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Question: A lift is moving down with a retardation of \[5m/{s^2}\]. Calculate the percentage change in the wei...

A lift is moving down with a retardation of 5m/s25m/{s^2}. Calculate the percentage change in the weight of the person in the lift (g=10m/s2g = 10m/{s^2})
(a) 25%
(b) 50%
(c) 75%
(d) 60%

Explanation

Solution

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: aeff=i=1nai{a_{eff}} = \sum\limits_{i = 1}^n {{a_i}} ……(1)
Where,
ai{a_i} is the ithi^{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=maW = ma ……(2)
where,
mm is the mass of the body
aa is acceleration in a downward direction.
3.Percentage change in weight: δW=WfWiWi×100\delta W = \dfrac{{{W_f} - {W_i}}}{{{W_i}}} \times 100 ……(3)
where,
Wi{W_i} is the initial weight
Wf{W_f} is the final weight

Complete step by step answer:
Given:
1. Retardation of lift a=5m/s2a = - 5m/{s^2}
2. Acceleration due to gravity g=10m/s2g = 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:
aeff=g+a{a_{eff}} = g + a
aeff=10m/s2+(5)m/s2{a_{eff}} = 10m/{s^2} + ( - 5)m/{s^2}
aeff=5m/s2{a_{eff}} = 5m/{s^2}
Therefore, the body is accelerated by 5m/s25m/{s^2} in a downward direction.

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

Step 3 of 3:
Calculate the percentage change in weight using eq (3):
δW=m(10)m(5)m(10)×100\delta W = \dfrac{{m(10) - m(5)}}{{m(10)}} \times 100
δW=12×100\delta W = \dfrac{1}{2} \times 100
δW=50%\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.