# Derive the expression kinetic energy, potential energy

Derive the expression kinetic energy, potential energy

Derivation for the equation of Potenstial Energy:
Let the work done on the object against gravity = W
Work done, W = force × displacement
Work done, W = mg × h
Work done, W = mgh

Since workdone on the object is equal to mgh, an energy equal to mgh units is gained by the object . This is the potential energy (Ep) of the object.
Ep = mgh

Derivation for the equation of Kinetic Energy:
The relation connecting the initial velocity (u) and final velocity (v) of an object moving with a uniform acceleration a, and the displacement, S is
v2 - u2 = 2aS
This gives
S = v 2 - u 2 2a

We know F = ma. Thus using above equations, we can write the workdone by the force, F as

``````                        W = ma × v 2 - u 2 2a
or
W = 1 2 m( v 2 - u 2 )
``````

If object is starting from its stationary position, that is, u = 0, then
W = 1 2 m v 2

It is clear that the work done is equal to the change in the kinetic energy of an object.

If u = 0, the work done will be W = 1 2 m v 2 .

Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is Ek = ½ mv2