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3 years ago

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In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 79.0 kg, and the collision on the floor lasts 0.0750 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

1 answer:

Travka [436]3 years ago

5 0

Answer:

A)247.4kgm/s b) 3299.0N

Explanation:

The potential energy of the victim = mgh where m is mass, g is acceleration due to gravity and h is height from where the victim fell

Pe = 79*9.81*0.5 = 1/2mv2 (kinetic energy of the motion) v is speed in m/s

v = √9.81 = 3.132m/s

Change in momentum of the victim = m(vi-vf)

Vi is zero and vf is the velocity of impact

Change in momentum = 79*(3.132-0) = 247kgm/s which also the impulse

B) average force x time = impulse = 247

Average force = 247/0.075 = 3299.04N

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