Energy Worksheet: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |  Go up
 - by Jacob Spindel, To Chong, 2002

1. A 1250 Kg car going 23 m/s can coast how far up a very tall hill if it loses no energy to friction? 

Our two points in time are going to be: 1. The car is going at 23 m/s and hasn't started up the hill yet. 2. The car has gone as far up the hill as it can go, and it has stopped moving. So, if you look at the formulas, the first one won't work just yet because you don't know height or PE, and the second one one work because this problem doesn't even have a spring in it. However, the third one will work since you know the initial velocity. KE=1/2*m*v2 Putting in numbers: KE=1/2*1250*232=330625 So now you know these quantities: m = 1250 Kg V0 = 23 m/s Vfinal = 0 (assumed) KE=330625 J Height = ? So again you look at the formulas. Remember, if the car stops moving because it moves upward, the kinetic energy will become potential energy. So now you can use the first problem, because kinetic energy becomes equal to potential energy: PE=m*g*h Plugging in numbers:

330625 J=(1250 kg)(9.8 m/s/s)h

so h (the height) = 26.9897 m, or 27 m with sig figs.


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2. How far will the car in the previous problem coast up the hill if it loses 150,000 J of energy to friction on the way up?

This problem is basically the same as the last one. We're going to transfer that kinetic energy to potential energy. However, this time, not all of the energy goes to potential energy. This time, some of it is lost to friction. We have to subtract this amount first.

Here's what we know: Amount of energy left to go to potential energy=KE - energy lost to friction

330625-150000=180625 J

180625 J are left to be transferred to potential energy. Set it equal to the potential
energy formula again.

PE=m*g*h

Plugging numbers into the formula:

180625=1250*9.8*h

So

h (Height) = 14.744 m, or 15 m with sig figs.

This problem is basically the same as the last one. We're going to transfer that kinetic energy to potential energy. However, this time, not all of the energy goes to potential energy. This time, some of it is lost to friction. We have to subtract this amount first.

Here's what we know:

Amount of energy left to go to potential energy=KE - energy lost to friction

330625-150000=180625 J

180625 J are left to be transferred to potential energy. Set it equal to the potential
energy formula again.

PE=m*g*h

Plugging numbers into the formula:

180625=1250*9.8*h

So

h (Height) = 14.744 m, or 15 m with sig figs.

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3. A 873 Kg car going 12 m/s at the top of a 6.2 m tall hill is going how fast at the bottom? (No loss to friction)

Our two points in time are:

1. The car is at the top of the hill, moving at 12 m/s.
2. The car is at the bottom of the hill, moving at a speed that we don't know yet.

At the first point in time, the car is off the ground, so it has potential energy. We can calculate that with the information we have:

PE=m*g*h

With numbers:
PE=873*9.8*6.2=53043.5 J

The car is also moving at the top of the hill, so it also has kinetic energy.

KE=1/2*m*v2

With numbers:
KE=1/2*873*122=62856 J

The total energy this car has is simply PE+KE=53043.5+62856=115899 J.

So now, we know:
Now, let's figure out what's going to happen to the car in the second point in time. It's not going to be on the hill anymore. This means that it won't have any potential energy because it's on the ground. The potential energy will be transferred to kinetic energy, which will add to the kinetic energy the car already had. All it's energy will be kinetic:

KE=115899 J.

We also know that KE=1/2*m*v2

So 115899=1/2*m*v2

Plugging in the number we know for the mass of the car gives us:

115899=1/2*873*v2

Solve this equation for v, and you'll find that v=16.3 m/s, or 16 m/s with sig-figs.
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4.A 312 Kg rocket ship in deep space fires an engine that produces 516 N of thrust, for a distance of 32 m. If the rocket ship was initially at rest, what is its final velocity?

The two points in time we know are:

1. The rocket ship is firing its engine to produce work.
2. The rocket ship has stopped firing its engine and is at the highest speed it will reach.

At the first point in time, there is only one kind of energy: work. At the second point in time, there is only one kind of energy: kinetic energy (the rocket is moving). This should be pretty easy.
(There is no potential energy anywhere because the rocket is in deep space and there is no gravity.)

We can figure out how much work the rocket is doing with the work formula:

W=F*d

With numbers:
W=516*32=16512 J

Let's check what we know: Remember, the work is going to transfer to kinetic energy, and the amount of kinetic energy we will have is equal to the amount of work we had because energy is conserved.

W = KE = 16512 J

Next, we use the formula we have for KE:

KE=1/2*m*v2=16512 J

We also know that m = 312 Kg
1/2*312*v2=16512
v = 10.2882 m/s, or 10. m/s with sig-figs.
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5. What is the final velocity of a .452 Kg object initially at rest if you exert a force of 6.50 N on it vertically for a distance of 12.5 m?

The two points in time we know in this problem are:
1. Work is being done on an object that is at rest and on the ground.
2. The work is done, and the object is now moving and in the air.

At the first point in time, the only energy is the work, just like the first point in time in the last problem. At the second point in time, the object is moving AND it's in the air. That means that the work has gone to kinetic energy AND potential energy. Sneaky crocodile!

Remember, of course, that if you add the kinetic and potential energies, it will still be equal to the work, because the total amount of energy from each point in time is the same.
In math terms,

W=PE+KE.

Here's what we already know:
We know that W=PE+KE. We also know that W=F*d, PE = m*g*h, and KE = 1/2*m*v2This means that:

F*d=m*g*h+1/2*m*v2

Plug in numbers:
6.5*12.5=0.452*9.8*12.5+1/2*0.452*v2
Solve for v and you'll find that it's 10.7011 m/s, or 10.7 m/s with sig-figs.
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6. A 100. Kg roller coaster has a speed of 8.0 m/s on the top of a hill that is 6.4 m tall. What is its speed on the top of a 2.4 m tall hill?

Our two points in time for this problem are:
1. The roller coaster is on a 6.4 m hill going 8.0 m/s.
2. The roller coaster is on a 2.4 m hill going at a speed we don't know yet.

Here is what you start with:


At the first point in time, the roller coaster has potential energy (it's on a hill) and kinetic energy (it's moving). At the second point in time, the roller coaster still has potential energy and kinetic energy. However, it has more kinetic energy and less potential energy than at the first point in time. The total energy from each point in time will be equal:

PE1+KE1 = PE2 + KE2

We can calculate the initial potential energy with the potential energy formula:

PE1 = m*g*h1
PE1 = 100*9.8*6.4 = 6272 J.

We also have enough information to calculate the initial kinetic energy:
KE1 = 1/2*m*v2
KE1 = 1/2*100*8.02 = 3200 J.

And we can calculate the final potential energy:
PE2 = m*g*h2
PE2 = 100*9.8*2.4 = 2352 J

Substitute the numbers we have into the formula PE1+KE2=PE2+KE2:
6272+3200 = 2352+KE2
Solving this reveals that KE2 = 7120 J.

We also know that KE=1/2*m*v2, so:
7120=1/2*m*v2
7120=1/2*100*v2
v = 11.9331 m/s, or 12 m/s with sig-figs.
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7. A coasting 980 Kg car going 21 m/s at the top of a 15 m tall hill is brought to rest by a crash barrier at the bottom of the hill in a distance of 12.4 m. What force did the barrier exert on the car, and what acceleration did the car undergo in stopping?

Here is what you start with:
Our two points in time will be:

1. The car is on top of the hill going 21 m/s.
2. The car is at the bottom of the hill and it has been stopped by the crash barrier.

The car has two different types of energy at the first point in time: it is on a hill, so it has potential energy, and it is moving, so it has kinetic energy. How much?

PE=m*g*h

With numbers:
PE=980*9.8*15=144060 J.

KE=1/2*m*v2

With numbers:
KE=1/2*980*212=216090 J.

Total energy=PE+KE=144060+216090=360150 J.

In the second frame, the car has come off the hill, and it has been stopped. That means that it no longer has potential or kinetic energy. That energy had to go somewhere. You can't just get rid of energy. Not even Congress is that wasteful. The wording of the problem makes it pretty obvious where this energy is going: the car is doing work on the crash barrier. The energy goes to work. The energy isn't going to any other source, so:

Total energy=W.
154350=F*d.

And we know the distance, d, so:
360150=F*12.4.F=29044.35 N, or 29000 N with sig-figs.

But wait! There's more!

Now that we know the force, it won't be hard to find the acceleration of the car. It's time to use a formula from the good old days:

F=m*a.

Force equals mass times acceleration. We know the force and the mass, and we want the acceleration.

29044.35=980*a.
a=29.637 m/s, or 30. m/s with sig-figs.

Here is what you start with:

Our two points in time will be:

1. The car is on top of the hill going 21 m/s.
2. The car is at the bottom of the hill and it has been stopped by the crash barrier.

The car has two different types of energy at the first point in time: it is on a hill, so it has potential energy, and it is moving, so it has kinetic energy. How much?

PE=m*g*h

With numbers:
PE=980*9.8*15=144060 J.

KE=1/2*m*v2

With numbers:
KE=1/2*980*212=216090 J.

Total energy=PE+KE=144060+216090=360150 J.

In the second frame, the car has come off the hill, and it has been stopped. That means that it no longer has potential or kinetic energy. That energy had to go somewhere. You can't just get rid of energy. Not even Congress is that wasteful. The wording of the problem makes it pretty obvious where this energy is going: the car is doing work on the crash barrier. The energy goes to work. The energy isn't going to any other source, so:

Total energy=W.
154350=F*d.

And we know the distance, d, so:
360150=F*12.4.F=29044.35 N, or 29000 N with sig-figs.

But wait! There's more!

Now that we know the force, it won't be hard to find the acceleration of the car. It's time to use a formula from the good old days:

F=m*a.

Force equals mass times acceleration. We know the force and the mass, and we want the acceleration.

29044.35=980*a.
a=29.637 m/s, or 30. m/s with sig-figs.

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8. A 120. Kg bicyclist going 5.60 m/s at the bottom of a 2.00 m tall hill exerts a forward force of 200. N for 10.0 m as he climbs the hill. What is their speed at the top of the hill?

Here is what you start with:
Our two points in time will be:

1. The biker is at the bottom of the hill, traveling at 5.60 m/s, exerting a force of 200. N, happily ringing the bell on his bike.
2. The biker is at the top of the hill and has finished exerting the force.

At the first point in time, the biker is moving, so he has kinetic energy:

KE=1/2*m*v2

With numbers:
KE=1/2*120.*5.602=1881.6 J.

The only other source of energy at the first point in time is the work done by the biker:

W=F*d

With numbers:
W=200.*10.0 m = 2000 J.

So the total energy equals W+KE=2000+1881.6 = 3881.6 J.

At the second point in time, the biker is on top of a hill, so he has potential energy. The problem says he has speed at the top of the hill (but we don't know how much yet), so the biker has some amount of kinetic energy. We can figure out how much potential energy he has:

PE=m*g*h

With numbers:
PE=120.*9.8*2=2352 J.

We know what the total energy is because we already found it at the first point in time in the problem. We also know, since the only types of energy at the second point in time are PE and KE, that PE+KE=Total Energy.

Plugging in numbers reveals:
2352+KE=3881.6
KE=1529.6 J.

And we know that KE=1/2*m*v2
So 1529.6=1/2*m*v2

m = 120. Kg, so:
1529.6=1/2*120.*v2
v=5.049 m/s, or 5.05 m/s with sig-figs.

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9. A 150 Kg roller coaster car is going 12 m/s at the top of a 12 m tall hill, and then rolls into the station at a height of 3.0 m where it is brought down to a speed of 6.0 m/s with a braking force of 8900 N. Over what distance must the force be exerted?

Here is what you start with:

Our two points in time in this problem will be:
1. The roller coaster is on the hill going 12 m/s.
2. The roller coaster is in the station at a lower altitude and speed.

At the first point in time, the coaster is moving and is elevated. That means it has potential energy and kinetic energy. Let's find out how much.

PE=m*g*h

With numbers:
PE=150*9.8*12=17640 J.

KE=1/2*m*v2

With numbers:
KE=1/2*150*122=10800 J.

Total energy=PE+KE=17640+10800=28440.

At the second point in time, the roller coaster is still moving and elevated, so it still has potential energy and kinetic energy (but not the same amounts as before). However, at the second point in time, some of the roller coaster's energy also goes into the work used to stop the coaster. We can figure out how much potential energy and kinetic energy the roller coaster has at the second point in time:

PE=m*g*h

With numbers:
PE=150*9.8*3=4410 J.

KE=1/2*m*v2With numbers:
KE=1/2*150*62=2700 J.

We also know that PE+KE+W=Total Energy, because there are no other sources of energy at this point.

4410+2700+W=28440
W=21330 J.

And W=F*d.
21330=8900*d
d=2.397 m, or 2.4 m with sig-figs.

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10. A coasting 1150 Kg car going 21 m/s hits a puddle that is 13 m long. It leaves the puddle going 18 m/s. What force did the puddle exert on the car?

Here is what you start with:

Our two points in time for this problem are:
1. The unsuspecting car is coasting toward the puddle but hasn't hit it yet.
2. The car has gone through the puddle and slowed down, and the driver is figuring out how he can use the mud on his car as an Area 4 Project.

At the first point in time,the car is moving, so it has kinetic energy. In fact, that's all it has. We can calculate how much kinetic energy using the KE formula:

KE=1/2*m*v2

With numbers:
KE=1/2*1150*212=253575 J.

Total energy=KE=253575 J because the car has no other types of energy.

At the second point in time, the car is still moving, so it still has kinetic energy. However, the car has slowed down, so it has less kinetic energy than it had before. We can find how much KE the car has at the second point in time:

KE=1/2*m*v2

With numbers:
KE=1/2*1150*182=186300 J.

The car has lost some of its kinetic energy. However, that energy had to go somewhere. In fact, it was transferred to the work done by the puddle. The two types of energy at the second point in time are KE and W.

Total Energy=KE+W

We know that the total energy is 253575 and the kinetic energy at the second point in time is 186300 J. So:

253575=186300+W
W=67275

We also know that W=F*d.

W=F*d
67275=F*13
F=5175 N, or 5200 N with sig-figs.

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11. The engines on a 45 Kg rocket fired vertically burn for a distance of 25 m generating a force of 620 N. (Assume for this problem that the rocket's mass remains constant.) What is the rocket's speed just after the engines quit burning? To what height does the rocket rise in the air before falling back to the earth?

Here's what you start with:

Our two points in time for this problem will be:
1. The rocket is on the ground and not moving, but it is doing work with its engines.
2. The rocket has stopped firing its engines and is in the air and moving.
3. The rocket is at the highest point it will reach before it starts falling back to the ground.
(Number of points in time in this problem calculated by Intel.)

Okay, okay. So we actually have three points in time in this problem. But that's because we actually have two problems. We're trying to answer two different questions in this problem, and each question will only deal with two of the points in time.

First, let's look at part A. We're trying to find the velocity of the rocket immediately after the engines stop firing. For this part of the question, we'll only use the point 1 in time and point 2 in time, because we are only comparing the rocket on the ground to the rocket immediately after the engines have stopped firing, and we don't care yet about what the rocket does after that.


At the first point in time, the only energy the rocket has is work. At the second point in time, this energy has transferred so that it has kinetic energy and potential energy but not work.

We can find the amount of work energy the rocket has at the first point in time:

W=F*d

With numbers:
W=620*25=15500

Because there is only one source of energy at the first point in time, Total Energy=W=15500 J.

At the second point in time, the rocket is moving and in the air, so it has potential and kinetic energy. We do know how much potential energy the rocket has. We don't know the height the rocket will have at point 3, but we do know that at point 2 the height is 25 m.
So we can calculate the PE at point 2:

PE=m*g*h

With numbers:
PE=45*9.8*25=11025 J.

With Roman numerals:
PE=VL*IX.IIX*XXV=MMMMMMMMMMMXXV J.

With unidentifiable symbols:
PE=#%*@.!*&/=!@#$% J.

I forget what we're doing.

Oh yeah. The rocket has potential and kinetic energy only at the second point in time, so we know that:

Total energy=PE+KE.

Now we know the total energy and the potential energy, so:

15500=11025+KE
KE=4475 J.

We also know that KE=1/2*m*v2, so:
4475=1/2*45*v2
v=14.102 m/s, or 14 m/s with sig-figs. This is the answer to part A.

Now, in part B, we need to figure out how high the rocket will be at point 3 in time - that is, we need to know how much potential energy it has. The rocket has the same total amount of energy as before - 15500 J.

Does the rocket have kinetic energy at point 3? When an object flying vertically reaches its highest point, it stops for an instant. The rocket is not moving. It does not have kinetic energy.

With kinetic energy ruled out, the only kind of energy the rocket has at point 3 is potential energy. So:

Total energy=PE

15500=PE

We also know that PE=m*g*h, so:
15500=45*9.8*h
h=35.147 m, or 35 m with sig-figs.

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12. A spring with a constant of 124 N/m is compressed 16.2 cm. How much potential energy does it store? How much is released if the spring is allowed to expand only from 16.2 cm to 12.5 cm?

Here is what we start with:


To find the initial spring potential energy (the answer to part A), we just use the spring potential energy formula:

PES=1/2*k*x2

 With numbers:
PES=1/2*124*0.1622=1.6271 J, or 1.63 J with sig-figs.

For part B, we want to find how much the energy has decreased after the spring has been partially released. To do this, we need to know how much energy the spring has after it has been partially released:

PES=1/2*k*x2

With numbers:
PES=1/2*124*0.1252=0.96875 J.

To figure out how much energy was lost, just subtract:
1.63-0.96875=0.66 J.

(Of course, the energy isn't really "lost." Energy is conserved! The energy probably went into doing work on the hand of the person holding the spring, or else it pushed on the air and gave the air some extra kinetic energy.)
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13. A 2.0 g penny is pushed down on a vertical spring, compressing the spring by 1.0 cm. The force constant of the spring is 40. N/m. How far above this original position will the penny fly if it is released?

Here's what we start with:
And here is the word "banana:" Our two points in time will be:
1. The penny is on the spring and the spring is compressed.
2. The spring has been released and the penny has reached the highest point in the air it will reach.

At the first point in time, there is only one type of energy: spring potential energy. Let's see how much.

PES=1/2*k*x2

With numbers:
PES=1/2*40*0.012=0.002 J.

Since there are no other types of energy at the first point in time, the total energy=PES=0.002 J.

At the second point in time, the penny is in the air, but it has stopped moving and the spring is no longer compressed. This means that it has potential energy, but it does not have any other type of energy. So, all of the energy goes to potential energy:

Total energy=PE
0.002 J=PE

We also know that PE=m*g*h, so:

0.002=m*g*h

0.002=0.002*9.8*h

h=0.10204 m, or 10. cm with sig-figs. (You could also answer 0.10 m.)

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14. A 745 Kg roller coaster car traveling at a speed of 3.50 m/s at an elevation of 11.25 m is stopped in the station at an elevation of 10.21 m by a spring over a distance of 67 cm. What is the spring constant of the spring?

Here is what we know:
Our two points in time in this problem will be:
1. The roller coaster is at 11.25 m and moving at a speed of 3.50 m/s.
2. The roller coaster is in the station, the spring has stopped it, and the passengers are hoping that the spring that stopped them doesn't expand again and fling them to their deaths.
(Don't worry, it won't.)

At the first point in time, the roller coaster is on a hill and it's moving. That means that it has potential energy and kinetic energy. How much?

PE=m*g*h

With numbers:
PE=745*9.8*11.25=82136.25 J

KE=1/2*m*v2

With numbers:
KE=1/2*745*3.502=4563.125 J.

The car doesn't have any other types of energy at the first point in time, so the total energy = PE+KE = 86699.375 J.

At the second point in time, the car has stopped, so it no longer has kinetic energy. It still has potential energy, but not the same amount as before. At the second point in time, there is also a compressed spring, so we do have spring potential energy. In fact, the only types of energy that the roller coaster has at the second point in time are regular potential energy and spring potential energy. How much regular potential energy does it have?

PE=m*g*h

With numbers:
PE=745*9.8*10.21=74543.21 J.

We also know that Total energy=PE+PES

So 86699.375=74543.21+PES
PES=12156.165 J.

We also know that PES=1/2*k*x2, so:

12156.165=1/2*k*0.672
k=54159.7906 J, or 5.4 * 104 with sig-figs.

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15. A spring-loaded marble gun shoots 12.5 gram marbles off of a (Frictionless!) horizontal table that is 91 cm high. The spring in the gun has a spring constant of 52 N/m. By how much do you need to compress the spring to land a marble on the floor 45 cm horizontally from the edge of the table?

Here is what we start with:


Before we do anything with energy, let's figure out what the initial velocity of the marble has to be just after the spring has released it. All we need to do this is linear kinematics - it's a cliff problem.

First, let's find out how long the marble will be falling before it hits the ground by using the formula X=X0+1/2*a*t2

with the information we know about how the marble will move vertically.

-0.91=0+1/2*-9.8*t2
t=0.431 s.

The horizontal velocity of the marble will be constant after the spring has released it, so we can figure out what velocity the marble needs in order to travel 0.45 m in 0.431 seconds:

x=v*t

With numbers:
0.45=v*0.431
v=1.04 m/s.

So now we know that the initial velocity of the marble has to be 1.04 m/s, so we just need to figure out how much the spring needs to be compressed to give it that initial velocity.

Now it's time to use energy.

Our two points in time for this problem will be:
1. The marble is at rest but is compressed on the spring.
2. The spring has just been released and the marble is moving.

At the first point in time, there is only one kind of energy: spring potential energy. At the second point in time, there is also only one kind of energy: kinetic energy. We can figure out how much KE the marble has at the second point in time:

KE=1/2*m*v2

With numbers:
KE=1/2*0.0125*1.042=0.00676 J.

Because there is only one source of energy at each point in time, PES=Total energy and KE=Total energy, so PES=KE.

0.00676=PES

We also know that PES=1/2*k*x2,so:
0.00676=1/2*52*x2
x=0.01612 m, or 1.6 cm with sig-figs.
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16. A 350.g mass moves around in a vertical circle on the end of 17.0 cm string. At the top of the circle, the tension in the string is 2.28N. What is the centripetal acceleration  of the mass at the top? What is the mass at the bottom of the circle?


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17. A 1.20 kg mass moves in a vertical circle on the end of a 45.0 cm string. It has a velocity of 5.00m/s at the bottom of its path. What is the speed at the top? What is the tension in the string at the top and at the bottom?

<solution goes here>
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18. A 32 g mass is moving in a vertical circle on the end of a 67 cm string such as it clears the level ground at the bottom by a whisker. When the mass is at the very top, the tension in the string is .047N. What is the speed of the mass at the top? What is the tension in the string when the mass is at the very bottom? If the string were vaporized when the mass was at the very top, what distance would it fly before hitting the ground?

<solution goes here>
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