A golf club strikes a 0053kg golf ball in order to launch it

A golf club strikes a 0.053-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 6250 N, and is in contact with the ball for a distance of 0.010 m. With what speed does the ball leave the club?

Solution

We first calculate the work done on the ball which is the kinetic energy, from the kinetic energy, we calculate the speed of the ball

we have

work done on the ball = force*displacement = 6250 * 0.010 = 62.5 J

work done =kinetic energy = 62.5 J

kinetic energy is given by

KE = mv²/2

or

v = sqrt(2*KE/m)

substituting the values, we get

v =sqrt(2*62.5/0.053) = sqrt(2358.49) = 48.56 m/s

hence the speed with which the ball leaves the club is 48.56 m/s