Work done by gravitational force!

Suppose a ball is $10m$ high from the ground. It will have $E_p = 10mg$. Now, if the ball falls freely then we know $E_k$ will be $0.5mv^2 = E_p$. But it is said that the work done by a force will transfer some energy to the forced object if the work is positive. Here work done by gravity will be positive. So the energy to the object should be $E_k = E_p$ (potential energy when it was up) $+ mgh$ ($F=mg$ and displacement is $h$) $= 2E_p$.

But it doesn't happen. Why?

Yes, it does indeed happen. The $mgh$ part, is the work done by gravity. Work done by a conservative force is "reversable" and can therefore be considered "stored". When you lift something to the clif, the work you do is stored as potential energy. And this energy is what the gravity can then use to do work on the object, as it falls.

More precisely, as gravity does work on an object, it can do up to $E_p$ of work on it. So when you set up the energy conservation equation where work done in this case equals the gain in kinetic energy

$$W=\Delta K=K_2-K_1$$

then this work $W$ equals the change in potential energy, so

$$-\Delta U=\Delta K$$

(The minus is because of the definition of work: $W=-\Delta U=-(U_2-U_1)=U_1-U_2$. Since the work done is positive, the potential energy loss must be positive too - and $U_1$ is bigger than $U_2$.)

What you call $E_p$ is the same as the work done by gravity. So $mgh$ should only appear once in your equation, and $E_k=E_p$, just as you say it should.

• The work is not called potential energy. The work performed during a process is the difference in potential energy before and after. – Sebastian Riese Oct 18 '15 at 15:01
• @SebastianRiese Thank you, agreed. I have rewritten a bit and added some parts to clear this out. – Steeven Oct 18 '15 at 15:36