Hooke's Law and Elastic Potential Energy - Free Educational videos for Students in K-12

Hooke's Law and Elastic Potential Energy - By The Organic Chemistry Tutor

Transcript


00:01	in this video , we're going to talk about hook's
00:03	law and springs . We're also going to go over
00:06	some problems associated with it . So what's the basic
00:10	idea behind Hook's law ? So let's see if we
00:14	have a horizontal spring . And so that's the natural
00:18	left of the horizontal spring . Now , what we're
00:22	gonna do is we're gonna apply a force to stretch
00:25	the spring towards the right . We're going to call
00:31	that force F . P . Since we're pulling the
00:36	string towards the right now , that force is positive
00:43	and the displacement , other spring is also positive .
00:50	Now we stretched it a distance which will call X
00:57	. It turns out that this force , the force
01:00	that's needed to stretch it by distance , X .
01:04	Is proportional to X . So F . P is
01:06	equal to K . X . Where K . Is
01:10	a proportionality constant . Also known as the spring constant
01:14	. Now this equation works up to a limit ,
01:18	so F and X a proportional up to limit once
01:21	you pass the elasticity region , then the spring can
01:26	elongate without much extra force . So we're going to
01:30	focus on the elastic region where F . And X
01:34	. Are proportional . Now it turns out that as
01:38	you stretch the spring once you at this point ,
01:41	and you apply a force F . P . If
01:45	you don't increase the force , it's not going to
01:46	stretch further . So right now it's an equilibrium ,
01:49	which means that there's another force that is pulling the
01:55	spring back to its original left . And that voice
01:59	is known as the restoring force , which you can
02:02	call it F . S . Or fr . And
02:05	the restoring force is in the negative X . Direction
02:10	. Now these two forces , they're equal in magnitude
02:14	but opposite in direction because once you apply force FP
02:21	that springs that rest , it's not moving to the
02:22	left or to the right . Once you let go
02:24	FS takes over and it's going to causes them to
02:28	snap back to its original life . So because FS
02:32	is negative , we could say that the restoring force
02:34	is negative K . X . And this equation is
02:38	associated with hooks law . So the magnitude of the
02:42	restoring force is proportional to how much you stretch or
02:47	compress the spring from its natural left . So now
02:52	that you've received a basic introduction into hooks along ,
02:57	let's go ahead and focus on Finishing this problem .
03:00	No one , A force of 200 Newton stretches of
03:04	spring by four m . What is the value of
03:08	the supreme constant ? So if the force that we
03:12	used to stretch it , which is F . P
03:14	. Is equal to K . X . Then the
03:16	spring constant . K . Is simply the ratio of
03:19	the force and the distance that you stressed the spring
03:24	. So we have a force of 200 newtons .
03:27	And it stretches the spring by a distance of four
03:29	m . So 200 divided by four is 50 .
03:33	And we can see the spring cost . That has
03:35	unit newtons per meter . So it's 50 newtons per
03:39	meter . So what this means is that in order
03:47	to stretch a spring by a distance of one m
03:49	, a force of 50 newtons is required In order
03:53	to stretch it by two m . You need the
03:55	force of 100 newtons to do so . And so
03:59	the spring constant , it tells you how much force
04:02	you need to excuse me , stretches spring by one
04:06	m . So let's say , for example , if
04:08	the spring constant was 300 newtons per meter , That
04:11	means a force of 300 newtons is required to stretch
04:15	a spring . Buy one m . Now looking at
04:20	these two spring constants , which one is more stiff
04:24	The spring ? That's $300 per meter or 15 years
04:27	per meter . This spring is going to be more
04:31	stiff . It's gonna be hard to stretch or compress
04:36	the spring because it has a higher spring constant .
04:40	This spring is going to be loose . It's one
04:41	of those light springs that you can easily pull apart
04:45	. So as K increases as the spring constant increases
04:49	in value , the stiffness of the spring increases as
04:54	well . Let's move on to number two . The
04:59	spring constant is 300 newtons per meter . What force
05:03	is required to stretch a spring By 45 cm .
05:11	So let's use this equation . F P is equal
05:13	to K . Ox . So our goal is to
05:17	find FP K is 300 newtons per meter . Now
05:25	we don't have X and meters , we have it
05:27	in centimeters . So we need to convert that two
05:29	m in order to convert centimeters to meters , You
05:33	need to divide by 100 one m Is 100 cm
05:38	. And so these units will cancel 45 divided by
05:42	100 is .45 m . So if we multiply the
05:48	spring constant by .45 m , We can see that
05:52	this is going to give us the Unit Newton's ,
05:54	which is the unit of force . So it's 300
05:59	Times .45 . And so that's going to be 135
06:04	mutants . So that's the force that's required to stretch
06:08	the spring By 45 cm . Keep in mind if
06:13	the spring constant is $300 per meter In order to
06:16	stretch it by one m , It requires a force
06:19	of 300 newtons . So thus to stretch it by
06:23	0.5 m will be 150 Nunes . So paint 45
06:27	m , which is just under half . What makes
06:30	sense that it's 1:35 Nunes . It's always good to
06:34	take a mental check of your answers . So let's
06:38	make a table between X and the four supply .
06:42	So the spring constant was 300 newtons per meter .
06:46	So that means at a distance of one m ,
06:48	a force of £300 is required . So if we
06:52	divide the distance by tune , We should divide the
06:54	force by two . So therefore I answer At .45
06:58	m is 135 , which makes sense . So this
07:04	is less than 25 and this is less than 150
07:07	. So I always try to see if your answer
07:09	is reasonable . Doesn't make sense . Now let's move
07:13	on to number three . A force of 250 newtons
07:18	is required to stretch a spring by 24 centimeters .
07:23	How far can the force of 900 and instruction ?
07:29	So we're given a force and a distance and were
07:32	given another force and we're trying to find another distance
07:35	. You can find a spring constant Kane or you
07:39	could come up with another equation to do this .
07:42	So I'm going to show you two ways of getting
07:43	the answer . The first is to calculate the supreme
07:46	constant cake . It's going to be the force divided
07:49	by the distance . So a force of 250 newtons
07:54	is required to stretch it by 24 cm , Which
07:58	is .24 m . If you divided by 100 .
08:01	So 250 divided by .24 will give us a spring
08:06	constant Of 1041 0.6 repeating now to find out how
08:16	far it's going to stretch , we need to calculate
08:18	the value of X and X is f divided by
08:23	K . So we have a new force of 900
08:27	and we're going to divide it by 10 41.6 ,
08:30	repeating . So this is going to be about point
08:40	864 m , which in cm is 86.4 cm If
08:46	you multiplied by 100 . So that's one way in
08:50	which you can get the answer . Now let me
08:52	show you another way . Let's take the second force
09:00	and let's divide it by the first force , The
09:04	second forces K times the second distance and the first
09:07	forces K times the first distance . So because K
09:12	is the same , we can cancel it . So
09:14	therefore the ratio of the forces is equal to the
09:19	ratio of the displacements and X . And it makes
09:26	sense because X and F . R proportional to each
09:28	other . So we're going to call this effluent And
09:32	this is going to be X one And this is
09:35	F two . And we're looking for X two .
09:37	Now the unit centimeters will cancel . So it doesn't
09:40	matter if you use meters or centimeters when dealing with
09:43	ratios , the ratio will still be the same .
09:47	So let's replace F two of 900 newtons and F
09:51	one With 250 Nunes . Our goal is to calculate
09:56	X two And X one is 24 cm . So
10:01	let's cross multiply . So first we have 900 tax
10:06	24 . So that's 21,600 with the units Newton's timecm
10:14	and that's equal to 250 Newtons Times X two .
10:21	Now , in order to get X two , we
10:23	gotta divide both sides by 250 newtons on the right
10:29	side . These two will cancel . Given Us X
10:33	two on the left side notice that the units new
10:37	into a council leaving behind centimeters , Which is going
10:41	to be the unit for X two . So it's
10:43	21,600 divided by 2:50 . And so you're gonna get
10:48	86.4 cm . So you have two ways in which
10:53	you can get the same answer . Consider this problem
10:57	. How much work is required To stretch a spring
11:00	by 75 cm and were given the spring constant .
11:06	So let's try spring . Now the force that's required
11:14	to stretch the spring above its natural left . Let's
11:17	say this is the natural for the spring is we're
11:23	going to call it sp . Now F . P
11:27	. Is equal two K axe . Now once you
11:32	stretch it and you stop stretching it , you still
11:35	need to apply a force to hold its position .
11:38	Now it's at rest . That means that there's an
11:40	equal and opposite force that let's snap back this spring
11:44	back to its natural left . And that is the
11:46	restoring force . Now this forces positive , it's going
11:50	in a positive X . Direction and this forces negative
11:54	. So according to Hook's law to restoring forces negative
11:57	K . Ox . But we're going to focus on
11:59	this force . How much work is done by this
12:03	force in order to stretch the spring by a displacement
12:08	. X . X . Is the distance between the
12:13	natural life of the spring . And it's currently so
12:19	it's how much it stretches or compresses by . Now
12:24	work is equal to force times displacement . So the
12:29	displacement is X . So work is equal to fx
12:34	. Now s it's not a constant value . F
12:38	depends on X . F . Is a function of
12:40	X . So if we were to make a graph
12:44	between force and displacement , it would look something like
12:47	this . Yeah . Now F is not a constant
12:55	value but it's a function of X . It's equal
12:58	to K . Ox . And so all we have
13:00	is a linear equation where the slope is came .
13:04	So this is going to be a straight line that
13:05	looks like this . Now , whenever you have a
13:09	forced displacement graph , we know that work is force
13:12	times displacement . But for such a graph , the
13:15	work is equal to the area under the curve .
13:20	So what is the area of the triangle ? The
13:22	area of a triangle is one half base times height
13:30	. In this example , the base of the triangle
13:32	is equal to X . And the height of the
13:35	triangle is equal to F . So therefore the area
13:40	is 1/2 F times X . Where F is the
13:44	maximum uh force value at a distance X . So
13:50	this equation holds true if the force is constant .
13:54	So if he had a graph where this is F
13:57	. And this is X . And the force Wisconsin
14:00	, the area will be the area of the rectangle
14:02	, which is left times with that would be simply
14:04	F times X . There won't be a half in
14:08	front of it . That's if the force is constant
14:12	. Now what we have is a variable force that
14:14	increases according to the displacement or how much you stretch
14:18	it . So therefore the work done by that variable
14:21	force is the area under curve , which is 1/2
14:25	f times x . Mhm . Yeah , So the
14:33	work required to stretch the spring is 1/2 times the
14:37	force , the maximum force times X . And based
14:41	on hooks along , we know that F . Is
14:42	equal to K . Ox . So what I'm gonna
14:45	do is replace F with chaos . And keep in
14:50	mind this axe is really and delta X . So
14:55	this is going to be cain delta X . And
14:58	this is supposed to be DELTA X . Because it's
15:00	really the change in the position . So let's say
15:06	if this is the natural life of the spring ,
15:08	it's at position X . A . And you stretch
15:12	it to a new position XP . This is the
15:17	difference between XB and exciting . So technically it's delta
15:21	X . So therefore The work required to stretch the
15:25	spring is going to be 1/2 Okay , times delta
15:30	X squared . So this is the equation that you
15:33	want to use . So Kay is 250 newtons per
15:43	meter And we want to stretch the spring by 75cm
15:47	. However , we need to convert that to meters
15:50	To convert cm to m divided by 100 . So
15:53	this is going to be .75 m squared . So
15:59	it's .75 squared times 2 50 times 25 . So
16:06	therefore the work required is 70 0.3 jewels of energy
16:16	. Now , what is the elastic potential energy stored
16:19	in the spring When it is stretched 25 cm by
16:23	a 450 new enforce . So let's come up with
16:26	the equation to calculate the elastic potential energy first .
16:32	So we said that The work required to stretch a
16:35	spring by the applied force is 1/2 cane delta X
16:40	squared . Mhm . Now the work done by such
16:46	a force , it's equal to the change in the
16:51	potential energy . If the acceleration is zero , which
16:55	it is the applied force and restoring force their equal
16:58	to each other . So work is equal to the
17:05	final potential energy minus the initial potential energy . Now
17:11	some textbooks may describe the elastic potential energy as U
17:15	. S . Capital . You can be described as
17:17	potential energy . So we see us . That's elastic
17:20	potential energy . Yugi gravitational potential energy . So just
17:25	keep that in mind . I like to use pE
17:27	because pe potential energy it makes sense . So what
17:31	I'm gonna do is replace the W . With this
17:34	expression . So 1/2 delta X squared is equal to
17:42	the stuff that we have on the right and delta
17:45	acts is basically the final position minus initial position .
17:48	So we have one half K X final squared minus
17:54	X initial square . So one half K X vital
18:01	square minus one half K X . Initial square is
18:06	equal to the final potential energy minus the initial potential
18:11	energy . So what does this all mean ? What
18:14	would you see that ? This is equal to the
18:16	final potential energy and this is equal to initial potential
18:21	energy . Therefore the elastic potential energy in general is
18:26	simply one half K X squared . So notice the
18:30	difference between this equation and this equation . So what
18:40	do you notice the work depends on the change of
18:44	position , whereas the potential energy , it depends on
18:48	X and X by itself . It has to be
18:52	relative to the natural length of the spring . So
18:55	the potential energy is based on the relative length between
18:59	where you are and the natural length of the spring
19:02	. The work required doesn't have to be associated with
19:06	the natural life for the spring . It's simply the
19:08	energy required to move it from position A . To
19:12	position . Being for this equation , Position A .
19:16	Has to be the natural for the spring . But
19:19	for this one position A . Or B does not
19:21	have to be the natural left of spring . So
19:23	keep that in mind . Mhm . Well , I
19:27	mean naturally you can also describe as the natural position
19:30	of the spring , Which is usually defined as x
19:33	equals zero . So let's focus on this equation .
19:38	What is the elastic potential energy stored in the spring
19:41	? So we're given X . And were given the
19:44	force F . Mhm . So we know that the
19:48	elastic potential energy is one half K . X .
19:52	Square . The only thing we're missing is king .
19:56	So we could find K . Using this equation F
19:58	. Equals K . Ox . So , okay ,
20:01	is the ratio between the force and the displacement relative
20:06	to the the origin or the natural position of the
20:09	spring ? So it's gonna be 450 newtons divided by
20:15	25 cm in m , which is .25 if you
20:18	divided by 100 . So 450 divided by .25 is
20:25	1800 . So that's the spring constant . So this
20:30	spring is very stiff as K increases the stiffness of
20:37	the spring increases . So just Stressed to spring by
20:43	a distance of one m Requested a force of 1800
20:48	. That's what the spring constant tells us . So
20:54	now that we have cane , let's calculate the potential
20:58	energy . So it's one half times came times X
21:04	squared Where X is .25 m . Mhm . So
21:10	.25 squared Times 1800 times .5 , That's equal to
21:18	56.25 jewels . And so that's the answer . How
21:25	much work is required to stretch of spring from 4.5
21:29	centimeters to 8.2 centimeters given a spring constant of 100
21:33	and 50 newtons per meter . So let me give
21:35	you a visual first of what we have here .
21:38	So let's say this is the spring at its natural
21:41	left , so at this point X is equal to
21:43	zero . Now the spring is currently stretch to this
21:50	position , let's call it exciting . That's when X
21:55	is equal to 4.5 . And we need to calculate
22:01	how much work is required to stretch it further to
22:05	position B Where X is April two cm . So
22:12	how can we find the work required to go from
22:14	position A to position B . Go ahead and try
22:23	that problem . The equation that you need is this
22:29	one one half K X k times delta X square
22:38	. And I need to be careful because there's something
22:42	you don't want to get mixed up with . When
22:48	I wrote this expression , I meant that delta X
22:51	squared Is equal to 1/2 times X final square minus
22:58	X initial squared . So not like this , not
23:03	1/2 came X final minus accident . Show squared .
23:10	If you do it this way you will get the
23:12	wrong answer . So it's the difference of the squares
23:16	of X . Not the square difference of the X
23:21	values if that makes sense . So this is the
23:25	right way to calculate it because then work is equal
23:27	to the change in potential energy . So if you
23:32	do it this way , you're not gonna get the
23:33	right answer . I just want to highlight that point
23:38	. So it's gonna be one half times spring constant
23:42	, which is 1:50 times the square of the final
23:46	position . Which We need to be in m .
23:49	So that's going to be playing 082 meters squared minus
23:54	the square of the initial position , which is point
23:57	045 meters squared . So go ahead and type this
24:02	in exactly the way you see it . So you
24:14	should get .352 jewels for the work required to stretch
24:20	it from position A to position B . Now ,
24:24	for those of you who may want another way to
24:27	verify this answer , calculate the potential energy at points
24:32	A and B . So the potential energy at point
24:35	A . It's going to be 1/2 K . X
24:39	square where X . Is that position A . So
24:42	it's one half times 1 50 times . Print 045
24:48	m square . So opposition eh The elastic potential energy
24:58	is .15 19 jewels . Now let's calculate the elastic
25:04	potential energy , opposition B . So it's 1/2 ex
25:13	B squared . That's 1/2 times 1 15 times 0.0
25:20	82 square . And so this will give you point
25:28	5043 jewels . So keep in mind you can always
25:34	calculate the work by calculating the change of potential energy
25:44	. So it's going to be the difference between these
25:46	two values . So it's .5043 -1519 . And so
25:58	you get the same answer .352 jewels . So the
26:05	work done by a force is equal to the change
26:08	in potential energy . Yes , Then the acceleration of
26:12	the system is zero . Now , there are a
26:15	few additional details that I want to mention regarding the
26:19	work required to pull a spring . As I mentioned
26:26	earlier in the video sp is the pulling force or
26:30	the force that we use to stretch a spring .
26:35	And based on the way the the drawing that we
26:39	have on the screen since it's directed to the right
26:42	, this is going to be defined as a positive
26:44	force . And we described Fs as the force exerted
26:50	by the spring , which is a restoring force .
26:53	That force tries to bring the spring back to equilibrium
26:58	as we stretch the spring to the right . The
27:01	displacement vector is towards the right . And we know
27:05	that work is force times displacement . Notice that these
27:11	two vectors are in the same direction . Therefore the
27:14	work that's required to stretch a spring , it's positive
27:20	. Now , the work that's done by the spring
27:23	itself as it pulls the spring back to equilibrium ,
27:27	That work is negative as you can see the force
27:32	and the displacement vectors , they're in opposite directions .
27:37	So therefore the work required to stretch of spring is
27:41	positive . One half K and delta X squared .
27:50	Now the work done by the elastic force that is
27:54	F . S . The work done by the spring
27:58	, that's gonna be negative . Yeah , one half
28:02	K delta X squared as the spring is being pulled
28:07	to the right Now , this makes sense because in
28:13	other videos , I mentioned that the work done by
28:16	conservative forces like gravity , the electric force or in
28:21	this case the elastic force negative Fs that work is
28:26	negative times to change in potential energy and the changing
28:31	potential energy is equal to this expression , it's one
28:35	half K delta X squared . This is for those
28:40	of you who may be wondering whether or not you
28:43	should have a positive or negative sign . It really
28:46	depends on which force you're talking about and where the
28:50	spring is being , whether it's being stretched or compressed
28:54	. So just remember the work is positive if the
28:56	force and the displacement vector are in the same direction
29:00	, but its negative if they're in opposite directions .

Summarizer

DESCRIPTION:

This video provides a basic introduction into Hooke's law. It explains how to calculate the elastic potential energy and how to determine the amount of work required to stretch a spring.

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