A student, standing on a scale in an elevator at rest, sees that his weight is 997 N. As the elevator rises, the scale reads 1226 N and then returns to normal. When the elevator slows to a stop at the top floor, the scale reads 650 N and then returns to normal. Determine the magnitude of the acceleration when the elevator is slowing to a stop. (no negative numbers)

Question

Physics

Twinkle

4 April, 15:29

A student, standing on a scale in an elevator at rest, sees that his weight is 997 N. As the elevator rises, the scale reads 1226 N and then returns to normal. When the elevator slows to a stop at the top floor, the scale reads 650 N and then returns to normal. Determine the magnitude of the acceleration when the elevator is slowing to a stop. (no negative numbers)

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Melanie · Answer 1

The total force F on the student is given by:

F = N - mg = ma

N normal force of the scale on to the student (scale readout)

m mass of the student

g acceleration due to gravity

a acceleration of the elevator

997 = mg

650 = m (g - a)

ma = 997 - 650

a = g (997 - 650) / 997

a = 3,4