A 100 kg elevator rises from rest in the basement to the fourth flor, a distance of 20 m. As it passes the fourth floor its speed is 4m/s. There is a constant frictional force of 500N. The work done by the lifting mechanism is (Take `g=10m//s^(2)`)

Options

A. `200xx10^(3)J`
B. `205xx10^(3)J`
C. `218xx10^(3)J`
D. `210xx10^(3)J`

Answer

C

Explanation

`v=5x^(3//2) and m=0.5`
At x=0, `v_(1)=5 xx 0=0`
and at x=2, `v_(2)=5xx (2)^(3//2)=5 xx sqrt8=10sqrt2`
`therefore` Work done =Increase in K.E. `=1/2 m[v_(2)^(2)-v_(1)^(2)]`
`=1/2 xx 0.5 (10sqrt(2))^(2)-0]`
`therefore W=1/2 xx 5/10 xx 200=50J`