Question

when a 0.045 kg ball takes off after being hit, its speed is 37 m/s. a) how much work is done on the ball by the club? b) assume that the force of the golf ball acts parallel to the motion of the ball and that the club is in contact with the ball for a distance of 0.013 meters ignore the weight of the ball and determine the average force applied to the ball by the club.​

Answer 1

• (a) Ke = (1/2) m v^2 = (.045/2)(41)(41)

• (b) Work in = increase in energy = Ke

• (c) F * d = work = same Ke

• (d) new Ke = (.045/2)(31)(31)

• so answer = (.045/2)(31)(31) - (.045/2)(41)(41)
• which is negative, force opposite to motion

• (e) again work done = change of energy = F (84)