Posted by **Lisa** on Monday, September 8, 2008 at 3:58pm.

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 -
**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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