question_answer

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)  \[\frac{{{w}_{1}}+{{w}_{2}}}{2}\]                            

B)  \[\frac{{{w}_{1}}-{{w}_{2}}}{2}\]

C)  \[2{{w}_{1}}\]                                  

D)  \[2{{w}_{2}}\]

Correct Answer: A

Solution :

                 When lift moves up with constant    acceleration a, then                  \[{{w}_{1}}-mg=ma\]                                                     ?(i) When lift moves down with constant acceleration a, then \[mg-{{w}_{2}}=ma\]                                                     ?(ii) From Eqs. (i) and (ii), we get \[{{w}_{1}}+{{w}_{2}}=2mg\]                                                     ?(iii) When lift moves up with constant speed, its acceleration is zero. So,          \[w-mg=0\] or            \[w=mg\]                                            (iv) From Eqs. (iii) and (iv) \[{{w}_{1}}+{{w}_{2}}=2w\]                 or\[w=\frac{{{w}_{1}}+{{w}_{2}}}{2}\]