Question

A lift is moving down with a retardation of \[5m/{s^2}\]. Calculate the percentage change in the weight of the person in the lift (\[g = 10m/{s^2}\])
(a) 25%
(b) 50%
(c) 75%
(d) 60%

Answer 1

Hint: Forces in the same direction can be added up to give net force in that direction, Similarly, when a body is experiencing acceleration due to different factors in the same direction, we can add them up to find the effective acceleration of the body.

Formula used:
1. Effective acceleration in a particular direction: ${a_{eff}} = \sum\limits_{i = 1}^n {{a_i}} $ ……(1)
Where,
${a_i}$ is the $i^{th}$ acceleration.
So, we are just adding all the accelerations in a particular direction just like we add force.
2. Weight of a body: $W = ma$ ……(2)
where,
$m$ is the mass of the body
$a$ is acceleration in a downward direction.
3.Percentage change in weight: $\delta W = \dfrac{{{W_f} - {W_i}}}{{{W_i}}} \times 100$ ……(3)
where,
${W_i}$ is the initial weight
${W_f}$ is the final weight

Complete step by step answer:
Given:
1. Retardation of lift $a = - 5m/{s^2}$
2. Acceleration due to gravity \[g = 10m/{s^2}\]

To find: Percentage change in weight of the man inside the lift.

Step 1 of 3:
Use eq (1) to find the effective acceleration:
${a_{eff}} = g + a $
${a_{eff}} = 10m/{s^2} + ( - 5)m/{s^2} $
${a_{eff}} = 5m/{s^2} $
Therefore, the body is accelerated by $5m/{s^2}$ in a downward direction.

Step 2 of 3:
Calculate the initial weight of the body (on the ground) using eq (2):
${W_i} = m \times 10$
Calculate the final weight of the body (in the lift) using eq (2):
${W_f} = m \times 5$

Step 3 of 3:
Calculate the percentage change in weight using eq (3):
$\delta W = \dfrac{{m(10) - m(5)}}{{m(10)}} \times 100 $
$\delta W = \dfrac{1}{2} \times 100 $
$\delta W = 50\% $

The percentage change in the weight of the person in the lift is 50%. So, Option (b) is correct.

Note:
When we are standing in a lift moving downwards, we feel lighter because of our apparent weight decrease. Similarly, when the lift is moving upwards, we feel heavier as our apparent weight increases.