Question

A person is standing on a weighing machine placed on the floor of an elevator. The elevator goes up with some acceleration, then moves with a uniform velocity and finally decelerates to stop. The maximum and minimum weights recorded are 720 N and 600 N. Assuming that the magnitude of the acceleration and deceleration are same, find
1) true weight of the person
2) the magnitude of acceleration

Answer 1

When the elevator is accelerating upwards, W = m(g+a)
When the elevator is moving with a uniform velocity, W = mg
When the elevator is decelerating, W = m(g-a)

As per the question, m(g+a) = 720 N ..........(1)
and m(g-a) = 600 N ..........(2)

Dividing equations (1) and (2), we get

$\frac{g+a}{g-a}=\frac{6}{5}\Rightarrow a=\frac{g}{11}=\frac{10}{11}m/s^2$

Substituting this in equation (1), we get m = 66 kg
So, true weight = mg = 660 N