Question 192: Laws of Motion

Question

A child stands on a weighing machine inside a lift. When the lift is going down with acceleration $\mathrm{g} / 3$, the machine shows a reading 20 N . When the lift goes upwards with acceleration $\mathrm{g} / 3$, the reading would be ( $\mathrm{g}=$ gravitational acceleration)

Accepted Answer

Given:
The reading of the weighing machine (apparent weight) inside the lift going down with acceleration $\dfrac{g}{3}$ is
20 N
.
We need to find the reading when the lift is going
up
with acceleration $\dfrac{g}{3}$.
Let actual weight of the child$=mg$.
Step 1: Find the true weight using the "down" scenario
When the lift is
going down
with acceleration $a = \dfrac{g}{3}$:
The apparent weight is given by:
$$ N_1 = m(g - a) = m\left(g - \frac{g}{3}ight) = m \left(\frac{2g}{3}ight) $$
Given $N_1 = 20\ 	ext{N}$,
$$ m \left(\frac{2g}{3}ight) = 20 $$
Step 2: Express $mg$ in terms of $N_1$
$$ m g = x\ (	ext{let}) $$
So,
$$ \frac{2}{3} x = 20 \implies x = \frac{20 	imes 3}{2} = 30 $$
Thus, the real weight $mg = 30\ 	ext{N}$.
Step 3: Find the reading when the lift is going UP
When the lift is
going up
with acceleration $a = \dfrac{g}{3}$:
The apparent weight is
$$ N_2 = m(g + a) = m\left(g + \frac{g}{3}ight) = m \left(\frac{4g}{3}ight) $$
We already found $mg = 30$ N:
$$ N_2 = 30 	imes \frac{4}{3} = 40\ 	ext{N} $$
Final Answer:
Option A
40 N

Answer

40 N

Answer

30 N

Answer

20 N

Answer

50 N

Laws of Motion

A child stands on a weighing machine inside a lift. When the lift is going down with acceleration \mathrm{g} / 3, the machine shows a reading 20 N . When the lift goes upwards with acceleration \mathrm{g} / 3, the reading would be ( \mathrm{g}= gravitational acceleration)

Choose the correct answer: