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

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

Step-by-step 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 = (v_(f)^(2))/(2d) = ((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!