Physics

Riding in an elevator, a man notices that he feel heavier or lighter based when the elevator speeds up or slows down. He stands on a bathroom scale in the elevator to take some measurements. Assume that his mass is 85 kg.

a. What is this man’s weight while at rest?

b. What does the scale read when the elevator accelerates upward at 2.35 m/s2?

c. What is the acceleration (magnitude AND direction) of the elevator if the scale reads 700 N?

d. What acceleration (magnitude AND direction) is required for the man to feel “weightless”? Prove it by showing all work.

asked by Madison
  1. Part A:
    weight at rest = 85 kg

    Part B:
    9.8 m/s² - 2.35 m/s² = 7.45 m/s²
    W = (7.45 m/s²)(85 kg) = 633.25 N

    Part C:
    a = F/m = (700 N)/(85 kg) = 8.24 m/s² (downward)

    Part D:
    to feel "weightless," acceleration must be 9.8 m/s²
    9.8 m/s² - 9.8 m/s² = 0 m/s²
    W = (0 m/s²)(85 kg) = 0 N

    posted by herp_derp
  2. Correction:
    In Part C, it should say upward, not downward.

    posted by herp_derp
  3. Herp: Weight is in newtons, not kg.
    a) weight at rest: mg=85*9.8 N
    b) weight=m(g+a)=85(9.8+2.35)=1033N
    c) Net weight= m(g+a), a positive direction is up
    745=m(g+a)
    745/85 = g+a
    745/85-g=a =-1.04m/s, direction negative means it is going downward
    d)
    to be weightless, the elevator must be falling at 9.8 m/s
    Net weight=m(g+a)
    a=-g

    posted by bobpursley

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