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

Answer:

$SR=949.2N$

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

$a=\frac{v}{t}$

Therefore acceleration for the first 0.80 sec is

$a=\frac{1.2}{0.80}$

$a=1.5m/s^2$

Therefore

Spring Reading=Normal force -Reaction

$SR=m(g+a)$

$SR=84(9.8+1.5)$

$SR=949.2N$