A student, standing on a scale in an elevator at rest, sees that his weight is 840 n. as the elevator rises, his weight increases to 1050 n, then returns to normal. when the elevator slows to a stop at the 10th floor, his weight drops to 588 n, and then returns to normal. determine the acceleration at the beginning and end of the trip.

Question

Anabella Ross

Physics

15 July, 20:06

A student, standing on a scale in an elevator at rest, sees that his weight is 840 n. as the elevator rises, his weight increases to 1050 n, then returns to normal. when the elevator slows to a stop at the 10th floor, his weight drops to 588 n, and then returns to normal. determine the acceleration at the beginning and end of the trip.

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Oliver · Answer 1

1) at rest his weight is 840 N

=> 840N = mass * g = > mass = 840 N / g = 840 N / 9.8 m/s^2 = 85.7 kg

2) as the elevator rises, his weight increases to 1050 N,

The reading of the scale is the norma force of it over the body of the person.

And the equation for the force is: Net force = mass * acceleration = normal force - weight at rest

=> mass * acceleration = 1050 N - 840 N = 210 N

acceleration = 210 N / mass = 210 N / 85.7 kg = 2.45 m/s^2 (upward)

3) when the elevator slows to a stop at the 10th floor, his weight drops to 588 N

=> mass * acceleration = 588 N - 840 N = - 252 N

=> acceleration = - 252 N / 85.71 kg = - 2.94 m / s^2 (downward)

Answer:

Acceleration at the beginning of the trip 2.45 m/s^2 upward

Acceleration at the end of the trip 2.94 m/s^2 downward