Question

Consult Concept Simulation 6.1 in preparation for this problem. A golf club strikes a 0.039-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 7810 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

72.15723 m/s

Step-by-step explanation:

F = Force = 7810

s = Displacement = 0.013 m

m = Mass of club = 0.039

v = Velocity

Here work done will be the change in kinetic energy

`Fs=(1)/(2)mv^2\\\Rightarrow v=\sqrt{(2Fs)/(m)}\\\Rightarrow v=\sqrt{(2* 7810* 0.013)/(0.039)}\\\Rightarrow v=72.15723\ m/s`

The speed at which the ball leaves the club is 72.15723 m/s