umping up before the elevator hits. After the cable snaps and the safety system
ID: 2260160 • Letter: U
Question
umping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 32.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of 9.00 m/s relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?Explanation / Answer
a) first we need to find the velocity at the bottom
conservation of energy
1/2 mv^2 = m gh
v = sqrt(2gh)
impulse = dp = m dv = m sqrt(2gh) = 90*sqrt(2*9.81*41)=2553
b) F = dp/dt = 2553/5E-3=5.11E5 N
c) impulse = dp = m dv = m( sqrt(2gh) - 6.8) = 90*(sqrt(2*9.81*41)-6.8)=1941
d) F = dp/dt =1041/5E-3=3.88E5 N
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