In a tennis serve, a 0.070 kg ball can be accelerated from rest to 41 m/s over a distance of 0.90 m. Find the magnitude fo the average force exerted by the racket on the ball during the serve.

I know it is a force problem, but I am not even sure how to set this up.

To solve this problem, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a).

In this case, the mass of the ball is given as 0.070 kg and the acceleration can be calculated using the kinematic equation: final velocity squared equals initial velocity squared plus 2 times acceleration times displacement (v^2 = u^2 + 2as).

Since the ball starts from rest (initial velocity, u = 0 m/s), the equation simplifies to v^2 = 2as. Rearranging the equation, we get a = v^2 / (2s).

Now, we can substitute the given values into the equation to find the acceleration:
a = (41 m/s)^2 / (2 * 0.90 m).

Calculating it gives a value of 953.6 m/s^2.

Finally, we can find the magnitude of the average force exerted by the racket on the ball by multiplying the mass of the ball and the acceleration:
F = m * a = 0.070 kg * 953.6 m/s^2.

Evaluating the expression, the magnitude of the average force exerted by the racket on the ball is approximately 66.8 N.