# Definition of Gravitational Potential Energy [closed]

Suppose,we throw a ball of mass,m upward with an acceleration, A such that A>gravitational acceleration (g).clearly,the ball must stop after reaching it's maximum height, h.so,HOW SHOULD THE POTENTIAL ENERGY HERE BE CALCULATED?,Here the definition of the gravitational potential energy proves wrong. Gravitational potential energy is defined as the amount of work done in raising it from the ground to that height,which is the weight of the object.I can prove this definition this way,The work done on the ball=F*displacement=mAh which is greater that the weight of the ball. Am I wrong somewhere,then please tell? Also,why do we define gravitational potential energy like that?(what is the reason behind that definition?)

## closed as unclear what you're asking by sammy gerbil, Jon Custer, ZeroTheHero, peterh, John RennieJun 21 '17 at 5:00

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• "Gravitational potential energy is defined as the amount of work done in raising it from the ground to that height,[...]" Agree. "[...] which is the weight of the object" What, no. How can energy equal weight (a force)? This is not in the definition of gravitational potential energy. – Steeven Jun 20 '17 at 19:36
• hyperphysics.phy-astr.gsu.edu/hbase/gpot.html as regards gravitational P.E. – user154420 Jun 20 '17 at 19:47

## 1 Answer

The definition of gravitional potential energy (at small heights) is $$U=mgh$$ Plug in the height $h$ and you have the $U$ associated with that point.

This potential energy is correctly equal to the amount of work done to bring the ball up there. But remember that the work is only being done while the force is throwing the ball. When throwing a ball, you accelerate it and then you let it go and it flies upwards, stops, and returns down. Your throwing force only works until you let go.

So when trying to calculate work with the formula $W=Fh$, then the $h$ is wrong. The force $F$ did not do work on the ball over the entire distance $h$. You don't know the distance (this distance would be the swing of the arm), so you can't use this formula in this case.

Of the same reason, the acceleration is not as you expect. The force might accelerate the ball with $a=A$, but as soon as you let go, and the force doesn't work anymore, this acceleration disappears. During the flight only gravity pulls, and therefore the ball only accelerates (downwards) because of gravity during the whole flight (and this acceleration is $a=-g$). This is why is slows down, stops and then speeds up downwards.

When you throw the ball, in the moment you let it go, you have done an amount of work on it, which has given it a lot of kinetic energy. This kinetic energy will now during the flight be converted into potential energy, while it flies up. When it reaches it's highest point, all kinetic energy is used up and no more can be converted into potential energy, so it doesn't go any higher.