Why is the scale reading 397 N when the elevator is slowing down?

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The discussion revolves around a scale reading of 397 N in an elevator that is slowing down, which confuses the poster regarding the person's weight. Initially, the scale reads 587 N when the elevator accelerates upward, indicating an increase in apparent weight due to acceleration. As the elevator decelerates, the scale reads 397 N, which does not represent the person's actual weight but rather the reduced apparent weight during deceleration. The key point is that the scale reading reflects the dynamic conditions of the elevator's motion rather than the person's true weight, which would be observed once the elevator comes to a complete stop. Understanding the difference between apparent weight during acceleration and deceleration is crucial for solving this problem.
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Homework Statement


this should be simple question, but i could not get the answer. hope someone could point out my mistake.

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 587 N. Later, as the elevator later stops, the scale reading is 397 N. Assume the magnitude of the acceleration is the same during starting and stopping.

The Attempt at a Solution


Determine the weight of the person
it should be 397N, since the scale shows 397N when the elevator stops. but my answer is wrong. why?
 
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The elevator is stopping, not stopped, when the scale reads 397 N.
 
As written, the question is poorly stated. They mean: While the elevator is speeding up, the scale reads 587 N; while it's slowing down, the scale reads 397 N.

Once it stops, of course, the reading will equal his normal weight.
 
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