Chapter 11, Problem 54A

To determine

The kinetic energy as it leaves the racket.

Answer to Problem 54A

  $168.75\text{J}$

Explanation of Solution

Given:

The mass of the ball, $m=0.060\text{kg}$

The force applied on the ball, $F=150.0\text{N}$

Time duration for which the force is applied on the object, $\Delta t=0.03\text{s}$

Since the ball starts at rest, thus its initial velocity, $v_i=0\text{m/s}$

Formula used:

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse}=F\Delta t=p_f-p_i$

where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

And, the kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $KE=\frac{1}{2}m{v}^2$

Calculation:

Using the given values in above formula of Impulse, the velocity with the ball leaves the racket is,

  $\begin{array}{c}F\Delta t=p_f-p_i\\ \left(150\right)\left(0.03\right)=m\left(v_f-v_i\right)\\ 4.5=0.06\left(v_f-0\right)\\ v_f=75\text{m/s}\end{array}$

Then, the kinetic energy of the ball as it leaves the racket is,

  $\begin{array}{c}KE=\frac{1}{2}m{v}_f^2\\ =\frac{1}{2}\left(0.06\right){\left(75\right)}^2\\ =168.75\text{J}\end{array}$

Conclusion:

Thus, the kinetic energy of the ball as it leaves the racket is $168.75\text{J}$ .