Q. The apparent weight of a person inside a lift is $ w_{1}$ when lift moves up with a certain acceleration and is $ w_{2}$ when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is
Punjab PMETPunjab PMET 2008Laws of Motion
Solution:
When lift moves up with constant acceleration a, then
$w_{1}-m g=m a \,\,\,\ldots(i)$
When lift moves down with constant acceleration a, then
$m g-w_{2}=m a \,\,\,\ldots(i i)$
From Eqs. (i) and (ii), we get
$w_{1}+w_{2}=2 m a \,\,\,\ldots(iii)$
When lift moves up with constant speed,
its acceleration is zero
So, $w-m g=0$
or $w=m g\,\,\, \ldots(i v)$
From Eqs. (iii) and (iv), we get
$w_{1}+w_{2}=2 m g $
or $w=\frac{w_{1}+w_{2}}{2}$
