What is the change in the weight of the body of mass 'm' in the lift, when the lift moves upwards without acceleration?

Question

Accepted Answer

The correct option is (3):

Let's understand the concept of apparent weight in a lift. The weight of a body is the force exerted on it due to gravity, which is given by $W = mg$, where 'm' is the mass of the body and 'g' is the acceleration due to gravity. When a body is in a lift, its apparent weight changes depending on the motion of the lift. Apparent weight is the normal force exerted by the surface of the lift on the body.

According to Newton's second law of motion, the net force ($F_{	ext{net}}$) acting on an object is equal to its mass (m) times its acceleration (a): $F_{	ext{net}} = ma$.

In this specific scenario, the lift is moving upwards without acceleration. This means the acceleration ($a$) of the lift is $0$.
Let $N$ be the normal force (apparent weight) acting upwards on the body, and $mg$ be the actual weight acting downwards.
Applying Newton's second law for the vertical motion:
$N - mg = ma$

Since the lift is moving without acceleration, $a = 0$.
So, the equation becomes:
$N - mg = m(0)$
$N - mg = 0$
$N = mg$

This means that the apparent weight ($N$) of the body is equal to its actual weight ($mg$). Therefore, there is no change in the weight of the body. The lift moving upwards at a constant velocity (zero acceleration) implies that the net force is zero, and thus the normal force supporting the body is exactly equal to its gravitational weight.

The incorrect option is (1): Increases.
Weight would increase if the lift were accelerating upwards. In that case, the apparent weight would be $N = m(g + a)$, which is greater than $mg$.

The incorrect option is (2): Decreases.
Weight would decrease if the lift were accelerating downwards. In that case, the apparent weight would be $N = m(g - a)$, which is less than $mg$. It would also decrease if the lift were moving downwards at a constant velocity, but the question specifies upward motion.

The incorrect option is (4): Weight increases 2 times.
This would imply a specific upward acceleration such that $N = 2mg$, which is not the case when there is no acceleration.