Consider a 77.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.08 m/s in 0.900 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 1.70 s and comes to rest.

a. What does the spring scale register before the elevator starts to move?

b. What does the spring scale register during the first 0.900 s?

c. What does the spring scale register while the elevator is traveling at a constant speed?

d. What does the spring scale register during the time it is slowing down?

Question

Physics

Deegan

22 August, 08:40

Consider a 77.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.08 m/s in 0.900 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 1.70 s and comes to rest.

a. What does the spring scale register before the elevator starts to move?

b. What does the spring scale register during the first 0.900 s?

c. What does the spring scale register while the elevator is traveling at a constant speed?

d. What does the spring scale register during the time it is slowing down?

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Answers (1)

Know the Answer?

Eden Rojas · Answer 1

The answers to the question are

a. Before the elevator starts to move the spring scale registers 755.37 N

b. During the first 0.900 s the spring scale registers 847.77 N

c. While the elevator is traveling at constant speed the spring scale registers the force due to gravity on the man = 755.37 N

d. During the time it is slowing down the spring scale registers 706.45 N

Explanation:

We solve each part as follows, knowing the velocity and time during each phase of the elevator motion

a. Before the elevator starts to move the spring scale registers the weight of the man which is

Weight = mass * gravity, where gravity or g = 9.81 m/s² and the mass of the man = 77.0 kg

Hence the scale registers 77 * 9.81 = 755.37 N

(b) During the first 0.900 s the spring scale registers

The wight of the man ad the force of the upward moving elevator on him thus we are required to calculate the acceleration of the elevator thus

v = u + at where v = final velocity = 1.08 m/s u = initial velocity = 0 and t = 0,900 s therefore,

1.08 = 0.9*a or a = 1.08/0.9 = 1.2 m/s²

Hence the force on the man by the elevator plus the weight of the man = 755.37 N + 1.2*77 N = 847.77 N

(c) While the elevator is traveling at constant speed the spring scale registers the force due to gravity on the man or the weight of the man thus

77 * 9.81 = 755.37 N

(d) During the time it is slowing down the spring scale registers

Since in this case v = u-at then

a = 1.08/1.7 = 0.635

Hence the spring registers 77 * (9.81 - 0.635) = 706.45 N