question_answer

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

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