4.8 Interconversion of Potential Energy and Kinetic Energy

4.8 Interconversion of Potential Energy and Kinetic Energy

Consider a body of mass $m$ at rest, at a height $h$ above the surface of the Earth as shown in Fig. 4.12. 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.

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

Now, height from the ground is $(h - x)$, so that:

and

Velocity $v_{B}$ at position $B$, can be calculated from the relation:

as $v_{i} = 0$, so $v_{B}^{2} = 2gx$

Therefore $\mathrm{K.E.} = \tfrac{1}{2}m(2gx) = mgx$

Total energy at position $B = \mathrm{P.E.} + \mathrm{K.E.}$

At position $C$, just before the body strikes the Earth, $\mathrm{P.E.} = 0$ and $\mathrm{K.E.} = \tfrac{1}{2}mv_{C}^{2}$, where $v_{C}$ can be found out by the following expression:

i.e., $\mathrm{K.E} = \tfrac{1}{2}mv_{C}^{2} = \tfrac{1}{2}m \times 2gh = mgh$

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. 4.13) that:

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

where $v_{1}$ and $v_{2}$ are the velocities of the body at the heights $h_{1}$ and $h_{2}$ 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 $fh$. The remaining P.E. $= mgh - fh$ is converted into K.E.

Hence $mgh - fh = \tfrac{1}{2}mv^{2}$

or

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

Conversely, Loss of K.E. = Gain in P.E. + Work done against friction

Example 4.5

A car weighing $1100\,\mathrm{kg}$ is moving with a velocity of $12\,\mathrm{m\,s^{-1}}$. When it is at point $P$, its engine stops. If the frictional force is $120\,\mathrm{N}$, what will be its velocity at point $Q$? How far beyond $Q$ will it go before coming to rest?

Solution

The kinetic energy possessed by the car at point $P$ will partly be converted into P.E. and partly used up in doing work against friction as it reaches point $Q$. Therefore:

Loss of K.E. = Gain in P.E. + Work against friction

Example 4.6

An object of mass $3\,\mathrm{kg}$ falls from a height of $15\,\mathrm{m}$. If it strikes the ground with a velocity of $16\,\mathrm{m/s}$, calculate the average frictional force of the air.

Solution

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

or