A 75.0 kg man stands on a spring scale in an elevator. Starting from rest, the e

Question

A 75.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.63 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 2.00 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?

Explanation / Answer

Part A)

F = mg

F = (75)(9.8) = 735 N

Part B)

We need the acceleration

vf = vo + at

1.63 = 0 + a(1.5)

a = 1.087 m/s2

F = mg + ma

F = (75)(9.8 + 1.087)

F = 816.5 N

Part C)

There is no acceleration at constant velocity, so F = 735 N

Part D)

We need the deceleration

vf = vo + at

0 = 1.63 + a(2)

a = -.815 m/s2

F = mg + ma

F = (75)(9.8 - .815)

F = 673.9 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.