A stroboscopic photo of a club hitting a golf ball, such as the photo shown in t

Question

A stroboscopic photo of a club hitting a golf ball, such as the photo shown in the figure below, was made by Harold Edgerton in 1933. The ball was initially at rest, and the club was shown to be in contact with the ball for about 0.0021 s. Also, the ball was found to end up with a speed of 2.2 102 ft/s. Assuming that the golf ball had a mass of 55 g, find the average force exerted by the club on the ball.

________ kN

Solution:

F = m(vfinal - vinital)/change in time
F = (55E-3kg)(2.2E2ft - 0)/ .0021s

I get it up to here, but then they multiply it by (1m/s / 3.281ft/s), getting 1.76 kN

Why do they multiply that?

Explanation / Answer

Speed Vf = 2.2x10^2 ft /s Speed is given in ft/s in FPS while other units are in SI units. conversion factor of ft to meter is 3.281 ft = 1m                = 2.2x10^2ft / (3.281 ft/m) = 67.052m/s The average force is             F = m (Vf- Vi) / t               = (0.055)[67.052 - 0 ] / 0.0021               =1.756x10^3 N               = 1.76kN

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