A 72.4 man steps off a platform 3.02 above the ground. He keeps his legs straigh

Question

A 72.4 man steps off a platform 3.02 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 0.620 before coming to rest.

calculate the force the ground exerts on him while he is slowing down. Express this force in newtons.

What is the magnitude of the reaction force to the force

Explanation / Answer

(a) v^2 = u^2 +2gh here u = 0 Hence the speed at the time his feet touch the ground v = sqrt (2gh) = sqrt (2 * 9.8 * 3.10) = 7.8 m/s (b) As he slows down, lets assume his deceleration is a here u = 7.8 m/s a = ? v = 0 as he will halt and s = 0.60 m v^2 = u^2 - 2 a s => 0 = 7.8^2 - 2 a * 0.60 => a = 50.7 m/ss it sign will be negative as it is opposing the motion. (c) Newton's law say that Impulse = change in momentum ------ eqn 1 First find out time to come to rest => v = u - at => 0 = 7.8 - 50.7 * t => t = 0.15 s so according to eqn 1 Impulse = F * t = m * u F * 0.15 = 75 * 7.8 => average force F = 3900 N or = 3900/10 kgF

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