Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height
of 30.0 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
1 answer:
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Answer:
616.223684211 N
Explanation:
= Resistive force on the wheel = 115 N
F = Force acting on sprocket
= Radius of sprocket = 4.75 cm
= Radius of wheel = 25 cm
Moment of inertia is given by
Torque
Torque is given by
The force on the chain is 616.223684211 N