Golf ball, average force, work done by the club on the ball?
Posted on
September 28th, 2009
by gcw
vampgrl92 asked:
I. When a 0.041-kg golf ball takes off after being hit, its speed is 42 m/s. How much work is done on the ball by the club?
I. When a 0.041-kg golf ball takes off after being hit, its speed is 42 m/s. How much work is done on the ball by the club?
II. When a 0.049-kg golf ball takes off after being hit, its speed is 36 m/s. Assume that the force of the golf club acts parallel to the motion of the ball and that the club is in contact with the ball for a distance of 0.007 m. Ignore the weight of the ball and determine the average force applied to the ball by the club.
golf swing analysis software
No related posts.
Related posts brought to you by Yet Another Related Posts Plugin.
One Response to “ Golf ball, average force, work done by the club on the ball? ”
Discussion Area - Leave a Comment
You must be logged in to post a comment.
Golf clubs sales
The work done on an object by a net force equals the change in kinetic energy of the object: W = KEf – KEi.
The kinetic energy ( KE ) of an object of mass m that is moving with velocity v is: KE = (1/2)mv^2.
Thus, for number 1, w=(1/2)(.041)(42^2), which is 36 Joules.
Work also equals force times distance: W=fd.
Thus, for number two, we can calculate the work through finding the final kinetic energy; w=(.5)(.049)(36^2), which is 32 Joules.
Use this same value for work and plug it into w=fd, and you will get 32=f(.007). Solve for F, and the answer is 4600 N.