Question

In serving, a tennis player accelerates a 59 g tennis ball horizontally from rest to a speed of 34 m/s Assuming that the acceleration is uniform when the racquet is applied over a distance of 0.36 m, what is the magnitude of the force exerted on the ball by the racquet

Answer 1

Answer:

The magnitude of the force exerted on the ball by the racquet is 94.73 N.

Explanation:

The force exerted on the ball is the following:

$F = ma$

Where:

m: is the mass of the ball = 59 g

a: is the acceleration

The acceleration of the ball can be found with the following kinematic equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

d: is the distance = 0.36 m

$v_{f}$: is the final speed = 34 m/s

$v_{0}$: is the initial speed = 0 (it start from rest)

Hence, the acceleration is:

$a = \frac{v_{f}^{2}}{2d} = \frac{(34 m/s)^{2}}{2*0.36 m} = 1605.6 m/s^{2$

Finally, the force is:

$F = ma = 59 \cdot 10^{-3} kg*1605.6 m/s^{2} = 94.73 N$    

Therefore, the magnitude of the force exerted on the ball by the racquet is 94.73 N.                                

                                                                 

I hope it helps you!