Consider a body of mass m at rest, at a height h above the surface of the earth as shown in Fig. 4.11.

At position A, the body has P.E. – mgh and K.E. = 0. We release the body and as it falls, we can examine how kinetic and potential energies associated with it interchange.

surface-of-the-earth

Fig. 4.11

Let us calculate P.E. and K.E. at position B when the body has fallen through a distance x, ignoring air friction.

Kinetic-Energy-equation-01

 

Velocity at B, can be calculated from the relation,

 Kinetic-Energy-02

Total energy at B = P.E. + K.E.

                        = mg (h – x) + mgx = mgh     …………..                  (4.18)

At position C, just before the body strikes the Earth, P.E. = 0

Kinetic-Energy-03

 Thus at point C, kinetic energy is equal to the original value of the potential energy of the body. Actually when a body falls, its velocity increases i.e., the body is being accelerated under the action of gravity. The increase in velocity results in the increase in its kinetic energy. On the other hand, as the body falls, its height decreases and hence, its potential energy also decreases. Thus we see (Fig. 4.12) that,

potential-energy -also - decreases

Fig. 40.12

Loss in P.E. = Gain in K.E.

Kinetic-Energy-04

Where V1 and V2 are velocities of the body at heights h1 and h2 respectively. This result is true only when frictional force is not considered.

If we assume that a frictional force f is present during the downward motion, then a part of P.E. is used in doing work against friction equal to f h. the remaining P.E. = mgh – fh is converted into K.E.

Kinetic-Energy-05

Thus,

Loss in P.E. = Gain in K.E. + work done against friction.