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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.

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• 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 ?
• 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
• 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?
• 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.
• in the brick lifting example why is the work done equal to 19,600J it should have been 500 x 4 = 2000 J ?
• 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
• 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?
• 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
• 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.
• 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!
• 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?
• 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.
(1 vote)
• what is elastic potential energy?
• 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.
• 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?
(1 vote)
• 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.