Question

A block of mass M kg kept on a weighing machine in an elevator.If elevator is retarding upward by ams^2 the reading of weighing machine is (in Kgf) (1)mg (2)m(g-a)(3)m(1-a/g) (4)m(g+a)

Answer 1

Motion in an elevator or lift :

1) If the man inside the lift falls freely under gravity without any upward force then the apparent weight is zero.

2) If he falls with an acceleration a less than g then he experiences an upward force of m(g - a) and hence his apparent weight is m(g-a).

3) If he moves up with an acceleration a he experiences an upward force of    m(g +a) and his apparent weight is equal to m(g + a).

From the explanations above the answer is :

4) m(g +a)

Answer 2

Answer is given below