Question

II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically starts from rest and is in contact with the tennis racquet for 30.00 ms. Assuming constant acceleration, what was the average force exerted on the tennis ball during this record serve, expressed in terms of the ball’s weight W?

Answer 1

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

$a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}$

Since we start counting at 0s and the ball departs from rest, this is just $a=\frac{v}{t}$

So we can write:

$F=ma=\frac{mv}{t}=\frac{gmv}{gt}$

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

$F=\frac{Wv}{gt}=\frac{v}{gt}W$

Substituting our values:

$F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N$

Where the average force exerted has been written it terms of the tennis ball's weight W.