# Work done by gravity

When an object is lifted up it gains potential energy equal to $mgh$. When it is dropped from this height, gravity does work on it (which is also $mgh$) and this work is converted into the kinetic energy.

But what happens to the potential energy that it initially possessed when it was at a height? Shouldn't the object now have $mgh+\frac{1}{2} mv^2$ as the total energy ( previous potential energy + work done by gravity) ?

It is not the object that has the potential energy but the object and the Earth.

When the object is dropped the object/Earth system loses potential energy and the object and the Earth gain kinetic energy.
Because the mass of the Earth is so much greater than that of the object the Earth gains very little kinetic energy compared with the kinetic energy gained by the object.

So it is very common to read that the potential energy of the object is converted to the kinetic energy of the object.

• I love this answer. I had never found this explanation about potential energy of the system before.
– user104909
Feb 15 '17 at 16:50

Remember the law of conservation of energy:

$$K_1+U_1=K_2+U_2$$

$_1$ is before and $_2$ is after. This law says that energy must come from somewhere. So the gain in potential energy must come from somewhere - it must be another amount of energy, which is just converted into potential energy.

So, if you have an object high up and potential energy is stored $U_1=mgh_1$, then you can put this together in your equation:

$$K_1+U_1=K_2+U_2\\ 0+U_1=K_2+0$$

The object starts from rest $K_1=0$ and ends down so that there is no more stored potential energy $U_2=0$. The end result is that all the stored potential energy is converted into kinetic energy. You don't add them up; rather one converts into the other.

Say at the very beginning the energy of the system

earth-"object of mass m"-"machine lifting the object"

is zero and the object lies on the ground. When the machine lifts the object to $s=h$ this costs energy $\approx mgh$, (e.g., a battery is used where chemical processes take place, reducing the energy of the battery), assuming the radius $R$ of the earth is much larger than $h$, $h\ll R$, and the change of the potential energy of the machine is negligible. Since the total energy has to be conserved, $mgh$ is the energy extracted from the battery (forget about losses due to friction).

Now, if the mass $m$ is dropped to the ground we have a two body system earth-"object of mass m" where the masses are falling towards each other with the center of mass close to the earth center if the mass of the earth is much larger than $m$. At some point, both objects collide. This collision can be something between elastic or perfectly plastic. In the first case both momentum and kinetic energy is conserved. In the case of a perfectly plastic collision the potential energy -gained from the battery- is converted into deformation (re-ordering of atoms in the many-body-potential, changing the surface of the object and by this the amount of dangling bonds...) and heat (phonons and photons are created).

We define the change in potential energy of an object in gravitational field as the negative of the work done by the gravity.

$$\Delta{U}=-W=-mgh\cos{\theta}$$

When the object is at height $h$:

$$U_1=mgh$$ $$\implies U_2=0$$ This is potential energy of object at just above the ground.

Then where the energy has gone?

It has been converted into kinetic energy of the object.

I.E.

$$K.E=mgh$$

But you would say that $K.E=\frac{1}{2}mv^2$. Ok see what happens to this.

When object is at height h and moves to just above the ground under the influence of gravity so that it gains a final velocity $v$:

$$2aS=v_f ^2 - v_i ^2$$ $$\implies 2gh=v^2 - 0=v^2$$ $$\implies K.E=\frac{1}{2}m(2gh)$$ $$\implies K.E=mgh$$

Now if your object hits the ground, some of its kinetic energy is converted into heat and some into sound and some into other forms therefore the object does not go back to the same height where it was in the very beginning.