Physics - NLM Question with Solution | TestHub

Let the mass of the man be $m$ and the acceleration of the lift be $a$.

When the lift moves upwards with acceleration $a$, the apparent weight is given by:

$W_1=m(g+a)=660$

When the lift moves downwards with acceleration $a$, the apparent weight is given by:

$W_2=m(g-a)=380$

Adding the two equations, we get:

$m(g+a)+m(g-a)=660+380$

$2mg=1040$

$mg=520$ N

The real weight of the man when the lift is at rest is $mg=520$ N.

Answer: $520$ N