Question

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)

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