Question

A 86.8-kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.72 m/s2, (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.67 m/s2?

(a) Number Units

(b) Number Units

(c) Number Units

Answer 1

Apparent weight will be given by Net force on person:

From Newton's 2nd law: (Upward is positive)

Fnet = m*a

Wa - Fg = m*a

Fg = Force due to gravity on person = m*g

Wa = apparent weight = ?

Wa = m*a + m*g = m*(g + a)

Part A

Now when acceleration is upward = +1.72 m/sec^2

Wa = 86.8*(9.81 + 1.72) = 1000.8 N

Part B

when constant speed, acceleration = 0 m/sec^2

Wa = 86.8*(9.81 + 0) = 851.5 N

Part C

Now when acceleration is downward = -1.67 m/sec^2

Wa = 86.8*(9.81 - 1.67) = 706.5 N

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