Falak Kinney
2021-02-11
Answered

A 95.0kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.80 $\frac{m}{{s}^{2}}$ , (b) moving upward at a constant speed and (c) accelerating downward with an acceleration of 1.30 $\frac{m}{{s}^{2}}$ ?

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Viktor Wiley

Answered 2021-02-12
Author has **84** answers

We have 2 forces acting on the person standing on thescale.

W = Mg the weight of the person due to gravity is actingdownward.

F(scale) = the reading of the scale is acting upward.

The resultant of these 2 forces provides the acceleration ofthe person on the scale.

So Mg - F(scale) = Ma

Solve for F(scale) from the values given. Obviously if a(acceleration of the elevator) = 0

then Ma = 0 and F(scale) = Mg.

a) Elevator is moving upward N = M( g + a) = 95( 9.8 + 1.8)$=1102\text{}kg-{m}^{2}/S$

b) Elevator moving downward N = M( g - a) = 95( 9.8 -1.3)$=807.5\text{}kg-{m}^{2}/S$

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