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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1 > Unit 12

Lesson 1: AP Physics 1 concept review- AP Physics 1 review of 1D motion
- AP Physics 1 review of 2D motion and vectors
- AP Physics 1 review of Forces and Newton's Laws
- AP Physics 1 review of Centripetal Forces
- AP Physics 1 review of Energy and Work
- AP Physics 1 review of Momentum and Impulse
- AP Physics 1 review of Torque and Angular momentum
- AP Physics 1 review of Waves and Harmonic motion
- AP Physics 1 Review of Charge and Circuits

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# AP Physics 1 review of Energy and Work

In this video David explains the concepts in Work and Energy and does an example problem for each concept. Link for document: https://www.dropbox.com/s/t1w6xlnkozzel17/Energy%20review.pdf?dl=0. Created by David SantoPietro.

## Want to join the conversation?

- at7:57, does the direction of the velocities not matter because the squares will make them positive anyways?(7 votes)
- Correct; one could also say that the direction of the velocities does not matter because we are looking for a scalar quantity, that is, kinetic energy.(12 votes)

- When a force is perpendicular to the direction of motion, why is there 0 work being done on the object?(2 votes)
- Think of an object already moving horizontally across the floor as a constant speed.

There is a string attached to the object, which you can pull straight up on.

If you pull straight up on the string with insufficient force to lift it, it is not moved. Since there is no displacement, there can be no work done (something times zero is always zero).

If you pull straight up on the string with MORE than enough force to lift it, the object moved off the ground and is given a vertical displacement. This gives something for the force to be multiplied by to calculate work.

I hope this made sense and helped you out!(4 votes)

- At11:56, wouldn't the time it takes for the box to fall the same because gravity is the only acceleration that makes the box fall. And the system is frictionless.(2 votes)
- No, because the distance traveled is not the same. The length of the ramp with the angle θ is longer than the ramp with angle 2θ, so if the boxes are accelerated at the same rate with no friction, the box on the right has farther to travel.(4 votes)

- at4:46

what if the object didn't get displaced because of friction would there be work?(2 votes)- Great question Taha!

No, if there is enough friction to stop any displacement from occurring, no work would be done. For work to be done on an object, it**must**be displaced.(1 vote)

- How does the earth perform work (MGH)on one of your incline problems if the box was not moving in the direction of gravity?(2 votes)
- Any path, inclined or curved, can be broken into small vertical and horizontal segments like the base and perpendicular of a right triangle with the short curved path segment substituting for the hypotenuse. As you move down the incline, you're moving both vertically and horizontally. The vertical displacements lead to a change in the gravitational potential energy while the horizontal displacements lead to zero change in U.(2 votes)

- at2:58why did you say that the velocity slows down(2 votes)
- If you recall that the energy in the system must remain constant, then we can say that the Kinetic Energy will always equal the Gravitation Potential Energy.

The the block goes up the ramp, the change in height is positive so GPE=mgh will increase.

The only way that GPE can increase is if Kinetic Energy Decreases.

The equation for Kinetic Energy is KE=1/2mV^2. So if (1/2) and (m) are both constants, the only way KE can decrease is if velocity decreases.

Another way of thinking about this is to visualize the system with a ball, the ball won't continue to maintain its speed as it goes up the hill as a result of gravity doing work on it.(1 vote)

- At11:45can you also use the formula work=force*displacement*cos theta? Would the times be the same if you were to do that?(1 vote)
- @Andrew M - I tried it:

P = W/T

P = F*d*cosθ / T

P = Fg*d*cosθ / t

P1 = (M*g)*H*cos2θ / T

P2 = (M*g)*H*cosθ / T

But in this case, unless I've done something wrong, the work on the second box seems to be greater as the cosine of a smaller angle is greater than that of a larger one. Even with the knowledge that the box on the steeper incline would reach the bottom in a shorter amount of time, I don't know how the power on that box would compare to the power on the second one due to the increased work on the second box.

If you could tell me where I messed up, it would be greatly appreciated.

EDIT: Did I perhaps forget to account for the fact that the net force on a box on an incline is across the parallel component?(3 votes)

- at7:57Does that mean that, if the final velocity of that object were -4 m/s (still going in the same direction) as opposed to 4 m/s as was seen in the video, the net work would still be the same? That doesn't really make sense to me, but mathematically it works out for whatever reason.

i mean, doesn't it take a lot of effort(energy) to slow down the object and then accelerate it again to 4m/s on the opposite direction rather than slow down it a little bit so that it moves 4m/s to the left?(2 votes) - When a force is perpendicular to the direction of motion, why is there 0 work being done on the object?(0 votes)
- because the equation for work includes cosine of the angle. the cosine of a 90 degree (perpendicular) angle is 0 so there is 0 work being done.(9 votes)

- At3:43, wouldn't there also be potential energy acting on the box?(0 votes)

## Video transcript

- [Teacher] Not only are
there many different kinds of energies, but both objects, and systems of objects, can have energy. Once they have that energy,
they can transfer it to another system or object or that energy could
transform to a different type of energy inside that system. When energy gets transferred,
we call that work, and the amount of work
that's done is the amount of energy that was transferred. You often hear people say,
"Energy is conserved," which really just means
that you can't create or destroy energy. You can simply transfer it
between objects or systems. So, what are all the
different types of energy? There's kinetic energy which is the energy due to something moving, and the formula's one-half the mass times the speed squared. There's gravitational potential energy which is the energy something
has due to its height, and the formula's the mass times the magnitude of
acceleration due to gravity times the height of the object. Height above what? Height above whatever you're choosing as the H equals zero reference line. Is that cheating? No, because all that really
matters is the change in gravitational potential energy, not the actual value itself. There's also spring potential energy which has to do with a
compressed or stretched spring, and the formula's one-half
times the spring constant times x, which is not
the length of the spring. X is the amount that the
spring has been compressed or stretched. These three types of
kinetic and potential energy constitute what we call mechanical energy. Mechanical energy's another
word for the kinetic energy plus gravitational potential energy plus spring potential energy in a system, and it's important to know
that mechanical energy does not include thermal energy. Thermal energy's the heat energy generated by dissipative forces like
friction and air resistance, and you can find the amount
of thermal energy generated by taking the size of the dissipated force times the distance through
which that force was acting. The unit of energy is Joules
and energy is not a vector. But maybe the most important
thing to remember about energy is if there's no external
work done on a system, then there's no change in
the energy of that system. In other words, if there's
no external work done on a system, the initial
energy of that system will equal the final
energy of that system, which is the way you solve many conservation of energy problems. So, what's an example problem
involving energy look like? Let's say a box started
with an initial speed and slides from one platform
up to another platform. We'll assume that frictional forces and air resistance are negligible. And for the system that's consisting of the mass and the Earth, what's happening to the
total mechanical energy in this system? So, you've got to pay special attention to what is in your system. Since my system includes the mass, which is going to be moving, my system's gonna have kinetic energy, and since my system has two objects that are interacting gravitationally, the mass and the Earth,
my system's also going to have gravitational potential energy. So, when I asked about the
total mechanical energy of the system, that's really just code for the total kinetic and
potential energy of the system. So, as this mass slides
up to a higher point on the ramp, the gravitational
potential energy increases, but the mass is gonna slow down, so the kinetic energy's gonna decrease. However, since the Earth and
the mass are in our system, and there's no dissipative forces, there's no external
work done on our system. Yes, the Earth is doing work on the box, but the Earth is part of our system so it can't do external work, and that means energy
just gets transferred from one form to another
within our system, and the total mechanical energy, here, is gonna remain the same
for the entire trip. Now, what if we asked this same question but we consider a system that
consists only of the box. In that case, our system
has a box that's moving, so it'll have kinetic energy. But, our system no longer
includes two objects interacting gravitationally
so our system will have no gravitational potential energy. What happens to the total
mechanical energy in this case? Well, the only energy that
I've got in my system, now, is kinetic energy and since
that kinetic energy decreased, the total mechanical energy of the box, as a system, decreases. How does it decrease? It decreases because
now the Earth is outside of our system and the
work that it is doing on the box is external work and it's taking away energy from the box. What does work mean? In physics, work is the
amount of energy transferred from one system, or object, to another. In other words, if a person lifted a box and gave it 10 Joules of
gravitational potential energy, we'd say that person did
positive 10 Joules of work on the box since that person gave the box 10 Joules of energy. But since the box took 10 Joules of energy from that person, we'd say that the box did negative 10 Joules
of work on the person since the box took 10 Joules of energy. So, you can find the work done if you can determine the amount of energy that was transferred. But, there's an alternative
formula to find the work done. If something's having work done to it, there's got to be a force on that object, and that object has to be displaced. So, if you take the force on the object times the displacement of the object, and multiply by the cosine of the angle between the force and the displacement, you'll also get the work done. In other words, one way
to find the work done is by finding the amount of energy that was transferred. But, another way to find the work done is by taking the magnitude of
force exerted on an object times the displacement of the object and then times cosine of the angle between the displacement and the force. Since work is a transfer of energy, it also has units of Joules. And even though work is not a vector, it can be positive or negative. If the force on an object has a component in the direction of motion, that force will do
positive work on the object and give the object energy. If the force on the object has a component in the opposite direction of the motion, the work done by that
force would be negative and it would take away
the object's energy. And, if the force on an
object is perpendicular to the motion of the object, that force does zero work on the object. It neither gives the object energy nor takes away the object's energy. So, what's an example problem
involving work look like? Let's say a box of mass M slides down a frictionless ramp of height,
H, and angle two-theta, as seen in this diagram, here, and a separate box of
mass two-M slides down another frictionless ramp of height, H, and angle theta, as seen
in this diagram, here. And, we want to know how
work done on the object by the Earth compares for each case? The easiest way to find the work done here is by finding the change in energy. The box will gain an
amount of kinetic energy equal to the amount of
potential energy that it loses. So, the work done by the
Earth is just gonna equal positive m-g-h. Both heights are the same. So, the H's are equivalent,
but one box has twice the mass. So, the work done by
gravity on the mass, two-M, is gonna be twice as great as the work done on the mass one-M. What's the work energy principle mean? The work energy principle states that the total work, or the
net work done on an object, is gonna be equal to the
change in kinetic energy of that object. So, if you add up all the
work done by all forces on an object, that's got
to be equal to the change in the kinetic energy of that object. In other words, one-half
m-v-final, squared, minus one-half m-v-initial, squared. So, this is a really handy
way to find how the speed of an object changes if you
can determine the net work on an object. In other words, if there's
multiple forces on an object, and you can find the work
done by each of those forces, you can determine how much kinetic energy that object gained or lost. So, what's an example of
the work energy principle? Let's say a four kilogram box started with a velocity of six meters
per second to the left. Some net amount of work
is done on that box, and it's now moving with
a velocity of four meters per second to the right. We want to know what was
the amount of new work done on the box? Without even solving it, we can say since this object's slowed
down, energy was taken from it, so the amount of net
work had to be negative which means it's either B or D. To figure out which one exactly, we could use the work energy principle which says that the net
work done is equal to the change in kinetic energy. So, if we take the final kinetic energy, which is one-half times four kilograms times the final speed squared, and we subtract the
initial kinetic energy, one-half times four kilograms
times the initial speed, squared, six meters per second, we get negative 40 Joules of net work. If you get a force versus position graph, the area under that graph
will represent the work done. So, when you see F versus x, you should think area equals work. But, be careful, area above
the x-axis is gonna count as positive work done, and
area underneath the x-axis is gonna count as negative work done, and make sure the x-axis
really is position. If you get a force versus time graph, the area's impulse, not work. So, what would an example
of work as area look like? Let's say a box started at x equals zero with a velocity of five
meters per second to the right and a net horizontal
force on the box is given by the graph below. We want to know, at what position
other than x equals zero, will the box, again, have a velocity of five meters per second to the right? Well, since the box will
end with the same speed that it began with, the change
in kinetic energy is gonna equal zero. But, that means the net
work would also equal zero, since the net work is equal to the change in kinetic energy. So, if the box starts at x equals zero, how far do we have to go in order for us to have no net work done. Between zero and three meters, the work done is gonna be negative, and the area of this triangle is gonna be one-half the base times the height which is one-half times three meters time negative six Newtons which is negative nine
Joules of work done, and the area under this triangle, between three and five seconds, would again be one-half base times height, which is one-half times two meters times height of four Newtons which is positive four
Joules of work done. So, by the time that the box
has made it to five meters, there's been a total amount of work done of negative nine plus four, which is negative five Joules of work. But, we want no net work done. So, we're gonna have to keep going until this positive area
contribution is gonna equal the negative area. In other words, if I can make it so that all of this negative area is equal to all of the positive area, my net work's gonna equal zero. My negative area is negative nine. My positive area, so
far, is positive four. If I continue on to the six meter mark, I've pick up another
positive four Joules of work since the height of this rectangle is four and the width is one meter, which means we're almost there. Four plus four is eight. I'd only need to pick up one more Joule, so I can't go all the way to seven meters. I'd only need to go one
more fourth of a meter to pick up one more Joule so that one plus four plus four is equal to negative nine. So, the net work would equal zero somewhere between x equals
six and x equals seven which would ensure that the change in kinetic energy is zero and we would end with the same speed that we began with. What does power mean? In physics power is the
amount of work done per time, which can also be thought of as the amount of energy
transferred per time. In other words, the amount
of Joules per second that are transferred, and the name given to a
Joule per second is a Watt. So, you can solve for the power by finding the work divided by the time or the change in energy
divided by the time. And you can increase the amount of power by increasing the work done or decreasing the amount of time it takes for that work to be done. And just like energy and
work, power is not a vector. So, what's an example problem
involving power look like? Let's say a box of mass M slid all the way down a frictionless ramp of height H and angle two-theta as
seen in this diagram, and a separate mass M slides all the way down a frictionless ramp of height H and angle theta as seen in that diagram, and we want to know how
the average power developed by the force of gravity
on the boxes compares for each incline? So, we use the formula for power, power's the work done per time. The work done on these boxes
is gonna equal the change in kinetic energy of these boxes which would equal the
change in potential energy of the boxes, but the mass
of the boxes are the same, the gravitational
acceleration is the same, and the height they fall from is the same. So, the work done on the boxes are equal, but the time it takes for these boxes to slide down the ramp is not equal. The mass on the steeper ramp
will reach the bottom faster which means it has a higher
rate of power being done compared to the mass
on the less steep ramp. So, even though the same
amount of work is being done, the rate at which that work
is being done is greater for the steeper ramp compared
to the more shallow ramp.