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:

We release the body and as it falls, we can examine how kinetic and potential energies associated with it interchange.

Energy at Position B

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

Velocity \( v_B \) at B can be calculated from the relation:

where \( v_i = 0 \) and \( S = x \):

Thus,

Total energy at B:

Energy at Position C

At position C, just before the body strikes the Earth:

where \( v_C \) can be found out by the following expression:

Thus,

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:

Key Takeaways

• In an ideal frictionless system, the total mechanical energy (P.E. + K.E.) remains constant
• Potential energy converts completely to kinetic energy during free fall
• When friction is present, some energy is lost to work against friction

Considering Frictional Forces

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 \( fh \). The remaining P.E. \( = mgh – fh \) is converted into K.E.

Hence,

or

Thus,