Question

Consider a 75.5-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.65 m/s in 0.675 s. What does the spring scale register in pounds during these 0.675 s?

Answer 1

208.08 pounds

Step-by-step explanation:

Given that,

Mass of a man, m = 75.5 kg

Starting from rest, the elevator ascends, attaining its maximum speed of 1.65 m/s in 0.675 s.

We need to find the value of spring scale during these 0.675 s.

As the elevator ascends, net force will be given by :

F = mg + ma

a can be calculated as follows :

`a=(1.65-0)/(0.675)\\\\a=2.45\ m/s^2`

So,

F = 75.5[(9.81) + (2.45)]

F = 925.63 N

We know that,

1 N = 0.224 Pounds

It means, 925.63 N = 208.08 pounds

So, the spring scale will register 208.08 pounds.