Question

An object of mass ‘m‘ is kept on the floor of a lift. What will be the apparent weight of the object when the lift moves with a constant speed ‘v’ downwards?

Answer 1

Answer:

(b) moving downward with uniform acceleration. (c) moving upward with constant speed.

Answer 2

Answer:

The apparent weight of the object will be "mg".

Explanation:

The apparent weight of an object when the lift is moving downwards is given as,

$W=m(g-a)$       (1)

Where,

W=apparent weight of the object

m=mass of the object

g=acceleration due to gravity

a=acceleration with which the object is moving downwards

From the question, we see that the lift moves with constant speed "v" downwards. And for acceleration, we know that it is the rate of change of velocity. So, if velocity is constant then acceleration will be zero. i.e.

$a=0$        (2)

By substituting the value of "a" in equation (1) we get;

$W=m(g-0)$

$W=mg$    (3)

Hence, the apparent weight of the object will be "mg".