5) A 90-kg man stands on a spring scale in an elevator. Starting from rest, the

Question

5)

A 90-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.72 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.5 s, and then comes to rest. What does the spring scale register in each of the following time intervals? before the elevator starts to move during the first 0.72 s of the elevator's ascent while the elevator is traveling at constant speed during the elevator's negative acceleration

Explanation / Answer

a)

When elevator is moving with constant speed, no net force acts on the man. Two forces acting on him, his weight, mg, downwards and normal reaction, N, from the sacle upwards balance each other.
=> N = mg = spring balance reading = 900N

b)

As we know that from first equation of motion

v = u + a*t

put the values in above equation

So,

1.2 = a * 0.72

a = 1.666667 m/s

According to Free body diagram of man

Reading of spring balance = N = mg + ma

= m(9.8 + 1.6667) N

= 1032 N

c)

When Elevator moves with constant speed so

N = mg = Reading of spring balance = 900 N

d)

By applying equation of motion

v = u + a*t

0 = 1.2 + a * 1.5

a = - 0.8 m/s

So,

N = mg + ma

N = reading of spring scale = 732 N

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