Question

A 95.1-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 2.48 m/s2, (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.84 m/s2

Answer 1

(a) 1168.78 N, (b) 932.93 N and (c) 757.95 N, approximately.

Step-by-step explanation:

In general, let's suppose the elevator is moving upward with a constant acceleration
`a`, then by applying Newton's second law on the person and writing for the normal force (which is the one the scale displays), we have

`ma=\sum F\\ma=N-mg\\N=m(a+g).`

(a)
`a = 2.48 m/s^2`

`N=95.1 kg*\left(2.48(m)/(s^2)+9.81(m)/(s^2)\right)\\N=1168.779 N`.

(b)
`a = 0 m/s^2`

`N=95.1 kg*\left(0(m)/(s^2)+9.81(m)/(s^2)\right)\\N=932.931 N`.

(c)
`a = -1.84 m/s^2`

`N=95.1 kg*\left(-1.84(m)/(s^2)+9.81(m)/(s^2)\right)\\N=757.947 N`.

Answer 2

Step-by-step explanation:

Mass of the person in elevator

M = 95.1kg

a. When the elevator is accelerating upward at 2.48m/s²

Fnet = ma

R — W = ma

R — mg = ma

R = ma+mg

R = m(a+g)

R = 95.1(2.48+9.81)

R = 95.1 × 12.2

R = 1168.78N

b. Elevator moving upward at a constant speed

This means that it is not accelerating. Therefore, the acceleration is zero

Fnet = ma

a = 0

Fnet = 0

R - W = 0

R = W

R = mg

R = 95.1 × 9.81

R = 932.931 N.

c. Elevator accelerating downward at 1.84m/s².

Fnet = ma

Since it is downward, a is moving in -ve direction of y axis

R — W = -ma

R = W—ma

R = mg —ma

R = m(g-a)

R = 95.1(9.81—1.84)

R = 95.1 × 7.97

R = 757.947 N.