contestada

A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the
ball for 0.020 s. What is the net force of the club on the ball?

Respuesta :

The ball accelerates from rest to 75.0 m/s in a matter of 0.020 s, so the drive applies an (average) acceleration of

a = (75.0 m/s - 0) / (0.020 s) = 3750 m/s²

Assuming this acceleration was constant, then the net force F on the ball is

F = (0.050 kg) (3750 m/s²) = 187.5 N ≈ 190 N