Question

Estimate the average force exerted by a short-putter on a 7.0 kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s.

Short answer

The average force exerted by the short-putter is 211.26 N.

Step-by-step solution

Step 1. Given data and assumption

During the shot-putter's motion, the short-putter gives it a maximum speed using the muscles of his hand.

Given data:

The mass of the shot is $m=7.0\mathrm{kg}$.

The initial speed of the shot is $v_{\mathrm{i}}=0$as it starts from rest.

The final speed of the car is $v_{\mathrm{f}}=13\frac{\mathrm{m}}{\mathrm{s}}$.

The shot moved through the distance $x-x_o=2.8\mathrm{m}$.

Assumption:

Let a be the average acceleration of the shot.

Step 2. Calculation of the average force

The average acceleration of the shot is

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2+2a\left(x-x_o\right)\\ {\left(13\frac{\mathrm{m}}{\mathrm{s}}\right)}^2=0^2+\left(2a\times 2.8\mathrm{m}\right)\\ a=30.18\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, from Newton's second law, you get that the average force on the car is

$\begin{array}{c}F=ma\\ =\left(7.00\mathrm{kg}\right)\times \left(30.18\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\\ =211.26\mathrm{N}\end{array}$

Hence, the average force exerted by the short-putter is 211.26 N.