Chapter 8.5, Problem 43E

To determine

To find: the total done work on the ball to make it go this fast.

Answer to Problem 43E

The total done work is $50\text{ft-lb}$ .

Explanation of Solution

Given information:

A $2-\text{oz}$ tennis ball was served at $160\text{ft/sec}$ .

The starting velocity of the tennis ball $v_1$ is $0\text{ft/sec}$ , and the ending velocity $v_2$ is $160\text{ft/sec}$ . First find the mass of the tennis ball.

As the weight of the tennis ball is $2-\text{oz}$ or it can be written as $\frac{2}{16}\text{pounds}$ and the acceleration due to gravity is $32{\text{ft/sec}}^{\text{2}}$ . So, the mass of the tennis ball can be written as,

  $\begin{array}{c}w=m\times a\\ m=\frac{2}{16\left(32\right)}\\ =\frac{2}{8\left(32\right)}\end{array}$

The work has done can be calculated as,

  $\begin{array}{c}\frac{1}{2}m\left(v_2^2-v_1^2\right)=\frac{1}{2}\left(\frac{1}{8\left(32\right)}\right)\left({\left(160\right)}^2-{\left(0\right)}^2\right)\\ =\frac{1}{2}\left(\frac{1}{8\left(32\right)}\right)\left(\left(160\right)\left(160\right)\right)\\ =\frac{1}{2}\left(\frac{1}{8\left(32\right)}\right)\left(\left(20\right)\left(5\right)\right)\\ =50\text{ft-lb}\end{array}$

Therefore, the total done work is $50\text{ft-lb}$ .