Question

1. Mass of a person sitting in a lift is 50 kg. If lift is coming down with a constant acceleration of 10 m/sec^2 Then the reading of spring balance will be(g= - 10 m/sec^2)
(a) o
(b) 1000N
(c) 100 N
(d) 10 N​

Answer 1

Given :

Mass of person = 50kg

Lift is moving down with a constant acceleration of 10m/s²$.$

To Find :

We have to find reading of spring balance (apparent weight)$.$

SoluTion :

The apparent weight of a man in a lift, when the lift moves downward with acceleration a is given by

• R = m(g - a)

➝ R = m(g - a)

➝ R = 50(10 - 10)

➝ R = 50 (0)

➝ R = 0N

Answer 2

Given ,

Mass (m) = 50 kg

Acceleration of lift (a) = 10 m/s²

Acceleration due to gravity (g) = 10 m/s²

We know that , the apparent weight of a man during downward acceleration of lift is given by

$\boxed{ \sf{R = m(g - a) }}$

Thus ,

R = 50(10 - 10)

R = 50 × 0

R = 0 N

Therefore ,

• The reading of spring balance will be 0 N