Question

A golf club strikes a 0.042-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of 6950 N, and is in contact with the ball for a distance of 0.013 m. With what speed does the ball leave the club

Answer 1

Explanation:

W = 6950N(0.013m) = 90.35J

KEf = W = 90.35J

90.35J = 1/2(0.042kg)v^2

simplify equation:

v = $\sqrt{\frac{2(90.35J)}{0.042kg}$

v = 65.6m/s approximately