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Scale in an elevator, time, mass, and velocity function given

  1. Nov 1, 2014 #1
    1. The problem statement, all variables and given/known data
    You are standing on a bathroom scale in an elevator in a small building. The elevator starts from rest and moves with a velocity given by v(t) = (3 m/s^2)t + (0.2 m/s^3)t^2. If your mass is 64 kg what is your weight at t=4.0 s as recorded by the scale (in N)

    2. Relevant equations
    v(t) = (3 m/s^2)t + (0.2 m/s^3)t^2
    The acceleration function is the derivative of the acceleration function so I guess: A(t) = .4t + 3
    I think this is needed: F = (ma)

    3. The attempt at a solution
    So I plug t=4 into the acceleration equation A(t) = .4t + 3 and get 4.6. Then F = m*a so 64*4.6 but that gives me 294N but the answers in the question are 627N, 1254N, 333N, 124N and 922N (the correct answer is 922N btw)

    I would really appreciate any help on this problem.
     
  2. jcsd
  3. Nov 2, 2014 #2

    ehild

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    You stand on the scales, that exerts a normal force on you, upward. The scale reads the normal force (divided by g). At the same time, gravity pulls downward with force mg. Ma= the sum of forces, N-mg. You accelerate upward, with acceleration 4.6 m/s2. Your apparent weight is equal to the normal force. How much is the normal force?
     
  4. Nov 2, 2014 #3
    294N + (64kg * 9.8m/s^2) = 921.2 N, thanks I got it
     
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