Chapter 6, Problem 62P

A shot-putter accelerates a 7.3-kg shot from rest to 14 m/s in 1.5 s. What average power was developed?

To determine

The average powerdeveloped.

Answer to Problem 62P

Solution:$\text{477 W}$

Explanation of Solution

Given:

Mass of shot Mass of shot $\text{=7}\text{.3}\text{kg}$

Initial speed $\left({\text{V}}_{\text{1}}\right)\text{=0 m/s}$

Final velocity $\left({\text{V}}_{\text{2}}\right)\text{=14 m/s}$

Time taken $\text{=1}\text{.5s}$

Concept used:

Work energy principle

The net work done on an object is equal to the change in the object’s kinetic energy. $W_{net}=\Delta K=\left(K{E}_2-K{E}_1\right)$

  $W_{net}=\frac{1}{2}m{V}_2{}^2-\frac{1}{2}m{V}_1{}^2$

Power $=\frac{W_{net}}{t}$

Calculation:

  $W_{net}=\frac{1}{2}m{V}_2{}^2-\frac{1}{2}m{V}_1{}^2$

  $\begin{array}{c}{W}_{net}\text{=}\frac{\text{1}}{\text{2}}\text{×7}{\text{.3×(14)}}^{\text{2}}\text{-}\frac{\text{1}}{\text{2}}m{\left(\text{0}\right)}^{\text{2}}\\ \text{=715}\text{.4J}\end{array}$

Power $=\frac{W_{net}}{t}$

  $\text{=}\frac{\text{715}\text{.4}\text{J}}{\text{1}\text{.5}\text{s}}$

  $\begin{array}{l}\text{=476}\text{.93watt}\\ \approx 477\text{watt}\end{array}$

Conclusion:

Average power required to accelerate a $\text{7}\text{.3 kg}$ shot from rest to $\text{14 m/s}$ in $\text{1}\text{.5s}$ $\text{=}\text{477 W }$ .