When a 0.045kg golf ball takes off after being hit, it\'s speed is41m/s. How muc

Question

When a 0.045kg golf ball takes off after being hit, it's speed is41m/s. How much work is being done on the ball by the club? Assume that the force of the club acts parallel to the motionof the ball and tha the club is in contact with the ball for adistance of 0.010m. Ignore the weight of the ball anddetermine that average force applied to the ball by the club.
m= .45kg W=? Vo= 0 W=F(s) -------> W= (Fcos(theta)s) s=.010 theta= 0 (is theta zero b/c it is parallel) Vf= 41m/s

m= .45kg W=? Vo= 0 W=F(s) -------> W= (Fcos(theta)s) s=.010 theta= 0 (is theta zero b/c it is parallel) Vf= 41m/s

Explanation / Answer

Thanks for the help!

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