this law is called Hooke's\nLaw, and it's named after-- I'll read it-- a physicist in\nthe 17th century, a British physicist. And he figured out\nthat the amount of force necessary to keep a spring\ncompressed is proportional to how much you've compressed it. And that's all that\nthis formula says. And that negative number,\nremember, this formula gives us the restorative force. So it says that the force is\nalways in the opposite direction of how much\nyou displace it. So, for example, if you were\nto displace this spring in this direction, if you were to\napply a force and x were a positive and you were to go in\nthat direction, the force-- no wait, sorry. This is where the\nspring rests. If you were to apply some force\nand take the spring out to here, this negative number\ntells us that the spring will essentially try to pull back\nwith the restorative force in the other direction. Let's do one more problem\nand I think this will be clear to you. So let's say I have a spring,\nand all of these problems kind of go along. So let's say when I apply a\nforce of 2 Newtons, so this is what I apply when I apply\na force of 2 Newtons. Well, let's say it this way. Let's say when I stretch\nthe spring. Let's say this is the spring,\nand when I apply a force of 2 Newtons to the right, the spring\ngets stretched 1 meter. So first of all, let's\nfigure out what K is. So if the spring is stretched\nby 1 meter, out here, its restorative force will be 2\nNewtons back this way, right? So its restorative force, this\n2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter,\nso then we multiply both sides by negative 1, and we\nget K is equal to minus 2. So then we can use Hooke's Law\nto note the equation for this-- to figure out the\nrestorative force for this particular spring, and\nit would be minus 2x. And then I said, well, how much\nforce would I have to apply to distort the\nspring by 2 meters? Well, it's 2 times\n2, it would be 4. 4 Newtons to displace it by 2\nmeters, and, of course, the restorative force will then be\nin the opposite direction, and that's where we get the\nnegative number. Anyway, I've run out of time. I'll see you in the\nnext video.","attrs_markdown":"[Skip to main content](https:\/\/www.khanacademy.org\/science\/physics\/v\/intro-to-springs-and-hooke-s-law#main-content)\n\nIf you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains \\*.kastatic.org and \\*.kasandbox.org are unblocked.\n\nExplore\n\nSelect a category to view its courses\n\nSearch\n\nDonate\n\n[Log in](https:\/\/www.khanacademy.org\/login?continue=%2Fscience%2Fin-in-class11th-physics%2Fin-in-class11th-physics-work-energy-and-power%2Fin-in-class11-spring-potential-energy-and-hookes-law%2Fv%2Fintro-to-springs-and-hooke-s-law)[Sign up](https:\/\/www.khanacademy.org\/signup?continue=%2Fscience%2Fin-in-class11th-physics%2Fin-in-class11th-physics-work-energy-and-power%2Fin-in-class11-spring-potential-energy-and-hookes-law%2Fv%2Fintro-to-springs-and-hooke-s-law)\n\n[Search for courses, skills, and videos](https:\/\/www.khanacademy.org\/search)\n\n[Skip to lesson content](https:\/\/www.khanacademy.org\/science\/physics\/v\/intro-to-springs-and-hooke-s-law#content-library-content-panel)\n\n## Welcome to Khan Academy\\!\nSo we can give you the right tools, let us know if you're a...\n\nLearner\n\nTeacher\n\nParent\n\n## NCERT Physics Class 11\n### [Course: NCERT Physics Class 11](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics) \\> [Unit 6](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power)\nLesson 5: Spring potential energy and Hooke's law\n\n- [Intro to springs and Hooke's law](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law)\n- [Potential energy stored in a spring](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/potential-energy-stored-in-a-spring)\n- [Spring potential energy example (mistake in math)](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/spring-potential-energy-example-mistake-in-math)\n- [Spring force and Hooke's law](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/e\/hooke-s-law-and-spring-force-ap1)\n- [Calculating elastic potential energy](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/e\/elastic-potential-energy-ap1)\n- [Spring potential energy and Hooke's law review](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/a\/spring-force-and-energy-ap1)\n- [Spring force and Hooke's law](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/a\/what-is-hookes-law)\n- [What is elastic potential energy?](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/a\/what-is-elastic-potential-energy)\n\n[Science](https:\/\/www.khanacademy.org\/science)\\>\n\n[NCERT Physics Class 11](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics)\\>\n\n[Work, energy and power](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power)\\>\n\n[Spring potential energy and Hooke's law](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law)\n\n© 2026 Khan Academy\n\n[Terms of use](https:\/\/www.khanacademy.org\/about\/docs\/khan-academy-terms-of-service)[Privacy Policy](https:\/\/www.khanacademy.org\/about\/privacy-policy)[Cookie Notice](https:\/\/www.khanacademy.org\/about\/cookie-policy)[Accessibility Statement](https:\/\/www.khanacademy.org\/about\/accessibility-statement)\n\n# Intro to springs and Hooke's law\n![](https:\/\/cdn.kastatic.org\/images\/google_classroom_color.png)\n\nGoogle Classroom\n\n![](https:\/\/cdn.kastatic.org\/khanacademy\/images\/4dc54a20c2c0c033-microsoft-teams-icon.svg)\n\nMicrosoft Teams\n\nShare link\n\nCurrent time:0:00Total duration:10:06\n\nAbout\n\nTranscript\n\nDiscover the phenomena of springs and Hooke's Law. Explore how force applied to a spring results in compression or elongation, and how this relationship is linear. Uncover the concept of restorative force and how it counteracts applied force, keeping our spring in equilibrium. Created by Sal Khan.\n\n[Skip to end of discussions](https:\/\/www.khanacademy.org\/science\/physics\/v\/intro-to-springs-and-hooke-s-law#skipped-discussion)\n\nQuestions\n\nTips & Thanks\n\n## Want to join the conversation?\n[Log in](https:\/\/www.khanacademy.org\/login?continue=%2Fscience%2Fin-in-class11th-physics%2Fin-in-class11th-physics-work-energy-and-power%2Fin-in-class11-spring-potential-energy-and-hookes-law%2Fv%2Fintro-to-springs-and-hooke-s-law%3FloggedInViaDiscussion%3Dtrue)\n\nSort by:\n\nTop Voted\n\n- [![starky ultimate style avatar for user Srijan Devnath](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/starky-ultimate.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_832399177870034845428480\/discussion)\n  \n  [Srijan Devnath](https:\/\/www.khanacademy.org\/profile\/kaid_832399177870034845428480\/discussion)\n  \n  [11 years ago Posted 11 years ago. Direct link to Srijan Devnath's post “well what does x stand fo...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfODMyMzk5MTc3ODcwMDM0ODQ1NDI4NDgwDAsSCEZlZWRiYWNrGICAgICKp4YKDA&qa_expand_type=question)\n  \n  more\n  \n  well what does x stand for?\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (7 votes)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![male robot hal style avatar for user Andrew M](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/robot_male_3.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_589024660794290662870780\/discussion)\n    \n    [Andrew M](https:\/\/www.khanacademy.org\/profile\/kaid_589024660794290662870780\/discussion)\n    \n    [11 years ago Posted 11 years ago. Direct link to Andrew M's post “x is the displacement of ...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNTg5MDI0NjYwNzk0MjkwNjYyODcwNzgwDAsSCEZlZWRiYWNrGICAgJiy15sKDA&qa_expand_type=answer)\n    \n    more\n    \n    x is the displacement of the spring's end from its equilibrium point - how much the spring is stretched (or, in the other direction, compressed)\n    \n    Comment\n    \n    Button navigates to signup page\n    \n    (7 votes)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n- [![blobby green style avatar for user nick c](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/blobby-green.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_963869368453958021908617\/discussion)\n  \n  [nick c](https:\/\/www.khanacademy.org\/profile\/kaid_963869368453958021908617\/discussion)\n  \n  [3 years ago Posted 3 years ago. Direct link to nick c's post “Almost missed the last pa...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfOTYzODY5MzY4NDUzOTU4MDIxOTA4NjE3DAsSCEZlZWRiYWNrGICAs-CPgNAKDA&qa_expand_type=question)\n  \n  more\n  \n  Almost missed the last part, the last five to ten seconds there's an assumption to the problem that's changed, very hard to capture that change in a matter of five seconds.\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (6 votes)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n- [![duskpin ultimate style avatar for user GumballUnited](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/duskpin-ultimate.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_4713909139147542986292979\/discussion)\n  \n  [GumballUnited](https:\/\/www.khanacademy.org\/profile\/kaid_4713909139147542986292979\/discussion)\n  \n  [3 years ago Posted 3 years ago. Direct link to GumballUnited's post “Im a little confused can ...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJBCxIIVXNlckRhdGEiHmthaWRfNDcxMzkwOTEzOTE0NzU0Mjk4NjI5Mjk3OQwLEghGZWVkYmFjaxiAgLPc_56qCQw&qa_expand_type=question)\n  \n  more\n  \n  Im a little confused can someone explain it to me\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (3 votes)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![female robot grace style avatar for user Sukhada](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/robot_female_3.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_3767784179946764959085306\/discussion)\n    \n    [Sukhada](https:\/\/www.khanacademy.org\/profile\/kaid_3767784179946764959085306\/discussion)\n    \n    [3 years ago Posted 3 years ago. Direct link to Sukhada's post “Totally understandable! ...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJBCxIIVXNlckRhdGEiHmthaWRfMzc2Nzc4NDE3OTk0Njc2NDk1OTA4NTMwNgwLEghGZWVkYmFjaxiAgPPgnOHkCww&qa_expand_type=answer)\n    \n    more\n    \n    Totally understandable!   \n    Here's what I could gather from this video:-  \n       \n    The main thing Sal stressed was putting the formula F=-kx in our intuitive thinking.   \n    Here, x refers to the displacement the spring undergoes upon applying force.   \n    The force you apply to a spring is linearly directly proportional to the displacement (x) the spring undergoes. Consider a spring to undergo a 5m displacement when a force of 1 N is applied. Intuitively, when a force of 2N is applied, the spring moves 10m.   \n    The sign convention further makes the Force be negative, considering that right is +ve direction.   \n      \n    There's another thing here- since upon applying force, the spring doesn't accelerate- there must be a force acting in the opposite direction (3rd Law).   \n    This opposing force is the restoration force.  \n      \n    This entire reasoning was first done by a scientist named Hooke, hence earning the name, 'Hooke's Law.'  \n      \n    In simple words, the displacement will always be negative in direction to the force applied, and will be linearly proportional.   \n      \n    I hope that helped\\!\n    \n    3 comments\n    \n    Comment on Sukhada's post “Totally understandable! ...”\n    \n    (4 votes)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n- [![leaf blue style avatar for user Blondike](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/leaf-blue.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_654141591198555631352017\/discussion)\n  \n  [Blondike](https:\/\/www.khanacademy.org\/profile\/kaid_654141591198555631352017\/discussion)\n  \n  [5 years ago Posted 5 years ago. Direct link to Blondike's post “At 9:44 you wrote that F...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNjU0MTQxNTkxMTk4NTU1NjMxMzUyMDE3DAsSCEZlZWRiYWNrGICAnYDKi40KDA&qa_expand_type=question)\n  \n  more\n  \n  At\n  \n  ![](https:\/\/cdn.kastatic.org\/images\/play.png) 9:44\n  \n  you wrote that F=-2x, but if k=-2, then F = -(-2)x = 2x\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (3 votes)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n- [![leaf green style avatar for user Helly Shah](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/leaf-green.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_283236246366930666996104\/discussion)\n  \n  [Helly Shah](https:\/\/www.khanacademy.org\/profile\/kaid_283236246366930666996104\/discussion)\n  \n  [11 years ago Posted 11 years ago. Direct link to Helly Shah's post “If we were given a questi...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfMjgzMjM2MjQ2MzY2OTMwNjY2OTk2MTA0DAsSCEZlZWRiYWNrGICAgICohoQKDA&qa_expand_type=question)\n  \n  more\n  \n  If we were given a question and asked to solve for the spring constant, would we use Hooke's Law: k=(mg)\/x?  \n  Suppose we were given the speed as well, would we still use Hooke's Law?\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (1 vote)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![leaf yellow style avatar for user Taylor Logan](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/leaf-yellow.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_262479746245361721495753\/discussion)\n    \n    [Taylor Logan](https:\/\/www.khanacademy.org\/profile\/kaid_262479746245361721495753\/discussion)\n    \n    [11 years ago Posted 11 years ago. Direct link to Taylor Logan's post “Yes, Hooke's Law will yie...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfMjYyNDc5NzQ2MjQ1MzYxNzIxNDk1NzUzDAsSCEZlZWRiYWNrGICAgMCWrJYKDA&qa_expand_type=answer)\n    \n    more\n    \n    Yes, Hooke's Law will yield k=F\/∆x  \n      \n    If you are given the speed and must find something about the spring constant, you must know the expressions for the kinetic energy and potential energy of the system and use energy conservation to solve for k. Also remember that U(x)=1\/2kx^2\n    \n    2 comments\n    \n    Comment on Taylor Logan's post “Yes, Hooke's Law will yie...”\n    \n    (4 votes)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n- [![aqualine ultimate style avatar for user Andres Saavedra](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/aqualine-ultimate.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_13260870726667511549879\/discussion)\n  \n  [Andres Saavedra](https:\/\/www.khanacademy.org\/profile\/kaid_13260870726667511549879\/discussion)\n  \n  [8 years ago Posted 8 years ago. Direct link to Andres Saavedra's post “Are the units for \"K\" kg\/...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI_CxIIVXNlckRhdGEiHGthaWRfMTMyNjA4NzA3MjY2Njc1MTE1NDk4NzkMCxIIRmVlZGJhY2sYgICAgNTwigoM&qa_expand_type=question)\n  \n  more\n  \n  Are the units for \"K\" kg\/s^2 ? In that case does the number have this units for an experimental reason or are the units in there to make the result in newtons.\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (2 votes)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![male robot hal style avatar for user Andrew M](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/robot_male_3.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_589024660794290662870780\/discussion)\n    \n    [Andrew M](https:\/\/www.khanacademy.org\/profile\/kaid_589024660794290662870780\/discussion)\n    \n    [8 years ago Posted 8 years ago. Direct link to Andrew M's post “F = kx. What MUST the un...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNTg5MDI0NjYwNzk0MjkwNjYyODcwNzgwDAsSCEZlZWRiYWNrGICAgPKi5JUKDA&qa_expand_type=answer)\n    \n    more\n    \n    F = kx. What MUST the units of k be if x is in meters and F has to be in Newtons?  \n    That's how constants work.\n    \n    1 comment\n    \n    Comment on Andrew M's post “F = kx. What MUST the un...”\n    \n    (1 vote)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n  \n  Show more...\n- [![blobby green style avatar for user adilm1904](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/blobby-green.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_229153247969440122137713\/discussion)\n  \n  [adilm1904](https:\/\/www.khanacademy.org\/profile\/kaid_229153247969440122137713\/discussion)\n  \n  [6 years ago Posted 6 years ago. Direct link to adilm1904's post “can the k be ever greater...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfMjI5MTUzMjQ3OTY5NDQwMTIyMTM3NzEzDAsSCEZlZWRiYWNrGICAzdCD5a4KDA&qa_expand_type=question)\n  \n  more\n  \n  can the k be ever greater than 1? i think it cant, think about the consequence\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (1 vote)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![hopper cool style avatar for user obiwan kenobi](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/cs-hopper-cool.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_710737471466935488319243\/discussion)\n    \n    [obiwan kenobi](https:\/\/www.khanacademy.org\/profile\/kaid_710737471466935488319243\/discussion)\n    \n    [6 years ago Posted 6 years ago. Direct link to obiwan kenobi's post “It can be greater than 1....”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNzEwNzM3NDcxNDY2OTM1NDg4MzE5MjQzDAsSCEZlZWRiYWNrGICAzfCzmsEJDA&qa_expand_type=answer)\n    \n    more\n    \n    It can be greater than 1. If you think about it, there isn't really a consequence. If you have a spring constant of 400 N\/m, it just means that for every meter you stretch the spring will exert a force of 400 N more. Hope this helps\\!\n    \n    1 comment\n    \n    Comment on obiwan kenobi's post “It can be greater than 1....”\n    \n    (3 votes)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n- [![purple pi purple style avatar for user FurqanAhmedJ](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/purple-pi.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_85222079336815698822844\/discussion)\n  \n  [FurqanAhmedJ](https:\/\/www.khanacademy.org\/profile\/kaid_85222079336815698822844\/discussion)\n  \n  [11 years ago Posted 11 years ago. Direct link to FurqanAhmedJ's post “I am sorry for not indica...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI_CxIIVXNlckRhdGEiHGthaWRfODUyMjIwNzkzMzY4MTU2OTg4MjI4NDQMCxIIRmVlZGJhY2sYgICAgJrxhAoM&qa_expand_type=question)\n  \n  more\n  \n  I am sorry for not indicating time... It was almost halfway between the video but the question is how the spring constant can be a negative?  \n  And even if it is... What does this signify?\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (1 vote)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![piceratops tree style avatar for user Chandana Deeksha](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/piceratops-tree.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_978443585833336236587800\/discussion)\n    \n    [Chandana Deeksha](https:\/\/www.khanacademy.org\/profile\/kaid_978443585833336236587800\/discussion)\n    \n    [11 years ago Posted 11 years ago. Direct link to Chandana Deeksha's post “the spring constant is ne...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfOTc4NDQzNTg1ODMzMzM2MjM2NTg3ODAwDAsSCEZlZWRiYWNrGICAgID31pQKDA&qa_expand_type=answer)\n    \n    more\n    \n    the spring constant is never negative, it is always positive  \n    the minus sign you are talking of might be the one in the equation `F = - Kx` and this one means that the restorative force \"F\" is always opposite in direction to the displacement.\n    \n    Comment\n    \n    Button navigates to signup page\n    \n    (3 votes)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n- [![male robot donald style avatar for user ammarthecoder007](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/robot_male_2.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_953908838062167145682954\/discussion)\n  \n  [ammarthecoder007](https:\/\/www.khanacademy.org\/profile\/kaid_953908838062167145682954\/discussion)\n  \n  [6 years ago Posted 6 years ago. Direct link to ammarthecoder007's post “i wonder how (who ever it...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfOTUzOTA4ODM4MDYyMTY3MTQ1NjgyOTU0DAsSCEZlZWRiYWNrGICAsafovOcIDA&qa_expand_type=question)\n  \n  more\n  \n  i wonder how (who ever it is...) managed to discover this law,known as hooke's law.  \n  if any one knows please tell.  \n  thanks.\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (2 votes)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n  \n  - [![hopper cool style avatar for user obiwan kenobi](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/cs-hopper-cool.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_710737471466935488319243\/discussion)\n    \n    [obiwan kenobi](https:\/\/www.khanacademy.org\/profile\/kaid_710737471466935488319243\/discussion)\n    \n    [6 years ago Posted 6 years ago. Direct link to obiwan kenobi's post “Hooke probably discovered...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNzEwNzM3NDcxNDY2OTM1NDg4MzE5MjQzDAsSCEZlZWRiYWNrGICAsaeQkrQKDA&qa_expand_type=answer)\n    \n    more\n    \n    Hooke probably discovered it experimentally. He could have set up a vertical spring and observed the lengths to which it is stretched when objects of known mass were placed on the end.\n    \n    Comment\n    \n    Button navigates to signup page\n    \n    (1 vote)\n    \n    - Upvote\n      \n      Button navigates to signup page\n    - Downvote\n      \n      Button navigates to signup page\n    - Flag\n      \n      Button navigates to signup page\n    \n    more\n- [![leaf green style avatar for user Sami](https:\/\/cdn.kastatic.org\/images\/avatars\/svg\/leaf-green.svg)](https:\/\/www.khanacademy.org\/profile\/kaid_143993863927959464369367\/discussion)\n  \n  [Sami](https:\/\/www.khanacademy.org\/profile\/kaid_143993863927959464369367\/discussion)\n  \n  [13 years ago Posted 13 years ago. Direct link to Sami's post “We know that Force is equ...”](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI6CxIIVXNlckRhdGEiHWthaWRfMTQzOTkzODYzOTI3OTU5NDY0MzY5MzY3DAsSCEZlZWRiYWNrGJFODA&qa_expand_type=question)\n  \n  more\n  \n  We know that Force is equal to Mass times Acceleration. So, how does that relate with the equation : Restorative Force of Spring is equal to negative Spring Constant times Distance ?\n  \n  Answer\n  \n  Button navigates to signup page\n  \n  •\n  \n  Comment\n  \n  Button navigates to signup page\n  \n  (1 vote)\n  \n  - Upvote\n    \n    Button navigates to signup page\n  - Downvote\n    \n    Button navigates to signup page\n  - Flag\n    \n    Button navigates to signup page\n  \n  more\n\nShow more comments\n\n## Video transcript\nLet's learn a little bit about springs. So let's say I have a spring. Let me draw the ground so that we know what's going on with the spring. So let me see, this is the floor. That's the floor, and I have a spring. It's along the floor. I'll use a thicker one, just to show it's a spring. Let's say the spring looks something like this. Whoops, I'm still using the line tool. So the spring looks like this. This is my spring, my amazingly drawn spring. Let's say at this end it's attached to a wall. That's a wall. And so this is a spring when I don't have any force acting on it, this is just the natural state of the spring. And we could call this, where it just naturally rests, this tip of the spring. And let's say that when I were to apply a force of 5 Newtons into the spring, it looks something like this. Redraw everything. So when I apply a force of 5 Newtons-- I'll draw the wall in magenta now. When I apply a force of 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every day or a sofa. So let's say it compresses to here. If this was the normal resting-- so this is where the spring was when I applied no force, but when I applied 5 Newtons in that direction, let's say that this distance right here is 10 meters. And so a typical question that you'll see, and we'll explain how to do it, is a spring compresses or elongates when you apply a certain force by some distance. How much will it compress when you apply a different force? So my question is how much will it compress when I apply a 10-Newton force? So your intuition that it'll compress more is correct, but is it linear to how much I compress it? Is it a square of how much I compress it? How does it relate? I think you probably could guess. It's actually worth an experiment. Or you could just keep watching the video. So let's say I apply a 10-Newton force. What will the spring look like? Well, it'll be more compressed. Drop my force to 10 Newtons. And if this was the natural place where the spring would rest, what is this distance? Well, it turns out that it is linear. What do I mean by linear? Well, it means that the more the force-- it's equally proportional to how much the spring will compress. And it actually works the other way. If you applied 5 Newtons in this direction, to the right, you would have gone 10 meters in this direction. So it goes whether you're elongating the spring or compressing the spring within some reasonable tolerance. We've all had this experience. If you compress something too much or you stretch it too much, it doesn't really go back to where it was before. But within some reasonable tolerance, it's proportional. So what does that mean? That means that the restoring force of the spring is minus some number, times the displacement of the spring. So what does this mean? So in this example right here, what was the displacement of the spring? Well, if we take positive x to the right and negative x to the left, the displacement of the spring was what? The displacement, in this example right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the restorative force is going to be equal to minus K times how much it's distorted times minus 10. So the minuses cancel out, so it equals 10K. What's the restorative force in this example? Well, you might say, it's 5 Newtons, just because that's the only force I've drawn here, and you would be to some degree correct. And actually, since we're doing positive and negative, and this 5 Newton is to the left, so to the negative x-direction, actually, I should call this minus 5 Newtons and I should call this minus 10 Newtons, because obviously, these are vectors and we're going to the left. I picked the convention that to the left means negative. So what's the restorative force? Well, in this example-- and we assume that K is a positive number for our purposes. In this example, the restorative force is a positive number. So what is the restorative force? So that's actually the force, the counteracting force, of the spring. That's what this formula gives us. So if this spring is stationary when I apply this 5-Newton force, that means that there must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring would keep compressing. And if the force was more than 5 Newtons, the spring would go back this way. So the fact that I know that when I apply a 5-Newton force to the left, or a negative 5-Newton force, the spring is no longer moving, it means that there must be-- or no longer accelerating, actually, it means that there must be an equal and opposite force to the right, and that's the restorative force. Another way to think about it is if I were to let-- well, I won't go in there now. So in this case, the restorative force is 5 Newtons, so we can solve for K. We could say 5 is equal to 10K. Divide both sides by 10. You get K is equal to 1\/2. So now we can use that information to figure out what is the displacement when I apply a negative 10-Newton force? When I push the spring in with 10 Newtons in the leftward direction? So first of all, what's the restorative force here? Well, if the spring is no longer accelerating in either direction, or the tip of the spring is no longer accelerating in either direction, we know that the restorative force must be counterbalancing this force that I'm compressing with, right? The force that the spring wants to expand back with is 10 Newtons, positive 10 Newtons, right? And we know the spring constant, this K for this spring, for this material, whatever it might be, is 1\/2. So we know the restorative force is equal to 1\/2 times the distance, right? And the formula is minus K, right? And then, what is the restorative force in this example? Well I said it's 10 Newtons, so we know that 10 Newtons is equal to minus 1\/2x. And so what is x? Well, multiply both sides by minus 1\/2, and you get minus 20. I'm sorry, multiply both sides by minus 2, you get minus 20 is equal to x. So x goes to the left 20 units. So that's all that it's telling us. And this law is called Hooke's Law, and it's named after-- I'll read it-- a physicist in the 17th century, a British physicist. And he figured out that the amount of force necessary to keep a spring compressed is proportional to how much you've compressed it. And that's all that this formula says. And that negative number, remember, this formula gives us the restorative force. So it says that the force is always in the opposite direction of how much you displace it. So, for example, if you were to displace this spring in this direction, if you were to apply a force and x were a positive and you were to go in that direction, the force-- no wait, sorry. This is where the spring rests. If you were to apply some force and take the spring out to here, this negative number tells us that the spring will essentially try to pull back with the restorative force in the other direction. Let's do one more problem and I think this will be clear to you. So let's say I have a spring, and all of these problems kind of go along. So let's say when I apply a force of 2 Newtons, so this is what I apply when I apply a force of 2 Newtons. Well, let's say it this way. Let's say when I stretch the spring. Let's say this is the spring, and when I apply a force of 2 Newtons to the right, the spring gets stretched 1 meter. So first of all, let's figure out what K is. So if the spring is stretched by 1 meter, out here, its restorative force will be 2 Newtons back this way, right? So its restorative force, this 2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter, so then we multiply both sides by negative 1, and we get K is equal to minus 2. So then we can use Hooke's Law to note the equation for this-- to figure out the restorative force for this particular spring, and it would be minus 2x. And then I said, well, how much force would I have to apply to distort the spring by 2 meters? Well, it's 2 times 2, it would be 4. 4 Newtons to displace it by 2 meters, and, of course, the restorative force will then be in the opposite direction, and that's where we get the negative number. Anyway, I've run out of time. I'll see you in the next video.\n\n[Creative Commons Attribution\/Non-Commercial\/Share-Alike](https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0)[Video on YouTube](https:\/\/www.youtube.com\/watch?v=ZzwuHS9ldbY)\n\n[Up next: video](https:\/\/www.khanacademy.org\/science\/in-in-class11th-physics\/in-in-class11th-physics-work-energy-and-power\/in-in-class11-spring-potential-energy-and-hookes-law\/v\/potential-energy-stored-in-a-spring)","attrs_readable_markdown":"## Video transcript\nLet's learn a little bit about springs. So let's say I have a spring. Let me draw the ground so that we know what's going on with the spring. So let me see, this is the floor. That's the floor, and I have a spring. It's along the floor. I'll use a thicker one, just to show it's a spring. Let's say the spring looks something like this. Whoops, I'm still using the line tool. So the spring looks like this. This is my spring, my amazingly drawn spring. Let's say at this end it's attached to a wall. That's a wall. And so this is a spring when I don't have any force acting on it, this is just the natural state of the spring. And we could call this, where it just naturally rests, this tip of the spring. And let's say that when I were to apply a force of 5 Newtons into the spring, it looks something like this. Redraw everything. So when I apply a force of 5 Newtons-- I'll draw the wall in magenta now. When I apply a force of 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every day or a sofa. So let's say it compresses to here. If this was the normal resting-- so this is where the spring was when I applied no force, but when I applied 5 Newtons in that direction, let's say that this distance right here is 10 meters. And so a typical question that you'll see, and we'll explain how to do it, is a spring compresses or elongates when you apply a certain force by some distance. How much will it compress when you apply a different force? So my question is how much will it compress when I apply a 10-Newton force? So your intuition that it'll compress more is correct, but is it linear to how much I compress it? Is it a square of how much I compress it? How does it relate? I think you probably could guess. It's actually worth an experiment. Or you could just keep watching the video. So let's say I apply a 10-Newton force. What will the spring look like? Well, it'll be more compressed. Drop my force to 10 Newtons. And if this was the natural place where the spring would rest, what is this distance? Well, it turns out that it is linear. What do I mean by linear? Well, it means that the more the force-- it's equally proportional to how much the spring will compress. And it actually works the other way. If you applied 5 Newtons in this direction, to the right, you would have gone 10 meters in this direction. So it goes whether you're elongating the spring or compressing the spring within some reasonable tolerance. We've all had this experience. If you compress something too much or you stretch it too much, it doesn't really go back to where it was before. But within some reasonable tolerance, it's proportional. So what does that mean? That means that the restoring force of the spring is minus some number, times the displacement of the spring. So what does this mean? So in this example right here, what was the displacement of the spring? Well, if we take positive x to the right and negative x to the left, the displacement of the spring was what? The displacement, in this example right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the restorative force is going to be equal to minus K times how much it's distorted times minus 10. So the minuses cancel out, so it equals 10K. What's the restorative force in this example? Well, you might say, it's 5 Newtons, just because that's the only force I've drawn here, and you would be to some degree correct. And actually, since we're doing positive and negative, and this 5 Newton is to the left, so to the negative x-direction, actually, I should call this minus 5 Newtons and I should call this minus 10 Newtons, because obviously, these are vectors and we're going to the left. I picked the convention that to the left means negative. So what's the restorative force? Well, in this example-- and we assume that K is a positive number for our purposes. In this example, the restorative force is a positive number. So what is the restorative force? So that's actually the force, the counteracting force, of the spring. That's what this formula gives us. So if this spring is stationary when I apply this 5-Newton force, that means that there must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring would keep compressing. And if the force was more than 5 Newtons, the spring would go back this way. So the fact that I know that when I apply a 5-Newton force to the left, or a negative 5-Newton force, the spring is no longer moving, it means that there must be-- or no longer accelerating, actually, it means that there must be an equal and opposite force to the right, and that's the restorative force. Another way to think about it is if I were to let-- well, I won't go in there now. So in this case, the restorative force is 5 Newtons, so we can solve for K. We could say 5 is equal to 10K. Divide both sides by 10. You get K is equal to 1\/2. So now we can use that information to figure out what is the displacement when I apply a negative 10-Newton force? When I push the spring in with 10 Newtons in the leftward direction? So first of all, what's the restorative force here? Well, if the spring is no longer accelerating in either direction, or the tip of the spring is no longer accelerating in either direction, we know that the restorative force must be counterbalancing this force that I'm compressing with, right? The force that the spring wants to expand back with is 10 Newtons, positive 10 Newtons, right? And we know the spring constant, this K for this spring, for this material, whatever it might be, is 1\/2. So we know the restorative force is equal to 1\/2 times the distance, right? And the formula is minus K, right? And then, what is the restorative force in this example? Well I said it's 10 Newtons, so we know that 10 Newtons is equal to minus 1\/2x. And so what is x? Well, multiply both sides by minus 1\/2, and you get minus 20. I'm sorry, multiply both sides by minus 2, you get minus 20 is equal to x. So x goes to the left 20 units. So that's all that it's telling us. And this law is called Hooke's Law, and it's named after-- I'll read it-- a physicist in the 17th century, a British physicist. And he figured out that the amount of force necessary to keep a spring compressed is proportional to how much you've compressed it. And that's all that this formula says. And that negative number, remember, this formula gives us the restorative force. So it says that the force is always in the opposite direction of how much you displace it. So, for example, if you were to displace this spring in this direction, if you were to apply a force and x were a positive and you were to go in that direction, the force-- no wait, sorry. This is where the spring rests. If you were to apply some force and take the spring out to here, this negative number tells us that the spring will essentially try to pull back with the restorative force in the other direction. Let's do one more problem and I think this will be clear to you. So let's say I have a spring, and all of these problems kind of go along. So let's say when I apply a force of 2 Newtons, so this is what I apply when I apply a force of 2 Newtons. Well, let's say it this way. Let's say when I stretch the spring. Let's say this is the spring, and when I apply a force of 2 Newtons to the right, the spring gets stretched 1 meter. So first of all, let's figure out what K is. So if the spring is stretched by 1 meter, out here, its restorative force will be 2 Newtons back this way, right? So its restorative force, this 2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter, so then we multiply both sides by negative 1, and we get K is equal to minus 2. So then we can use Hooke's Law to note the equation for this-- to figure out the restorative force for this particular spring, and it would be minus 2x. And then I said, well, how much force would I have to apply to distort the spring by 2 meters? Well, it's 2 times 2, it would be 4. 4 Newtons to displace it by 2 meters, and, of course, the restorative force will then be in the opposite direction, and that's where we get the negative number. Anyway, I've run out of time. I'll see you in the next video.","meta_canonical":null}

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Meta Description	Discover the phenomena of springs and Hooke's Law. Explore how force applied to a spring results in compression or elongation, and how this relationship is linear. Uncover the concept of restorative force and how it counteracts applied force, keeping our spring in equilibrium.
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Boilerpipe Text	Video transcript Let's learn a little bit about springs. So let's say I have a spring. Let me draw the ground so that we know what's going on with the spring. So let me see, this is the floor. That's the floor, and I have a spring. It's along the floor. I'll use a thicker one, just to show it's a spring. Let's say the spring looks something like this. Whoops, I'm still using the line tool. So the spring looks like this. This is my spring, my amazingly drawn spring. Let's say at this end it's attached to a wall. That's a wall. And so this is a spring when I don't have any force acting on it, this is just the natural state of the spring. And we could call this, where it just naturally rests, this tip of the spring. And let's say that when I were to apply a force of 5 Newtons into the spring, it looks something like this. Redraw everything. So when I apply a force of 5 Newtons-- I'll draw the wall in magenta now. When I apply a force of 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every day or a sofa. So let's say it compresses to here. If this was the normal resting-- so this is where the spring was when I applied no force, but when I applied 5 Newtons in that direction, let's say that this distance right here is 10 meters. And so a typical question that you'll see, and we'll explain how to do it, is a spring compresses or elongates when you apply a certain force by some distance. How much will it compress when you apply a different force? So my question is how much will it compress when I apply a 10-Newton force? So your intuition that it'll compress more is correct, but is it linear to how much I compress it? Is it a square of how much I compress it? How does it relate? I think you probably could guess. It's actually worth an experiment. Or you could just keep watching the video. So let's say I apply a 10-Newton force. What will the spring look like? Well, it'll be more compressed. Drop my force to 10 Newtons. And if this was the natural place where the spring would rest, what is this distance? Well, it turns out that it is linear. What do I mean by linear? Well, it means that the more the force-- it's equally proportional to how much the spring will compress. And it actually works the other way. If you applied 5 Newtons in this direction, to the right, you would have gone 10 meters in this direction. So it goes whether you're elongating the spring or compressing the spring within some reasonable tolerance. We've all had this experience. If you compress something too much or you stretch it too much, it doesn't really go back to where it was before. But within some reasonable tolerance, it's proportional. So what does that mean? That means that the restoring force of the spring is minus some number, times the displacement of the spring. So what does this mean? So in this example right here, what was the displacement of the spring? Well, if we take positive x to the right and negative x to the left, the displacement of the spring was what? The displacement, in this example right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the restorative force is going to be equal to minus K times how much it's distorted times minus 10. So the minuses cancel out, so it equals 10K. What's the restorative force in this example? Well, you might say, it's 5 Newtons, just because that's the only force I've drawn here, and you would be to some degree correct. And actually, since we're doing positive and negative, and this 5 Newton is to the left, so to the negative x-direction, actually, I should call this minus 5 Newtons and I should call this minus 10 Newtons, because obviously, these are vectors and we're going to the left. I picked the convention that to the left means negative. So what's the restorative force? Well, in this example-- and we assume that K is a positive number for our purposes. In this example, the restorative force is a positive number. So what is the restorative force? So that's actually the force, the counteracting force, of the spring. That's what this formula gives us. So if this spring is stationary when I apply this 5-Newton force, that means that there must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring would keep compressing. And if the force was more than 5 Newtons, the spring would go back this way. So the fact that I know that when I apply a 5-Newton force to the left, or a negative 5-Newton force, the spring is no longer moving, it means that there must be-- or no longer accelerating, actually, it means that there must be an equal and opposite force to the right, and that's the restorative force. Another way to think about it is if I were to let-- well, I won't go in there now. So in this case, the restorative force is 5 Newtons, so we can solve for K. We could say 5 is equal to 10K. Divide both sides by 10. You get K is equal to 1/2. So now we can use that information to figure out what is the displacement when I apply a negative 10-Newton force? When I push the spring in with 10 Newtons in the leftward direction? So first of all, what's the restorative force here? Well, if the spring is no longer accelerating in either direction, or the tip of the spring is no longer accelerating in either direction, we know that the restorative force must be counterbalancing this force that I'm compressing with, right? The force that the spring wants to expand back with is 10 Newtons, positive 10 Newtons, right? And we know the spring constant, this K for this spring, for this material, whatever it might be, is 1/2. So we know the restorative force is equal to 1/2 times the distance, right? And the formula is minus K, right? And then, what is the restorative force in this example? Well I said it's 10 Newtons, so we know that 10 Newtons is equal to minus 1/2x. And so what is x? Well, multiply both sides by minus 1/2, and you get minus 20. I'm sorry, multiply both sides by minus 2, you get minus 20 is equal to x. So x goes to the left 20 units. So that's all that it's telling us. And this law is called Hooke's Law, and it's named after-- I'll read it-- a physicist in the 17th century, a British physicist. And he figured out that the amount of force necessary to keep a spring compressed is proportional to how much you've compressed it. And that's all that this formula says. And that negative number, remember, this formula gives us the restorative force. So it says that the force is always in the opposite direction of how much you displace it. So, for example, if you were to displace this spring in this direction, if you were to apply a force and x were a positive and you were to go in that direction, the force-- no wait, sorry. This is where the spring rests. If you were to apply some force and take the spring out to here, this negative number tells us that the spring will essentially try to pull back with the restorative force in the other direction. Let's do one more problem and I think this will be clear to you. So let's say I have a spring, and all of these problems kind of go along. So let's say when I apply a force of 2 Newtons, so this is what I apply when I apply a force of 2 Newtons. Well, let's say it this way. Let's say when I stretch the spring. Let's say this is the spring, and when I apply a force of 2 Newtons to the right, the spring gets stretched 1 meter. So first of all, let's figure out what K is. So if the spring is stretched by 1 meter, out here, its restorative force will be 2 Newtons back this way, right? So its restorative force, this 2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter, so then we multiply both sides by negative 1, and we get K is equal to minus 2. So then we can use Hooke's Law to note the equation for this-- to figure out the restorative force for this particular spring, and it would be minus 2x. And then I said, well, how much force would I have to apply to distort the spring by 2 meters? Well, it's 2 times 2, it would be 4. 4 Newtons to displace it by 2 meters, and, of course, the restorative force will then be in the opposite direction, and that's where we get the negative number. Anyway, I've run out of time. I'll see you in the next video.
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Explore how force applied to a spring results in compression or elongation, and how this relationship is linear. Uncover the concept of restorative force and how it counteracts applied force, keeping our spring in equilibrium. Created by Sal Khan. [Skip to end of discussions](https://www.khanacademy.org/science/physics/v/intro-to-springs-and-hooke-s-law#skipped-discussion) Questions Tips & Thanks ## Want to join the conversation? [Log in](https://www.khanacademy.org/login?continue=%2Fscience%2Fin-in-class11th-physics%2Fin-in-class11th-physics-work-energy-and-power%2Fin-in-class11-spring-potential-energy-and-hookes-law%2Fv%2Fintro-to-springs-and-hooke-s-law%3FloggedInViaDiscussion%3Dtrue) Sort by: Top Voted - [![starky ultimate style avatar for user Srijan Devnath](https://cdn.kastatic.org/images/avatars/svg/starky-ultimate.svg)](https://www.khanacademy.org/profile/kaid_832399177870034845428480/discussion) [Srijan Devnath](https://www.khanacademy.org/profile/kaid_832399177870034845428480/discussion) [11 years ago Posted 11 years ago. Direct link to Srijan Devnath's post “well what does x stand fo...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfODMyMzk5MTc3ODcwMDM0ODQ1NDI4NDgwDAsSCEZlZWRiYWNrGICAgICKp4YKDA&qa_expand_type=question) more well what does x stand for? Answer Button navigates to signup page • Comment Button navigates to signup page (7 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![male robot hal style avatar for user Andrew M](https://cdn.kastatic.org/images/avatars/svg/robot_male_3.svg)](https://www.khanacademy.org/profile/kaid_589024660794290662870780/discussion) [Andrew M](https://www.khanacademy.org/profile/kaid_589024660794290662870780/discussion) [11 years ago Posted 11 years ago. Direct link to Andrew M's post “x is the displacement of ...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNTg5MDI0NjYwNzk0MjkwNjYyODcwNzgwDAsSCEZlZWRiYWNrGICAgJiy15sKDA&qa_expand_type=answer) more x is the displacement of the spring's end from its equilibrium point - how much the spring is stretched (or, in the other direction, compressed) Comment Button navigates to signup page (7 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![blobby green style avatar for user nick c](https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg)](https://www.khanacademy.org/profile/kaid_963869368453958021908617/discussion) [nick c](https://www.khanacademy.org/profile/kaid_963869368453958021908617/discussion) [3 years ago Posted 3 years ago. Direct link to nick c's post “Almost missed the last pa...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfOTYzODY5MzY4NDUzOTU4MDIxOTA4NjE3DAsSCEZlZWRiYWNrGICAs-CPgNAKDA&qa_expand_type=question) more Almost missed the last part, the last five to ten seconds there's an assumption to the problem that's changed, very hard to capture that change in a matter of five seconds. Answer Button navigates to signup page • Comment Button navigates to signup page (6 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![duskpin ultimate style avatar for user GumballUnited](https://cdn.kastatic.org/images/avatars/svg/duskpin-ultimate.svg)](https://www.khanacademy.org/profile/kaid_4713909139147542986292979/discussion) [GumballUnited](https://www.khanacademy.org/profile/kaid_4713909139147542986292979/discussion) [3 years ago Posted 3 years ago. Direct link to GumballUnited's post “Im a little confused can ...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJBCxIIVXNlckRhdGEiHmthaWRfNDcxMzkwOTEzOTE0NzU0Mjk4NjI5Mjk3OQwLEghGZWVkYmFjaxiAgLPc_56qCQw&qa_expand_type=question) more Im a little confused can someone explain it to me Answer Button navigates to signup page • Comment Button navigates to signup page (3 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![female robot grace style avatar for user Sukhada](https://cdn.kastatic.org/images/avatars/svg/robot_female_3.svg)](https://www.khanacademy.org/profile/kaid_3767784179946764959085306/discussion) [Sukhada](https://www.khanacademy.org/profile/kaid_3767784179946764959085306/discussion) [3 years ago Posted 3 years ago. Direct link to Sukhada's post “Totally understandable! ...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJBCxIIVXNlckRhdGEiHmthaWRfMzc2Nzc4NDE3OTk0Njc2NDk1OTA4NTMwNgwLEghGZWVkYmFjaxiAgPPgnOHkCww&qa_expand_type=answer) more Totally understandable! Here's what I could gather from this video:- The main thing Sal stressed was putting the formula F=-kx in our intuitive thinking. Here, x refers to the displacement the spring undergoes upon applying force. The force you apply to a spring is linearly directly proportional to the displacement (x) the spring undergoes. Consider a spring to undergo a 5m displacement when a force of 1 N is applied. Intuitively, when a force of 2N is applied, the spring moves 10m. The sign convention further makes the Force be negative, considering that right is +ve direction. There's another thing here- since upon applying force, the spring doesn't accelerate- there must be a force acting in the opposite direction (3rd Law). This opposing force is the restoration force. This entire reasoning was first done by a scientist named Hooke, hence earning the name, 'Hooke's Law.' In simple words, the displacement will always be negative in direction to the force applied, and will be linearly proportional. I hope that helped\! 3 comments Comment on Sukhada's post “Totally understandable! ...” (4 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![leaf blue style avatar for user Blondike](https://cdn.kastatic.org/images/avatars/svg/leaf-blue.svg)](https://www.khanacademy.org/profile/kaid_654141591198555631352017/discussion) [Blondike](https://www.khanacademy.org/profile/kaid_654141591198555631352017/discussion) [5 years ago Posted 5 years ago. Direct link to Blondike's post “At 9:44 you wrote that F...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNjU0MTQxNTkxMTk4NTU1NjMxMzUyMDE3DAsSCEZlZWRiYWNrGICAnYDKi40KDA&qa_expand_type=question) more At ![](https://cdn.kastatic.org/images/play.png) 9:44 you wrote that F=-2x, but if k=-2, then F = -(-2)x = 2x Answer Button navigates to signup page • Comment Button navigates to signup page (3 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![leaf green style avatar for user Helly Shah](https://cdn.kastatic.org/images/avatars/svg/leaf-green.svg)](https://www.khanacademy.org/profile/kaid_283236246366930666996104/discussion) [Helly Shah](https://www.khanacademy.org/profile/kaid_283236246366930666996104/discussion) [11 years ago Posted 11 years ago. Direct link to Helly Shah's post “If we were given a questi...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfMjgzMjM2MjQ2MzY2OTMwNjY2OTk2MTA0DAsSCEZlZWRiYWNrGICAgICohoQKDA&qa_expand_type=question) more If we were given a question and asked to solve for the spring constant, would we use Hooke's Law: k=(mg)/x? Suppose we were given the speed as well, would we still use Hooke's Law? Answer Button navigates to signup page • Comment Button navigates to signup page (1 vote) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![leaf yellow style avatar for user Taylor Logan](https://cdn.kastatic.org/images/avatars/svg/leaf-yellow.svg)](https://www.khanacademy.org/profile/kaid_262479746245361721495753/discussion) [Taylor Logan](https://www.khanacademy.org/profile/kaid_262479746245361721495753/discussion) [11 years ago Posted 11 years ago. Direct link to Taylor Logan's post “Yes, Hooke's Law will yie...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfMjYyNDc5NzQ2MjQ1MzYxNzIxNDk1NzUzDAsSCEZlZWRiYWNrGICAgMCWrJYKDA&qa_expand_type=answer) more Yes, Hooke's Law will yield k=F/∆x If you are given the speed and must find something about the spring constant, you must know the expressions for the kinetic energy and potential energy of the system and use energy conservation to solve for k. Also remember that U(x)=1/2kx^2 2 comments Comment on Taylor Logan's post “Yes, Hooke's Law will yie...” (4 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![aqualine ultimate style avatar for user Andres Saavedra](https://cdn.kastatic.org/images/avatars/svg/aqualine-ultimate.svg)](https://www.khanacademy.org/profile/kaid_13260870726667511549879/discussion) [Andres Saavedra](https://www.khanacademy.org/profile/kaid_13260870726667511549879/discussion) [8 years ago Posted 8 years ago. Direct link to Andres Saavedra's post “Are the units for "K" kg/...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI_CxIIVXNlckRhdGEiHGthaWRfMTMyNjA4NzA3MjY2Njc1MTE1NDk4NzkMCxIIRmVlZGJhY2sYgICAgNTwigoM&qa_expand_type=question) more Are the units for "K" kg/s^2 ? In that case does the number have this units for an experimental reason or are the units in there to make the result in newtons. Answer Button navigates to signup page • Comment Button navigates to signup page (2 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![male robot hal style avatar for user Andrew M](https://cdn.kastatic.org/images/avatars/svg/robot_male_3.svg)](https://www.khanacademy.org/profile/kaid_589024660794290662870780/discussion) [Andrew M](https://www.khanacademy.org/profile/kaid_589024660794290662870780/discussion) [8 years ago Posted 8 years ago. Direct link to Andrew M's post “F = kx. What MUST the un...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNTg5MDI0NjYwNzk0MjkwNjYyODcwNzgwDAsSCEZlZWRiYWNrGICAgPKi5JUKDA&qa_expand_type=answer) more F = kx. What MUST the units of k be if x is in meters and F has to be in Newtons? That's how constants work. 1 comment Comment on Andrew M's post “F = kx. What MUST the un...” (1 vote) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more Show more... - [![blobby green style avatar for user adilm1904](https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg)](https://www.khanacademy.org/profile/kaid_229153247969440122137713/discussion) [adilm1904](https://www.khanacademy.org/profile/kaid_229153247969440122137713/discussion) [6 years ago Posted 6 years ago. Direct link to adilm1904's post “can the k be ever greater...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfMjI5MTUzMjQ3OTY5NDQwMTIyMTM3NzEzDAsSCEZlZWRiYWNrGICAzdCD5a4KDA&qa_expand_type=question) more can the k be ever greater than 1? i think it cant, think about the consequence Answer Button navigates to signup page • Comment Button navigates to signup page (1 vote) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![hopper cool style avatar for user obiwan kenobi](https://cdn.kastatic.org/images/avatars/svg/cs-hopper-cool.svg)](https://www.khanacademy.org/profile/kaid_710737471466935488319243/discussion) [obiwan kenobi](https://www.khanacademy.org/profile/kaid_710737471466935488319243/discussion) [6 years ago Posted 6 years ago. Direct link to obiwan kenobi's post “It can be greater than 1....”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNzEwNzM3NDcxNDY2OTM1NDg4MzE5MjQzDAsSCEZlZWRiYWNrGICAzfCzmsEJDA&qa_expand_type=answer) more It can be greater than 1. If you think about it, there isn't really a consequence. If you have a spring constant of 400 N/m, it just means that for every meter you stretch the spring will exert a force of 400 N more. Hope this helps\! 1 comment Comment on obiwan kenobi's post “It can be greater than 1....” (3 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![purple pi purple style avatar for user FurqanAhmedJ](https://cdn.kastatic.org/images/avatars/svg/purple-pi.svg)](https://www.khanacademy.org/profile/kaid_85222079336815698822844/discussion) [FurqanAhmedJ](https://www.khanacademy.org/profile/kaid_85222079336815698822844/discussion) [11 years ago Posted 11 years ago. Direct link to FurqanAhmedJ's post “I am sorry for not indica...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI_CxIIVXNlckRhdGEiHGthaWRfODUyMjIwNzkzMzY4MTU2OTg4MjI4NDQMCxIIRmVlZGJhY2sYgICAgJrxhAoM&qa_expand_type=question) more I am sorry for not indicating time... It was almost halfway between the video but the question is how the spring constant can be a negative? And even if it is... What does this signify? Answer Button navigates to signup page • Comment Button navigates to signup page (1 vote) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![piceratops tree style avatar for user Chandana Deeksha](https://cdn.kastatic.org/images/avatars/svg/piceratops-tree.svg)](https://www.khanacademy.org/profile/kaid_978443585833336236587800/discussion) [Chandana Deeksha](https://www.khanacademy.org/profile/kaid_978443585833336236587800/discussion) [11 years ago Posted 11 years ago. Direct link to Chandana Deeksha's post “the spring constant is ne...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfOTc4NDQzNTg1ODMzMzM2MjM2NTg3ODAwDAsSCEZlZWRiYWNrGICAgID31pQKDA&qa_expand_type=answer) more the spring constant is never negative, it is always positive the minus sign you are talking of might be the one in the equation `F = - Kx` and this one means that the restorative force "F" is always opposite in direction to the displacement. Comment Button navigates to signup page (3 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![male robot donald style avatar for user ammarthecoder007](https://cdn.kastatic.org/images/avatars/svg/robot_male_2.svg)](https://www.khanacademy.org/profile/kaid_953908838062167145682954/discussion) [ammarthecoder007](https://www.khanacademy.org/profile/kaid_953908838062167145682954/discussion) [6 years ago Posted 6 years ago. Direct link to ammarthecoder007's post “i wonder how (who ever it...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfOTUzOTA4ODM4MDYyMTY3MTQ1NjgyOTU0DAsSCEZlZWRiYWNrGICAsafovOcIDA&qa_expand_type=question) more i wonder how (who ever it is...) managed to discover this law,known as hooke's law. if any one knows please tell. thanks. Answer Button navigates to signup page • Comment Button navigates to signup page (2 votes) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![hopper cool style avatar for user obiwan kenobi](https://cdn.kastatic.org/images/avatars/svg/cs-hopper-cool.svg)](https://www.khanacademy.org/profile/kaid_710737471466935488319243/discussion) [obiwan kenobi](https://www.khanacademy.org/profile/kaid_710737471466935488319243/discussion) [6 years ago Posted 6 years ago. Direct link to obiwan kenobi's post “Hooke probably discovered...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJACxIIVXNlckRhdGEiHWthaWRfNzEwNzM3NDcxNDY2OTM1NDg4MzE5MjQzDAsSCEZlZWRiYWNrGICAsaeQkrQKDA&qa_expand_type=answer) more Hooke probably discovered it experimentally. He could have set up a vertical spring and observed the lengths to which it is stretched when objects of known mass were placed on the end. Comment Button navigates to signup page (1 vote) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more - [![leaf green style avatar for user Sami](https://cdn.kastatic.org/images/avatars/svg/leaf-green.svg)](https://www.khanacademy.org/profile/kaid_143993863927959464369367/discussion) [Sami](https://www.khanacademy.org/profile/kaid_143993863927959464369367/discussion) [13 years ago Posted 13 years ago. Direct link to Sami's post “We know that Force is equ...”](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/intro-to-springs-and-hooke-s-law?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXI6CxIIVXNlckRhdGEiHWthaWRfMTQzOTkzODYzOTI3OTU5NDY0MzY5MzY3DAsSCEZlZWRiYWNrGJFODA&qa_expand_type=question) more We know that Force is equal to Mass times Acceleration. So, how does that relate with the equation : Restorative Force of Spring is equal to negative Spring Constant times Distance ? Answer Button navigates to signup page • Comment Button navigates to signup page (1 vote) - Upvote Button navigates to signup page - Downvote Button navigates to signup page - Flag Button navigates to signup page more Show more comments ## Video transcript Let's learn a little bit about springs. So let's say I have a spring. Let me draw the ground so that we know what's going on with the spring. So let me see, this is the floor. That's the floor, and I have a spring. It's along the floor. I'll use a thicker one, just to show it's a spring. Let's say the spring looks something like this. Whoops, I'm still using the line tool. So the spring looks like this. This is my spring, my amazingly drawn spring. Let's say at this end it's attached to a wall. That's a wall. And so this is a spring when I don't have any force acting on it, this is just the natural state of the spring. And we could call this, where it just naturally rests, this tip of the spring. And let's say that when I were to apply a force of 5 Newtons into the spring, it looks something like this. Redraw everything. So when I apply a force of 5 Newtons-- I'll draw the wall in magenta now. When I apply a force of 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every day or a sofa. So let's say it compresses to here. If this was the normal resting-- so this is where the spring was when I applied no force, but when I applied 5 Newtons in that direction, let's say that this distance right here is 10 meters. And so a typical question that you'll see, and we'll explain how to do it, is a spring compresses or elongates when you apply a certain force by some distance. How much will it compress when you apply a different force? So my question is how much will it compress when I apply a 10-Newton force? So your intuition that it'll compress more is correct, but is it linear to how much I compress it? Is it a square of how much I compress it? How does it relate? I think you probably could guess. It's actually worth an experiment. Or you could just keep watching the video. So let's say I apply a 10-Newton force. What will the spring look like? Well, it'll be more compressed. Drop my force to 10 Newtons. And if this was the natural place where the spring would rest, what is this distance? Well, it turns out that it is linear. What do I mean by linear? Well, it means that the more the force-- it's equally proportional to how much the spring will compress. And it actually works the other way. If you applied 5 Newtons in this direction, to the right, you would have gone 10 meters in this direction. So it goes whether you're elongating the spring or compressing the spring within some reasonable tolerance. We've all had this experience. If you compress something too much or you stretch it too much, it doesn't really go back to where it was before. But within some reasonable tolerance, it's proportional. So what does that mean? That means that the restoring force of the spring is minus some number, times the displacement of the spring. So what does this mean? So in this example right here, what was the displacement of the spring? Well, if we take positive x to the right and negative x to the left, the displacement of the spring was what? The displacement, in this example right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the restorative force is going to be equal to minus K times how much it's distorted times minus 10. So the minuses cancel out, so it equals 10K. What's the restorative force in this example? Well, you might say, it's 5 Newtons, just because that's the only force I've drawn here, and you would be to some degree correct. And actually, since we're doing positive and negative, and this 5 Newton is to the left, so to the negative x-direction, actually, I should call this minus 5 Newtons and I should call this minus 10 Newtons, because obviously, these are vectors and we're going to the left. I picked the convention that to the left means negative. So what's the restorative force? Well, in this example-- and we assume that K is a positive number for our purposes. In this example, the restorative force is a positive number. So what is the restorative force? So that's actually the force, the counteracting force, of the spring. That's what this formula gives us. So if this spring is stationary when I apply this 5-Newton force, that means that there must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring would keep compressing. And if the force was more than 5 Newtons, the spring would go back this way. So the fact that I know that when I apply a 5-Newton force to the left, or a negative 5-Newton force, the spring is no longer moving, it means that there must be-- or no longer accelerating, actually, it means that there must be an equal and opposite force to the right, and that's the restorative force. Another way to think about it is if I were to let-- well, I won't go in there now. So in this case, the restorative force is 5 Newtons, so we can solve for K. We could say 5 is equal to 10K. Divide both sides by 10. You get K is equal to 1/2. So now we can use that information to figure out what is the displacement when I apply a negative 10-Newton force? When I push the spring in with 10 Newtons in the leftward direction? So first of all, what's the restorative force here? Well, if the spring is no longer accelerating in either direction, or the tip of the spring is no longer accelerating in either direction, we know that the restorative force must be counterbalancing this force that I'm compressing with, right? The force that the spring wants to expand back with is 10 Newtons, positive 10 Newtons, right? And we know the spring constant, this K for this spring, for this material, whatever it might be, is 1/2. So we know the restorative force is equal to 1/2 times the distance, right? And the formula is minus K, right? And then, what is the restorative force in this example? Well I said it's 10 Newtons, so we know that 10 Newtons is equal to minus 1/2x. And so what is x? Well, multiply both sides by minus 1/2, and you get minus 20. I'm sorry, multiply both sides by minus 2, you get minus 20 is equal to x. So x goes to the left 20 units. So that's all that it's telling us. And this law is called Hooke's Law, and it's named after-- I'll read it-- a physicist in the 17th century, a British physicist. And he figured out that the amount of force necessary to keep a spring compressed is proportional to how much you've compressed it. And that's all that this formula says. And that negative number, remember, this formula gives us the restorative force. So it says that the force is always in the opposite direction of how much you displace it. So, for example, if you were to displace this spring in this direction, if you were to apply a force and x were a positive and you were to go in that direction, the force-- no wait, sorry. This is where the spring rests. If you were to apply some force and take the spring out to here, this negative number tells us that the spring will essentially try to pull back with the restorative force in the other direction. Let's do one more problem and I think this will be clear to you. So let's say I have a spring, and all of these problems kind of go along. So let's say when I apply a force of 2 Newtons, so this is what I apply when I apply a force of 2 Newtons. Well, let's say it this way. Let's say when I stretch the spring. Let's say this is the spring, and when I apply a force of 2 Newtons to the right, the spring gets stretched 1 meter. So first of all, let's figure out what K is. So if the spring is stretched by 1 meter, out here, its restorative force will be 2 Newtons back this way, right? So its restorative force, this 2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter, so then we multiply both sides by negative 1, and we get K is equal to minus 2. So then we can use Hooke's Law to note the equation for this-- to figure out the restorative force for this particular spring, and it would be minus 2x. And then I said, well, how much force would I have to apply to distort the spring by 2 meters? Well, it's 2 times 2, it would be 4. 4 Newtons to displace it by 2 meters, and, of course, the restorative force will then be in the opposite direction, and that's where we get the negative number. Anyway, I've run out of time. I'll see you in the next video. [Creative Commons Attribution/Non-Commercial/Share-Alike](https://creativecommons.org/licenses/by-nc-sa/4.0)[Video on YouTube](https://www.youtube.com/watch?v=ZzwuHS9ldbY) [Up next: video](https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-work-energy-and-power/in-in-class11-spring-potential-energy-and-hookes-law/v/potential-energy-stored-in-a-spring)
Readable Markdown	## Video transcript Let's learn a little bit about springs. So let's say I have a spring. Let me draw the ground so that we know what's going on with the spring. So let me see, this is the floor. That's the floor, and I have a spring. It's along the floor. I'll use a thicker one, just to show it's a spring. Let's say the spring looks something like this. Whoops, I'm still using the line tool. So the spring looks like this. This is my spring, my amazingly drawn spring. Let's say at this end it's attached to a wall. That's a wall. And so this is a spring when I don't have any force acting on it, this is just the natural state of the spring. And we could call this, where it just naturally rests, this tip of the spring. And let's say that when I were to apply a force of 5 Newtons into the spring, it looks something like this. Redraw everything. So when I apply a force of 5 Newtons-- I'll draw the wall in magenta now. When I apply a force of 5 Newtons, the spring looks like this. It compresses, right? We're all familiar with this. We sit on a bed every day or a sofa. So let's say it compresses to here. If this was the normal resting-- so this is where the spring was when I applied no force, but when I applied 5 Newtons in that direction, let's say that this distance right here is 10 meters. And so a typical question that you'll see, and we'll explain how to do it, is a spring compresses or elongates when you apply a certain force by some distance. How much will it compress when you apply a different force? So my question is how much will it compress when I apply a 10-Newton force? So your intuition that it'll compress more is correct, but is it linear to how much I compress it? Is it a square of how much I compress it? How does it relate? I think you probably could guess. It's actually worth an experiment. Or you could just keep watching the video. So let's say I apply a 10-Newton force. What will the spring look like? Well, it'll be more compressed. Drop my force to 10 Newtons. And if this was the natural place where the spring would rest, what is this distance? Well, it turns out that it is linear. What do I mean by linear? Well, it means that the more the force-- it's equally proportional to how much the spring will compress. And it actually works the other way. If you applied 5 Newtons in this direction, to the right, you would have gone 10 meters in this direction. So it goes whether you're elongating the spring or compressing the spring within some reasonable tolerance. We've all had this experience. If you compress something too much or you stretch it too much, it doesn't really go back to where it was before. But within some reasonable tolerance, it's proportional. So what does that mean? That means that the restoring force of the spring is minus some number, times the displacement of the spring. So what does this mean? So in this example right here, what was the displacement of the spring? Well, if we take positive x to the right and negative x to the left, the displacement of the spring was what? The displacement, in this example right here, x is equal to minus 10, right? Because I went 10 to the left. And so it says that the restorative force is going to be equal to minus K times how much it's distorted times minus 10. So the minuses cancel out, so it equals 10K. What's the restorative force in this example? Well, you might say, it's 5 Newtons, just because that's the only force I've drawn here, and you would be to some degree correct. And actually, since we're doing positive and negative, and this 5 Newton is to the left, so to the negative x-direction, actually, I should call this minus 5 Newtons and I should call this minus 10 Newtons, because obviously, these are vectors and we're going to the left. I picked the convention that to the left means negative. So what's the restorative force? Well, in this example-- and we assume that K is a positive number for our purposes. In this example, the restorative force is a positive number. So what is the restorative force? So that's actually the force, the counteracting force, of the spring. That's what this formula gives us. So if this spring is stationary when I apply this 5-Newton force, that means that there must be another equal and opposite force that's positive 5 Newtons, right? If there weren't, the spring would keep compressing. And if the force was more than 5 Newtons, the spring would go back this way. So the fact that I know that when I apply a 5-Newton force to the left, or a negative 5-Newton force, the spring is no longer moving, it means that there must be-- or no longer accelerating, actually, it means that there must be an equal and opposite force to the right, and that's the restorative force. Another way to think about it is if I were to let-- well, I won't go in there now. So in this case, the restorative force is 5 Newtons, so we can solve for K. We could say 5 is equal to 10K. Divide both sides by 10. You get K is equal to 1/2. So now we can use that information to figure out what is the displacement when I apply a negative 10-Newton force? When I push the spring in with 10 Newtons in the leftward direction? So first of all, what's the restorative force here? Well, if the spring is no longer accelerating in either direction, or the tip of the spring is no longer accelerating in either direction, we know that the restorative force must be counterbalancing this force that I'm compressing with, right? The force that the spring wants to expand back with is 10 Newtons, positive 10 Newtons, right? And we know the spring constant, this K for this spring, for this material, whatever it might be, is 1/2. So we know the restorative force is equal to 1/2 times the distance, right? And the formula is minus K, right? And then, what is the restorative force in this example? Well I said it's 10 Newtons, so we know that 10 Newtons is equal to minus 1/2x. And so what is x? Well, multiply both sides by minus 1/2, and you get minus 20. I'm sorry, multiply both sides by minus 2, you get minus 20 is equal to x. So x goes to the left 20 units. So that's all that it's telling us. And this law is called Hooke's Law, and it's named after-- I'll read it-- a physicist in the 17th century, a British physicist. And he figured out that the amount of force necessary to keep a spring compressed is proportional to how much you've compressed it. And that's all that this formula says. And that negative number, remember, this formula gives us the restorative force. So it says that the force is always in the opposite direction of how much you displace it. So, for example, if you were to displace this spring in this direction, if you were to apply a force and x were a positive and you were to go in that direction, the force-- no wait, sorry. This is where the spring rests. If you were to apply some force and take the spring out to here, this negative number tells us that the spring will essentially try to pull back with the restorative force in the other direction. Let's do one more problem and I think this will be clear to you. So let's say I have a spring, and all of these problems kind of go along. So let's say when I apply a force of 2 Newtons, so this is what I apply when I apply a force of 2 Newtons. Well, let's say it this way. Let's say when I stretch the spring. Let's say this is the spring, and when I apply a force of 2 Newtons to the right, the spring gets stretched 1 meter. So first of all, let's figure out what K is. So if the spring is stretched by 1 meter, out here, its restorative force will be 2 Newtons back this way, right? So its restorative force, this 2 Newtons, will equal minus K times how much I displaced it. Well I, displaced it by 1 meter, so then we multiply both sides by negative 1, and we get K is equal to minus 2. So then we can use Hooke's Law to note the equation for this-- to figure out the restorative force for this particular spring, and it would be minus 2x. And then I said, well, how much force would I have to apply to distort the spring by 2 meters? Well, it's 2 times 2, it would be 4. 4 Newtons to displace it by 2 meters, and, of course, the restorative force will then be in the opposite direction, and that's where we get the negative number. Anyway, I've run out of time. I'll see you in the next video.
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