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Physics Question on Newtons Laws of Motion

The apparent weight of a person inside a lift is w1 when lift moves up with a certain acceleration and is w2 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{{w}_{1}}-mg=ma ?(i) When lift moves down with constant acceleration a, then mg−w2=mamg-{{w}_{2}}=ma ?(ii) From Eqs. (i) and (ii), we get w1+w2=2mg{{w}_{1}}+{{w}_{2}}=2mg ?(iii) When lift moves up with constant speed, its acceleration is zero.So, w−mg=0w-mg=0 or w=mgw=mg (iv) From Eqs. (iii) and (iv) w1+w2=2w{{w}_{1}}+{{w}_{2}}=2w or w=w1+w22w=\frac{{{w}_{1}}+{{w}_{2}}}{2}