91Ó°ÊÓ

In this problem, the work done in accelerating the shot put is equivalent to the variation in kinetic energy.

You can use the relation of work done and power over a period of time to calculate the average power.

Given data:

The mass of shot put is\(m = 7.3\;{\rm{kg}}\).

The final velocity of the shot put is\(v = 14\;{\rm{m/s}}\).

The time is\(t = 1.5\;{\rm{s}}\).

The relation of work done is given by:

\(\begin{aligned}W &= {E_{\rm{f}}} - {E_{\rm{i}}}\\W &= \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2}\end{aligned}\)

Here,\({E_{\rm{i}}}\)and\({E_{\rm{f}}}\)are the initial and final kinetic energies andu is the initial velocity of the shot put whose value will be zero because the shot put is at rest.

On plugging the values in the above relation, you get:

\(\begin{aligned}W &= \frac{1}{2}m{v^2} - \frac{1}{2}m{\left( 0 \right)^2}\\W &= \frac{1}{2}\left( {7.3\;{\rm{kg}}} \right){\left( {14\;{\rm{m/s}}} \right)^2}\\W &= 715.4\;{\rm{J}}\end{aligned}\)

The relation of work and power is given by:

\(P = \frac{W}{t}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}P &= \left( {\frac{{715.4\;{\rm{J}}}}{{1.5\;{\rm{s}}}}} \right)\\P &= 476.9\;{\rm{W}}\end{aligned}\)

Thus, \(P = 476.9\;{\rm{W}}\) is the average power developed.