What is the Apparent Weight in an Accelerating Elevator?

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SUMMARY

The discussion focuses on calculating apparent weight changes in an accelerating elevator. When stationary, the weight is 550N; during upward acceleration, it increases to 590N, and during downward acceleration, it decreases to 510N. The correct acceleration experienced while moving up is 0.713 m/s², and while moving down, it is -0.713 m/s². The confusion arises from comparing these results to incorrect values of 10.5 m/s² and 9.01 m/s² provided by others.

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Zach Knight
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Homework Statement


While stationary on Earth you have a weight of 550N. When in an elevator that accelerates upward your weight temporarily becomes 590N. When descending, your weight temporarily becomes 510N. Find a) the acceleration you experience as the elevator moves up and b) the acceleration you experience as the elevator moves down.

Homework Equations


\SigmaF=m*a


The Attempt at a Solution


There are two forces acting on you in the elevator: the force due to gravity Fg, which acts downward, and the force of the floor on you Ffloor, which acts upward. According to Newton's 2nd law, the sum of these forces will equal m*a, so we can write
Ffloor - Fg = m*a
We know that Fg equals m*g, where g = 9.8m*s-2, so we can find the mass by dividing Fg by g, which yields about 56.1kg. Substituting in all of the known values and solving for a, we find that
for a)
a = \frac{(590 - 550)N}{56.1kg} = .713m*s-2
and for b)
a = \frac{(510 - 550)N}{56.1kg} = -.713m*s-2
The answer key and the others students got 10.5m*s-2 for (a) and 9.01m*s-2 for (b), which are the answers I got, plus g.
I'm pretty sure I'm the one who's wrong here, but I don't understand where I messed up. Help would be greatly appreciated.
 
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When the elevator is not moving, your acceleration is zero. Therefore, I like your answers better.
 
i agree with lewando :smile:
 

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