A 75.0 kg man steps off a platform 3.02 m above the ground. He keeps his legs st

Question

A 75.0 kg man steps off a platform 3.02 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.650 m before coming to rest.

Part F

Use Newton's laws and the results of part to calculate the force the ground exerts on him while he is slowing down. Express this force as a multiple of the man's weight.

Part G

What is the magnitude of the reaction force to the force you found in part E? PART E the answer is 4150N

Explanation / Answer

Just at impact, his velocity is v = (2gh) = (2 * 9.8m/s² * 3.02m) = 7.7 m/s
Then he reduces his velocity to 0 while moving through 0.65 m:
u² = 0 = v² + 2as = (7.7m/s)² + 2a*0.65m a = -45.6 m/s²
This acceleration is in addition to gravity (in other words, he exerts some force merely by standing); so Favg = m(g + a) = 75kg * (-9.8 - 45.6)m/s² = 4155 N

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