Posted by Lisa on Monday, September 8, 2008 at 3:58pm.
74.0kg 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  GK, Monday, September 8, 2008 at 8:20pm
(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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