Question

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 37.0 m. During the collision at the bottom of the elevator shaft, a 92.0 kg passenger is stopped in 3.00 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)?

Answer 1

Answer 2

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

Answer 3

Consider the given data. Height, H 37 m Mass of the passenger, m -92 kg Acceleration due to gravity g -9.8 m/s Passenger upward speed, v-9m's Partl: Consider the expression of motion When the elevator cabs is free fall from height the initial speed of the elevator u is From the equation (1) -V2x9.8x37 725.2 v 26.92 m/s Consider the expression of impulse 92x 26.92 I-2476.64 kgm/s Hence magnitude of the impulse is 2476.6 kgm/s Part2

Consider the expression of average force is 2476.64 f 825.54 N Hence average force on the passenger during the collision is |f-825.54 N Part3 To tind the impulse, when the passenger relating velocity is -9 m/s Consider the expression of impulse, 92 x9 I -882 kgm/s Hence, the impulse when the passenger jump upward is 882 kgm/s Part4: Consider the expression of average force, 882 f 294 N

Part A Normal reaction force v -mg cose - m Rc ะโ (800x9.81x cos26.57°) -800 | 223.61 - 6729.67N