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

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

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

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

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