Asked by Emily
A golf club strikes a 0.061-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 7760 N, and is in contact with the ball for a distance of 0.014 m. With what speed does the ball leave the club?
Answers
Answered by
Francisca
W= change in Kinetic Energy
F*s = 1/2mv^2
Solve by v:
v= sqrt((2Fs)/m))
substitution:
v= sqrt((2*7760N*0.014m)/0.061kg)
v=59.68 m/s
F*s = 1/2mv^2
Solve by v:
v= sqrt((2Fs)/m))
substitution:
v= sqrt((2*7760N*0.014m)/0.061kg)
v=59.68 m/s
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