Question

A 84-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 0.73 as. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.4 s, and then comes to rest.

What does the spring scale register During the first 0.80s of the elevator’s ascent?

Answer 1

`SR=949.2N`

Step-by-step explanation:

From the question we are told that:

Mass
`M=84kg`

Speed
`V=1.2m/s`

Acceleration Time
`t_a=0.73`

Constant speed Time
`t_s=5.0s`

Deceleration time
`t_d=1.4s`

Generally the equation for Acceleration is mathematically given by

`a=(v)/(t)`

Therefore acceleration for the first 0.80 sec is

`a=(1.2)/(0.80)`

`a=1.5m/s^2`

Therefore

Spring Reading=Normal force -Reaction

`SR=m(g+a)`

`SR=84(9.8+1.5)`

`SR=949.2N`