Physics - NLM Question with Solution | TestHub

Let $m$ be the mass of the child.

When the lift is going down with acceleration $a=\frac{g}{3}$, the apparent weight is given by $W_1=m(g-a)$.

$W_1=m\left(g-\frac{g}{3}\right)=\frac{2mg}{3}$

Given $W_1=20$ N, we have:

$\frac{2mg}{3}=20\Rightarrow mg=30$ N

When the lift is going upwards with acceleration $a=\frac{g}{3}$, the apparent weight is given by $W_2=m(g+a)$.

$W_2=m\left(g+\frac{g}{3}\right)=\frac{4mg}{3}$

Substituting $mg=30$ N:

$W_2=\frac{4}{3}\times 30=40$ N

Answer: 40 N