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TiliK225 [7]
3 years ago
15

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?
Physics
1 answer:
wel3 years ago
7 0

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 Timet_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

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