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Question

Physics Question on laws of motion

The apparent weight of a person inside a lift is w1 w_{1} when lift moves up with a certain acceleration and is w2 w_{2} when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

A

w1+w22\frac{{{w}_{1}}+{{w}_{2}}}{2}

B

w1−w22\frac{{{w}_{1}}-{{w}_{2}}}{2}

C

2w12{{w}_{1}}

D

2w22{{w}_{2}}

Answer

w1+w22\frac{{{w}_{1}}+{{w}_{2}}}{2}

Explanation

Solution

When lift moves up with constant acceleration a, then w1−mg=ma   …(i)w_{1}-m g=m a \,\,\,\ldots(i) When lift moves down with constant acceleration a, then mg−w2=ma   …(ii)m g-w_{2}=m a \,\,\,\ldots(i i) From Eqs. (i) and (ii), we get w1+w2=2ma   …(iii)w_{1}+w_{2}=2 m a \,\,\,\ldots(iii) When lift moves up with constant speed, its acceleration is zero So, w−mg=0w-m g=0 or w=mg   …(iv)w=m g\,\,\, \ldots(i v) From Eqs. (iii) and (iv), we get w1+w2=2mgw_{1}+w_{2}=2 m g or w=w1+w22w=\frac{w_{1}+w_{2}}{2}