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### Course: MCAT > Unit 9

Lesson 5: Work and energy- Work and energy questions
- Introduction to work and energy
- Work and energy (part 2)
- Work and the work-energy principle
- Work as the transfer of energy
- Work example problems
- Work can be negative!
- Conservation of energy
- Work/energy problem with friction
- Intro to springs and Hooke's law
- Potential energy stored in a spring
- Spring potential energy example (mistake in math)
- Conservative forces
- Power
- Introduction to mechanical advantage

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# Work as the transfer of energy

Learn how to calculate work as the way that forces transfer energy to and from objects. See two ways to calculate the amount of work done on an object - the formula Fd cosine theta, and the method of calculating the amount of energy an object gains or loses by examining several examples. Created by David SantoPietro.

## Want to join the conversation?

- In the skateboarder example, W = 1/2 m.v^2 = 2500 J. But in formula : W= Fd = m.a.d = 0 because a= 0 ( velocity not change ) . Why ? What is difference between those formulas ?(24 votes)
- Its a great question. why? because it shows that you are really trying to understand your physics in detail.

OK; here is my answer (with a little challenge for you too...)

Think about WHEN the work is being done on the skateboard... at what point did it GAIN the energy?

Seems to me it gained the energy when he pushed it. (Flicked it with his finger) THAT was when the work was being done on it and it DID accelerate at that point. So all of your equations are correct!

AFTER the initial acceleration, he travelled at a constant speed with his KE of 2500J. Zero acceleration, zero work done.

(We have assumed zero friction and no energy loss as he was moving by the way...)

Using this idea (if you agree with it) can you apply it to the situation where he hit the pile of bricks and lost his energy?

Good luck(27 votes)

- in regards to the example where the stack of bricks is lifted into the air, is net work different from work done on the object. In other words, although the object did gain gravitational PE, we had to lift the stack of bricks with the same force as its weight. Therefore the forces cancel and no work is done on the object. Does this make sense?(12 votes)
- Yes, net work is different from the work exerted on the object by a particular force. The net work is the change in kinetic energy of the object, which in this case would be zero because the velocity does not change (remember KE is 1/2mv^2). But the work that we have done on the object is not zero because we are referring only the force we exerted, not the net force. We could also ask how much work gravity did on the object, and that would be the same magnitude of the work we have done but negative, making the net work 0.(10 votes)

- in the brick lifting example why is the work done equal to 19,600J it should have been 500 x 4 = 2000 J ?(4 votes)
- the work done will be equal to the amount the gravitational potential energy that the force has given to the bricks which is equal to

m.g.h

500kg. 9.8m/s2. 4m=19,600J

or 500kg. 10m/s2.4m=20,000J(8 votes)

- in the example where the bricks where lift straight up, if we used the formula W=F*d*cos(theta) wouldnt the work done be zero since the cos(pi/2)=0?(3 votes)
- You need to look at the angle between the force and the displacement. That angle is 0 if you lift the bricks up. W= F*d*cos(0) = F*d(7 votes)

- Following from the "Work example problems" video, does that mean the net work done on the stack of bricks is 0 J because gravity will do work equal and opposite to the work that we did to lift the bricks?

Or was the previous example from the other video because we were moving the trash can at a constant velocity?(5 votes)- Yes, the net force on the bricks would be zero (assuming they start and end at rest). We did positive work, gravity did negative work, and the total work is zero. The requirement for having the same velocity before and after is just so we don't have to worry about any change in kinetic energy.(3 votes)

- In thermodynamics we have three laws and I'm having a bit trouble understanding the point of having the second law of thermodynamics. " Energy can be converted to pure work with 100% efficiency ONLY AT 0 K." Similarly the third law says, "Yeah, we can convert it with 100% efficiency, but you're not going to reach 0 K EVER!". Aren't these laws totally ideal and no way realistic ?(1 vote)
- Restate the second law: You CAN'T convert with 100% efficiency. That's pretty important, and realistic.

Restate your version of the third law: You can't ever get all the energy out of anything. That's pretty important, and realistic.(7 votes)

- The skateboard hits the wall. There is a force exerted from the wall (pointing to the left) onto the skateboard.

This force doesn't move. Therefore the work done by the wall onto the skateboard is:

W = Fd cos theta = F (0)(cos 0 degees) = 0 joules.

So: the wall does no work on the skateboard.

Next: the skateboard: The center of mass of the skateboard does move a distance, but the point of contact between the skateboard doesn't move. The force on the wall from the skateboard must be pointing to the right and is acting from the stationary point fo contact: it is equal and opposite to the the force of the wall on the skateboard. So this second force doesn't move either.

W = Fd cos theta = -F (0)(cos 0 degees) = 0 joules.

Please explain how work is done if neither of the forces does any work (neither of the forces move). Thanks!(4 votes) - what is elastic potential energy?(3 votes)
- the energy possessed by a body due to change in its dimensions or we can say the change in its shape.For example the energy possessed by a stretched spring.(3 votes)

- So when we say W=mgh or PE=mgh, g here acts in the opposite direction as we move the brick wall in the upward direction, i.e opposite to the gravitational force acting on the wall so shouldn't the g have a negative sign implying the equation should ideally become PE=-mgh?(2 votes)
- The correct equation is

delta PE = mg*delta h

There is no such thing as an absolute amount of PE. PE is relative.

Raise the ball, PE increases. Lower it, PE decreases.(4 votes)

- At1:23when skater hits the wall and lose his energy - it's kind of applying force to the wall, right? and just because of some force opposite - similar to friction and friction itself - the bricks don't move. How do we know what was that force? is it possible to calculate it with the laws we know to this moment?

Even knowing the work done by bricks is 2500J and that W = FdcosA which will be in this case W=-Fd, we can't calculate because d is 0. How do we find the force?(2 votes)- You need to know how long it took to come to a stop after contacting the wall.

Then you can use F*t = m*v

Or you need to know how far the skater's center of mass traveled after contacting the wall - while the skater was getting smashed. d is not zero because the skater does not stop instantaneously. Then you can use W/d = F

You can make reasonable estimates for either of those items.(4 votes)

## Video transcript

One way to find the
amount of work done is by using the formula
Fd cosine theta. But this number for
the amount of work done represents the amount of energy
transferred to an object. For instance, if you
solve for the work done and you get positive
200 joules, it means that the force gave
something 200 joules of energy. So if you have a
way of determining the amount of energy that
something gains or loses, then you have an alternate
way of finding the work done, since the work done on an
object is the amount of energy it gains or loses. For instance, imagine a
50-kilogram skateboarder that starts at rest. If a force starts the
skateboarder moving at 10 meters per
second, that force did work on the skateboarder
since it gave the skateboarder energy. The amount of kinetic energy
gained by the skateboarder is 2,500 joules. That means that the work done
by the force on the skateboarder was positive 2,500 joules. It's positive because the
force on the skateboarder gave the skateboarder
2,500 joules. If a force gives
energy to an object, then the force is doing
positive work on that object. And if a force takes away
energy from an object, the force is doing negative
work on that object. Now imagine that the
skateboarder, who's moving with 10
meters per second, gets stopped because he
crashes into a stack of bricks. The stack of bricks
does negative work on the skateboarder because
it takes away energy from the skateboarder. To find the work done
by the stack of bricks, we just need to figure out
how much energy it took away from the skateboarder. Since the skateboarder
started with 2,500 joules of kinetic energy and ends with
zero joules of kinetic energy, it means that the work done by
the bricks on the skateboarder was negative 2,500 joules. It's negative because the
bricks took away energy from the skateboarder. Let's say we instead
lift the bricks, which are 500 kilograms, upwards
a distance of four meters. To find the work that
we've done on the bricks, we could use Fd cosine theta. But we don't have to. We could just figure
out the amount of energy that we've given to the bricks. The bricks gain energy here. And they're gaining
gravitational potential energy, which is given by
the formula mgh. If we solve, we
get that the bricks gained 19,600 joules of
gravitational potential energy. That means that the work
we did on the bricks was positive 19,600 joules. It's positive because our
force gave the bricks energy. This idea doesn't just work with
gravitational potential energy and kinetic energy. It works for every
kind of energy. You can always find the work
done by a force on an object if you could determine the
energy that that force gives or takes away from that object. [MUSIC PLAYING]