Question

A boy of $50\,{\text{kg}}$ is standing in a lift moving down with an acceleration of $9.8\,{\text{m/}}{{\text{s}}^2}$. The apparent weight of the boy is:
A. $\dfrac{{50}}{{9.8}}\,{\text{N}}$
B. $50 \times 9.8\,{\text{N}}$
C. $50\,{\text{N}}$
D. Zero

Answer 1

Use the formula for the apparent weight of an object. This formula gives the relation between actual weight of the object, mass of the object and the acceleration of the system of the object.

Formula used:
The weight of an object is
$W = mg$ …… (1)
Here, $W$ is the weight of the object, $m$ is the mass of the object and $g$ is the acceleration due to gravity.
The expression for the apparent weight of an object is
${W_{app}} = {W_{real}} - ma$
Here, ${W_{app}}$ is the apparent weight of the object, ${W_{real}}$ is the real weight of the object, $m$ is the mass of the object and $a$ is the acceleration of the system of the object.

Complete step by step answer:
The acceleration of the lift and hence the boy in the lift is $9.8\,{\text{m/}}{{\text{s}}^2}$.
Calculate the apparent weight of the boy in the lift.
${W_{app}} = {W_{real}} - ma$ …… (2)
Rewrite equation (1) for the real weight of the boy.
${W_{real}} = mg$
Here, $m$ is the mass of the boy.
Substitute $mg$ for ${W_{real}}$ in equation (2).
${W_{app}} = mg - ma$
Substitute $50\,{\text{kg}}$ for $m$, $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$ and $9.8\,{\text{m/}}{{\text{s}}^2}$ for $a$ in the above equation.
${W_{app}} = \left( {50\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right) - \left( {50\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)$
$\Rightarrow {W_{app}} = 0\,{\text{N}}$
Therefore, the apparent weight of the boy is zero.
Hence, the correct option is D.

Note: Newton’s second law of motion can also be used to determine the apparent weight of the boy in the lift. The solution using this law also ends with the same zero apparent weight of the boy.