Physics

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74.0-kg person stands on a scale in an elevator. What is the apparent weight when the elevator is

(a) accelerating upward with an acceleration of 1.10 m/s2,


(b) moving upward at a constant speed, and


(c) accelerating downward with an acceleration of 1.00 m/s2?

  • Physics -

    (a) accelerating upward with an acceleration of 1.10 m/s2,
    F(net) = F(up) + F(down)
    ma = Fup + Fg
    (74.0kg)(1.10m/s^2) = Fup + (74.0kg)(-9.8m/s^2)
    Find Fup to get the scale reading
    (The scale reading is the value of the upward force of the spring inside the scale)
    (b) moving upward at a constant speed,
    F(net) = F(up) + F(down)
    0 = F(up) + F(down)
    F(up) = -F(down)
    (c) accelerating downward with an acceleration of 1.00 m/s2
    (74.0kg)(-1.00m/s^2) = Fup + (74.0kg)(-9.8m/s^2)
    Find F(up)

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