Hi can someone explain what is it they want exactly in this question? Also how do I solve this?

Question

A 75.0 kg man stands on a platform scale in an elevator. Starting from rest, the elevator  ascends, attaining its maximum speed of 1.20 m/s in 1.00 s. It travels with this constant speed for  the next 10.00 s. The elevator then undergoes a uniform acceleration in the negative y direction  for 1.70 s and comes to rest. What does the scale register: 
(a) before the elevator starts to move? 
(b) during the first 1.00 s? 
(c) while the elevator is traveling at constant speed? 
(d) during the time it is slowing down? Take g = 10 ms-2

Ash · Accepted Answer

(a) Before the elevator starts to move, the scale will register steady state reading of 75.0 kg

(b) For 1.00s , the elevator accelerates upwards
a = (v - u)/t = (1.20 - 0)/1.00 = 1.20 m/s²

Normal force on scale, n = m(g+a)

Reading on scale = n/g = m(g+a)/g = (75.0 kg)(9.81 m/s² + 1.20 m/s²)/(9.81 m/s²) = 84.2 kg

During the first 1.00s , the reading on scale will be 84.2 kg

(c) While elevator is travelling at constant speed, the only acceleration on the person will be due to gravity

so reading on scale will be 75.0 kg

(d) During slowing down, the acceleration of elevator will be
a =  (0 - 1.20)/(1.70) = -0.706 m/s²

Reading on scale = n/g = m(g+a)/g = (75.0 kg)(9.81 m/s² - 0.706 m/s²)/(9.81 m/s²) = 69.6 kg
While elevator is slowing down, the reading on scale will be 69.6 kg

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Hi can someone explain what is it they want exactly in this question? Also how do I solve this?

1 Answer