Conservation of Energy Physics Problems - Free Educational videos for Students in K-12

Conservation of Energy Physics Problems - By The Organic Chemistry Tutor

Transcript


00:01	Let's work on this problem . A block slides down
00:04	a 150 m inclined plane , as shown in the
00:07	picture below , starting from us , what is the
00:10	speed of the block when it reaches the bottom of
00:13	the incline ? So we're gonna use conservation of energy
00:17	to solve this problem . So the initial mechanical energy
00:23	has to equal the final mechanical energy . The only
00:27	forces acting on the block , Our conservative forces like
00:32	gravity . So mechanical energy is can serve at point
00:37	A . The only form of energy that we have
00:39	is potential energy . And that point B . Choosing
00:43	this as the ground level , there's no potential energy
00:45	A point B . However , the block will have
00:48	kinetic energy at point B . So we can set
00:51	these two equal to each other As the block slides
00:54	down , potential energy is being converted to kinetic energy
00:59	. The potential energy of the block is MGH and
01:02	the kinetic energy of the block is one half mv
01:05	squared so we could cancel them . Therefore we don't
01:09	need to master block . Now , I'm going to
01:12	multiply both sides by two , so on the left
01:15	have to G . H . Is equal to two
01:18	times a half these will cancel . So that's one
01:21	and that's going to equal the square , taking the
01:24	square root of both sides . The final speed is
01:27	going to be the square root of two G .
01:29	H . So let's go ahead and plug everything that
01:34	we have . So it's two times the gravitational acceleration
01:39	of 9.8 Time is the height of the block ,
01:42	which is 150 m high . So by the way
01:46	, this is the height , So that's 150 m
01:52	. So if you type this in your calculator ,
01:58	you should get 54 0.22 m/s . So without friction
02:06	, that's how fast this block is going to be
02:08	moving when it features position be number two . An
02:15	eight kg Block compresses a horizontal spring By 2.5 m
02:20	beyond its natural length . As shown in the figure
02:23	below , what is the speed of the block ?
02:26	As soon as it's released from the spring ? So
02:29	let's call this position A an opposition B when it's
02:32	still on level ground , you want to find the
02:35	speed as soon as it's release . So opposition eh
02:41	The block spring system has potential energy . Elastic potential
02:45	energy energy is stored in the spring and what's its
02:50	release ? That energy is going to be converted to
02:52	kinetic energy . Now the height doesn't change . So
02:56	there's no gravitational potential energy . So we can say
03:00	the elastic potential energy will be equal to the kinetic
03:03	energy of the block . Energy is going to be
03:06	transferred from the spring to the block . The elastic
03:10	potential energy is 1/2 KX Square . The kinetic energy
03:14	is 1/2 M . V . Squared . If we
03:17	multiply both sides by two , we can get rid
03:19	of the fraction . So now we're going to start
03:23	for the speed . So we have K . Which
03:26	is 300 And X . is the amount the spring
03:30	is compressed by which is 2.5 m . The mass
03:34	of the block is eight . To let's calculate the
03:36	final speed 300 times 2.5 Squared is 1875 . And
03:43	if we divide that by eight , That's going to
03:46	be 234.375 . And that's equal to V squared .
03:53	Now let's take the square root of both sides .
04:00	So the speed is going to be 15.31 meters per
04:05	second . So this is the answer to part eight
04:10	. That's gonna be the speed of the block as
04:14	soon as it's released opposition B now part B .
04:18	How high up the hill with a black . Oh
04:21	so let's see if it gets up to that height
04:23	. How can we calculate H . Now ? Opposition
04:27	be the block has kinetic energy opposition . See where
04:31	it comes to rest . It's no longer going to
04:33	have any kinetic energy . The only energy is going
04:36	to have is gravitational potential energy . Mhm . So
04:41	what we need to do is that the kinetic energy
04:46	equal to the gravitational potential energy . So this is
04:49	gonna be one half Mv squared and that's going to
04:52	equal M . G . H . So once again
04:55	we could cancel em . Now let's go ahead and
04:59	plug in everything that we have . So it's one
05:01	half times v . Squared which is 15.31 square and
05:06	that's equal to G . Times H . So 15.31
05:13	Squared times 0.5 Divided by 9.8 will give us a
05:20	height of about 12 m if you wanted to nurse
05:24	whole number . So that's how high the block is
05:29	going to go before it comes to rest . Number
05:33	three , A 10 kg Block slides down a hill
05:38	That is 200 m tall . As shown in the
05:41	figure below With an initial speed of 12 m/s .
05:47	What is the speed of the block when it reaches
05:49	the bottom of the hill ? At let's say position
05:52	be assuming there's no friction . So opposition . A
06:01	Let's call that the original position . The block has
06:05	potential energy because it's above position be . It has
06:11	the capability of falling now because the block is in
06:14	motion . It has kinetic energy . Opposition being It's
06:19	only going to have kinetic energy so opposition A .
06:25	We have potential energy and kinetic energy . Opposition be
06:31	We only have kinetic energy . So all of the
06:34	potential energy opposition came is going to go to the
06:38	kinetic energy of the object . The object speed will
06:41	increase beyond 12 m/s . So let's go ahead and
06:45	calculate that speed . So this is going to be
06:48	MGH plus one half and the initial square . So
06:54	the initial speed is 12 . We're looking for the
06:56	final speed And that's equal to the final kinetic energy
06:59	of 1/2 MV Final Square . So we could divide
07:04	everything by em and multiply everything by two . So
07:08	it's gonna be to G . H . Plus the
07:12	initial squared and that's equal to the final square .
07:15	So basically that looks like this equation . The final
07:18	squared is equal to the initial squared plus two A
07:21	. D . That's a formula that you've seen in
07:24	schematics . So let's calculate the final speed . So
07:31	the final squared is equal to the initial squared which
07:33	is 12 square plus two times the gravitational acceleration Times
07:39	A height of 200 . So it's 12 square plus
07:45	two times 9.8 times 200 . And that's 4064 and
07:50	then take the square root of that number . And
07:53	you should get a final speed of 63 .75 m/s
08:01	. So this is the speed of the object opposition
08:04	being . If there's no friction , now let's move
08:08	on to part being what is the final speed of
08:11	the block opposition B After traveling a total distance of
08:14	500 m , given a coefficient of kinetic friction Of
08:19	.21 between the block and the surface . Now I
08:22	do want to modify this . So let's say that
08:25	the distance is only for the horizontal portion of the
08:30	incline . So let's say that distance is 500 m
08:35	, so not the total part . So let's say
08:38	this portion , it's frictionless And the 500 m is
08:42	only for the horizontal portion of the surface . Knowing
08:49	that how can we calculate the final speed of the
08:51	block ? A friction only acts on the bottom of
08:54	the incline . So how can we modify this equation
09:02	? Opposition A . We're still going to have potential
09:05	and kinetic energy . Now once you add friction to
09:09	the mix the final speed will no longer be 63.75
09:13	, it's going to be less . So some of
09:17	the energy from the left side is not all going
09:19	to go into kinetic energy , some of it is
09:22	going to go into the work done by friction .
09:25	Now if you put the work done by friction on
09:27	the left side it's going to be negative W because
09:29	friction is going to decrease um the total energy of
09:33	the object . And if it's negative W . On
09:36	the left side then it's positive W . On the
09:38	right side . So this represents the work done by
09:41	friction frictions . Job is to take away the mechanical
09:46	energy of the object and to convert it to thermal
09:49	energy . So the energy loss from the object it's
09:54	going to be W . And that amount of energy
09:56	is going to be converted to heat energy . So
10:01	let's go ahead and calculate w . 1st . So
10:14	the work done by friction is equal to the kinetic
10:18	friction of force . Times the distance that the force
10:21	acts over distance and displacement are the same if the
10:26	object doesn't change direction . So this work is force
10:30	times displacement . Now the frictional force is equal to
10:36	U . K . Times the normal force . So
10:40	we have the coefficient of kinetic friction and the normal
10:44	force on a horizontal surface is MG . Now if
10:49	this part had friction that would complicate the problem because
10:52	the normal force on let's say an incline is MG
10:57	co signed data and this is not a straight incline
11:01	so the angle changes . So that's just going to
11:03	make it more complicated . So let's keep the problem
11:06	simple . So now let's go ahead and calculate the
11:11	work done by friction . So it's um UK .
11:13	Which is 0.21 Multiplied by the mass of the object
11:17	which is 10 times gene Times the distance that the
11:21	object acts over which is 500 m . So let's
11:26	multiply point to one by 10 By 9.8 and by
11:31	500 . So you should get 10,000 290 jewels .
11:41	So that's the work done by friction . Now let's
11:47	use that to calculate the final kinetic energy and also
11:51	the final speed . So the potential energy is M
11:55	. G . H . And the kinetic energy opposition
11:58	A . Is one half and the initial square And
12:02	the final Kinetic energy is 1/2 and the final square
12:06	and then plus w . So let's plug in the
12:23	values that we have . So the potential energy is
12:25	gonna be 10 Time is 9.8 Times the height of
12:29	200 . And the initial kinetic energy is gonna be
12:33	1/2 times 10 times the initial speed which is 12
12:40	m/s . And then that's going to be the final
12:44	kinetic energy . Plus the work done by friction which
12:50	we can replace that with 10,290 jewels . So 10
12:56	times 9.8 times 200 That's 19,600 jewels . So there's
13:03	enough potential energy to satisfy the work done by friction
13:07	. Which tells us that The final speed is going
13:11	to be greater than 12 . If this was less
13:15	than friction , the final speed will be less than
13:16	12 . Now , let's multiply .5 x 10 by
13:21	12 square . So the kinetic energy of the object
13:24	is very low . It's 720 . Make sure that's
13:29	right . And yeah , that's it . So now
13:40	let's add 19,600 plus 7:20 Unless subtract it by 10,290
13:48	. So the final kinetic energy is 10,030 jewels .
13:54	So now let's set that equal to 1/2 MV final
13:58	square . So we have one half times a massive
14:04	10 times the final square . So half of tennis
14:09	, five and 10,030 divided by five is 2006 .
14:15	So the square of the final speed is 2006 .
14:19	Now let's take the square root of that number .
14:22	So the final speed is going to be 44 0.8
14:27	meters per second , which was less than the final
14:31	speed in part a a 12 kg block , moving
14:34	at a speed of 15 m per second , crashes
14:38	into a wall and comes to a complete stop .
14:41	How much thermal energy was produced during the collision .
14:46	So let's draw a picture . So , imagine you
14:48	have a a block that sliding , It's moving at
14:52	a speed of 15 m/s And it's a 12 kg
14:56	mass , and then it collides with a wall .
15:02	So once it hits the wall it comes to a
15:04	stop . So it's no longer moving . Where did
15:07	all of the kinetic energy go during collisions whenever you
15:12	have an object in motion and then it's no longer
15:15	in motion . You need to realize that the kinetic
15:18	energy of that object was transformed into thermal energy ,
15:22	it's lost due to heat , and that heat just
15:25	radiates outward into the surroundings . Think of let's say
15:31	if you rub your hands quickly , you're going to
15:33	feel that your hands start to get hot . So
15:36	all of that kinetic energy that you are that your
15:39	hands had as was moving back and forth . A
15:42	lot of it was converted to thermal energy and so
15:44	he was generated and that's what's going to happen here
15:48	when that block collides with the wall and both objects
15:52	where the wall is not moving , but the block
15:54	comes to a stop . All of the kinetic energy
15:57	that the bloc had was transferred to thermal energy .
16:03	So let's calculate the initial kinetic energy of the block
16:06	. So it's one half MV squared . So that's
16:10	one half times a massive 12 kg Time to speed
16:16	of 15 m/s . So half of 12 , 6
16:20	and 15 squared is to 25 Six times 2 25
16:28	is 1350 . So 13 15 jewels of kinetic energy
16:35	was transformed into thermal energy . So that's the answer
16:40	. No , let's move on to this problem .
16:43	A 1500 kg car Moving east at 35 m/s ,
16:49	Crashes head on into another 1800 kg car . Moving
16:53	west at 30 m/s , causing both cars to come
16:57	to a complete stop . So let's turn this into
17:00	a picture . So this is the 1500 kg object
17:07	. It's a car , but I'm just going to
17:08	draw a box and it's moving east At 35 m/s
17:14	. And then we have another Object which is 1800
17:18	kg and this vehicle is moving west At 30 m/s
17:30	. So they collide and they come to a complete
17:32	stop . So all of the kinetic energies of these
17:36	two objects that they once had , all of that
17:39	was converted to thermal energy . So to calculate the
17:42	total thermal energy produced , we need to calculate the
17:46	total kinetic energy which is the kinetic energy of the
17:51	first object . I'm going to call the object A
17:53	. Plus the kinetic energy of the . Excuse me
17:57	, The second object object B . So let's call
17:59	this A . And B . So this is going
18:03	to be 1/2 M . V . Squared plus one
18:07	half M . V . Squared . So the mass
18:12	of the first object is 1500 And it has a
18:14	speed of 35 m/s . The mass of the second
18:18	object is 1800 And it has a speed of 30
18:23	m/s . So .5 times 1500 times 35 squared .
18:31	That's 918,000 750 jewels . And then .5 times 1800
18:40	times 30 . Swearing Is 810,000 jewels . So if
18:46	we add these two numbers , This will give us
18:51	one million 728,000 750 jewels . So that's how much
18:59	thermal energy was produced during this collision . Consider this
19:07	problem . So we have a roller coaster that's released
19:10	from rest at .8 . How fast is it moving
19:15	at point B . So how can we find the
19:19	speed at point B . We need to use conservation
19:23	of energy at 0.8 . The roller coaster has potential
19:27	energy at point B . All of that potential energy
19:31	is converted to kinetic . So let's set the potential
19:35	energy , appoint a equal to the kinetic energy of
19:38	the roller coaster at point B . So we have
19:41	M . G . H . Will equal 1/2 Mv
19:45	swear . So we could divide both sides by em
19:50	. And then if we multiply both sides by tune
19:53	, we're gonna have to G . H . Is
19:55	equal to One half times two is once . So
19:58	it's just going to be v . Squared . So
20:02	the velocity , it's going to be the square root
20:05	Of two G . H . So it's going to
20:09	be the square root of two Times 9.8 . Multiplied
20:14	by the height which is 50 . So let's go
20:19	ahead and plug this in . So you should get
20:27	31.3 meters per second at point B . So that's
20:33	how fast the roller coaster is moving at point B
20:37	. If there's no friction in this problem now no
20:39	friction was mentioned . So we're assuming there's no friction
20:55	. Now it looks like I forgot to write part
20:57	B . So we're going to move on to part
21:00	Scene . How high is point C . Relative to
21:03	point B . And I forgot to give you the
21:05	speed at point C . To point seen . The
21:10	speed Is 20 m/s . So with that information calculate
21:18	the height of the roller coaster appoint scene , Point
21:25	B . Is ground level . So how high is
21:27	the point C . Feel free to pause the video
21:30	. So I'm going to focus on point A .
21:33	As my initial point . So the .8 all we
21:37	had to begin with for its potential energy appoint scene
21:41	. The object is above ground level . So it
21:44	has potential energy now it's still moving at point C
21:49	. So it also has kinetic energy . So we
21:53	need to start with this expression . So we have
21:55	M . G . H . This is gonna be
21:57	a church initial is equal to MGH final plus one
22:03	half M . V . Final square . Now we
22:08	can divide every turn by em . And so we
22:11	have this expression G . H . It's equal to
22:15	G . H . Final plus one half the final
22:20	square . So now let's plug in the numbers .
22:24	It's going to be 9.8 Times the initial height which
22:28	is 50 . And that's equal to 9.8 times the
22:32	final height which is what we're looking for . And
22:37	then plus 1/2 the final square the final at Point
22:43	c . s . 20 . So 9.8 times 50
22:48	. That's 490 . And so that's equal to 9.8
22:52	times h . And 20 squared is 400 . Half
22:57	of that is 200 Now 4 90 -200 is 2
23:01	90 . So let's take 2 90 . Unless divided
23:06	by 9.8 . Still 29 0.59 is equal to age
23:14	. So that's how high it is at point C
23:21	. Now let's move on to part D . How
23:24	fast is the roller coaster moving ? Uh pointy .
23:28	Mhm . So we're still gonna use part I mean
23:33	.8 as our initial point . So at point A
23:38	we still have initial potential energy at Point Dean .
23:43	We still have potential energy because it's above ground level
23:46	has the ability to fall . And that point D
23:50	. It's still moving so it has kinetic energy .
23:53	So we're gonna have the same equation . M .
23:56	G . H . Initial is equal to MGH final
24:02	plus one half M . V . Final square .
24:07	And just like before we could cancel them . So
24:11	G . is 9.8 . H . initial is still
24:15	50 Now H Final is now 15 . So the
24:20	difference between part D . And part C . Is
24:25	that in part D . We're looking for the final
24:27	speed now . So 9.8 times 15 . That's still
24:33	4 90 . And that's equal to 9.8 times 15
24:39	Which is 1 47 . Yeah . Now let's take
24:48	4 90 and subtracted by 147 . So that's going
24:54	to be 343 Is equal to 1/2 the final square
24:59	Now multiply both sides by two . So 3 43
25:03	times two is 686 , 1/2 times two is 1
25:07	. So this is equal to the square of the
25:10	speed . And now let's take the square root of
25:14	both sides . So the square root of 686 is
25:21	26.19 meters per second . So that's how fast the
25:27	roller coaster is moving at pointing . At that point
25:33	. It was moving at 20 and that point ,
25:35	but it was like 30 something . So that's how
25:40	you could solve the roller coaster problem using conservation of
25:43	energy .

Summarizer

DESCRIPTION:

This physics video tutorial explains how to solve conservation of energy problems with friction, inclined planes and springs.

OVERVIEW:

Conservation of Energy Physics Problems is a free educational video by The Organic Chemistry Tutor.

This page not only allows students and teachers view Conservation of Energy Physics Problems videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.