I am riding in the physics department elevator, you will recall from class, and

Question

I am riding in the physics department elevator, you will recall from class, and standing on a set of scales. Just after the doors close the scales read "690 N". Then, briefly, the scales read "740N". The reading returns to "690 N" for a little while, before jumping briefly down to "590 N" and, finally, returning to "690 N". The doors open. We'll call those 5 different readings I, II, III, IV and V. [In order to answer this series of questions, draw free body diagrams, which should also include your MODEL.

drawing graphs of acceleration, velocity and position vs time for the elevator.]

What was the magnitude of the acceleration (2 s.f.) of the elevator during reading I ("690 N", just after the doors close)?

Explanation / Answer

When elevator is moving with constant velocity, reading will be 690 N
This is your weight
m*g = 690
m*9.81 = 690
m = 70.34 Kg

When elevator is accelerating upward with acceleration a, reading will be more than 690 N given by,
Reading = m*(g+a)

When elevator is accelerating downward with acceleration a, reading will be less than 690 N given by,
Reading = m*(g-a)

Your question:
SInce reading is 690 N, you are moving with uniform velocity
Hence acceleration will be 0
Answer: 0

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