Question

An elevator undergoes constant upward acceleration, taking it from rest to 10.6m/s  in 4.0 s. An 64kg  man stands on a scale in the elevator.

part b

Nearing the top floor, the elevator slows from 10.6m/s  to rest in the last 22.4 m of upward travel. Assuming constant acceleration, what's the scale reading during this period?

Answer 1

Part a

Aceeleration upwards = 10.6 / 4 m/s^2 = 2.65 m/s^2

Net acceleration of man = 9.8 m/s^2 (down) + 2.65 m/s^2 (up) = 9.8 - 2.65 = 7.15 m/s^2

Scale reads = 64 kg * 7.5 m/s^2 = 457.60N

Part b

Aceeleration upwards = - (10.6 ^2) /( 2 * 22.4 )  m/s^2 = - 2.508 m/s^2 (negative sign implies downward acceleration)

Net acceleration of man = 9.8 m/s^2 (down) + 2.508m/s^2 (down) = 9.8 + 2.508 = 12.308 m/s^2

Scale reads = 64 kg * 12.308 m/s^2 = 787.712 N