# Why does the GPE of an object always equal the work done?

I had a question that said, "If you do 100 J of work to lift an object over your head, what is the gravitational potential energy relative to its starting position? What would be its gravitational potential energy if it were lifted twice as high?"

I understand the answers are 100 Joules and 200 Joules, but I dont get why. I thought work measured the change in energy. So why would it just equal the GPE if the object also has KE?

• You mean joules? – Pietro Oliva Jan 15 at 21:58

The statement applies to the situation after you have lifted the object, when it is not moving, and its KE is zero.

When you have lifted the object half way and it is still moving, you are done more than half the work, because you have increased its GPE and also given it some KE. During the second half, as it slows down, you do less than half the work and the KE is converted into GPE.

That situation is assuming a condition that is usually omitted: we are supposed to do that movement so slowly that kinetic energy is negligible. (That's extremely slowly, to the extent that it is like a huge set of fixed photographies). That's caleld quasi-static process.

If you performed the movement with some velocity, you'd need some extra work to increase the KE.

Motion in a conservative field obey the conservation of mechanical energy law, so the quantity K+U=const. where K is the kinetic energy $$\frac{1}{2}mv^2$$ and $$U=mgh$$.

A material point at ground and still has $$U=0$$ and $$K=0$$. After if you move it and bring it to altitude $$h$$ and stop it, it has $$U=mgh$$ and again $$K=0$$. You don't need to pass through kinetic energy if someone told you you spent a work $$W$$ to reach $$h$$. This energy is now stored in potential energy form and it's ready to be released again in form of kinetic energy if you let it fall. Eventually, after the system hit the ground and stops again, all yuor precious work done to lift will turn into heat.

P.S. If you need to understand how in general a change is driven, you can image that there is a "charge" $$q$$ (think to the electric charge) which is an extensive quantity (it sums up if you add charges). Its "current'' is of course $$I=\frac{\mathrm{d}q}{\mathrm{d}t}$$. The associated intensive quantity will then be the electric potential $$V$$ which represent the "driving strength" for the change (so that the work is $$W=qV$$ and the energy flow will be its time derivative $$P=IV$$).

In our gravitational case you have the "charge" that is the mass $$m$$, an extensive quantity. The intensive associated quantity being $$gh$$ (such that the work is charge multiplied by the intensive quantity $$U=mgh$$ and finally the energy flow, the power, its time derivative $$\dot{U}$$).

The book answer assumes that when you lift the object you bring it to a stop so that the increase in kinetic energy you gave it when first starting to lift the object by applying an upward force greater than $$mg$$ is cancelled by a decrease of the same amount of kinetic energy when you decrease the force to be less than $$mg$$ causing the mass to decelerate (gravity doing negative work).

Hope this helps