Aperson stands on a scale in an elevator. as the elevator starts, the scale has a constant reading of 591 n. as the elevator later stops, the scale reading is 391 n. assuming the magnitude of the acceleration is the same during starting and stopping, determine (a) the weight of the person, (b) the person’s mass, and (c) the acceleration of the elevator.

(a) 391 N

(b) 39.89 kg

(c) 5 m/s^2

Explanation:

Reading of scale when escalator is moving up = 591 N

Reading of scale when escalator is at rest = 391 N

Le a be the acceleration of the escalator.

(a) when the escalator is at rest, it gives the weight of person.

So, mg = 391 N

Thus, the weight of person is 391 N.

(b) Weight = m g

m = Weight / g = 391 / 9.8 = 39.89 kg

Thus, the mass of the person is 39.89 kg.

(c) As it is moving up, by using the Newton's second law

R - mg = m a

591 - 391 = 39.89 a

a = 5 m/s^2

Thus, the acceleration of the elevator is 5 m/s^2.

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