A 77.0 kg man stands on a spring scale in an elevator.?

Question

Starting from rest, the elevator ascends, attaining its maximum speed of 1.24 m/s in 1.50 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.80 s and comes to rest.
a) What does the spring scale register before the elevator starts to move?

(b) What does it register during the first 1.50 s?

(c) What does it register while the elevator is traveling at constant speed?

(d) What does it register during the time it is slowing down?

Answer 1

A man weights 77 kg (force) therefore his mass is 77/9.81 kg, where g = 9.81 m/s^2 is gravitational constant.
a. at rest the scale shows man's weight 77 kg.
b. Elevator accelerates with constant acceleration a = 1.24/1.5 m/s^2 and man's acceleration force ma adds to the gravitational force. Therefore scale will show: mg + ma = 77*(1 + a/g) = 83.49 kg.
c. There is no additional acceleration therefore scale show only gravitational weight 77 kg.
d. Elevator undergoes uniform deceleration a = 1.24/1.8 m/s^2 that subtracts from the gravitational force. Therefore the scale will show mg - ma = 77*(1 - a/g) = 71.593 kg.