Chapter 4, Problem 8P

To determine

The average force exerted by shot-putter on a shot.

Answer to Problem 8P

Solution:

The force exerted by shot-putter is $211.1\text{ N}$.

Explanation of Solution

The concept we must use to solve this problem is average acceleration and force. The average acceleration is the change in velocity with respect to the elapsed time. So, by using Newton’s equation of motion, the expression for the average acceleration is given as below:

$\begin{array}{l}{v}^2-u^2=2a_{\text{avg}}s\\ a_{\text{avg}}=\frac{v^2-u^2}{2s}\end{array}$………… (1)

Here, $v$ is the final velocity, $u$ is the initial velocity and $s$ is the distance.

The force on an object is calculated by using the Newton’s second law of motion as given below:

$F_{\text{avg}}=m{a}_{\text{avg}}$………… (2)

Given:

Mass of the shot $m=7.0\text{ kg}$

Distance $s=2.8\text{ m}$

Initial speed $u=0\text{ m/s}$

Final speed $v=13\text{ m/s}$

Formula used:

$a_{\text{avg}}=\frac{v^2-u^2}{2s}$

$F_{\text{avg}}=m{a}_{\text{avg}}$

Calculation:

Substitute, $v=13\text{ m/s}$ and $s=2.8\text{ m}$ in the equation (1)so the average acceleration is given as below:

$a_{\text{avg}}=\frac{v^2-u^2}{2s}$

        $=\frac{{\left(13\text{ m/s}\right)}^2\text{-0}}{2\left(2.8\text{ m}\right)}$

        =30.17 m/s²

Now we can find the force exerted by shot-putter on shot substituting

$m=7.0\text{ kg}$ and $a_{\text{avg}}=30.17{\text{ m/s}}^2$ in equation (2)as given below:

F_avg=ma_avg

       =(70 kg)(30.17 m/s²)

       =211.1N