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

Answer:

72.15723 m/s

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

$Fs=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2Fs}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 7810\times 0.013}{0.039}}\\\Rightarrow v=72.15723\ m/s$

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