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A 1000 kg elevator rises from rest in th...

A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is ` 4 m s^(-1)` . There is a constant frictional force of 500 N. The work done by the lifting mechanism is

A

`196 xx 10^(3)`J

B

` 204 xx 10^(3) J`

C

` 214 xx 10^(3) J `

D

`203 xx 10^(5) J`

Text Solution

Verified by Experts

The correct Answer is:
C
Work done againt gravitational force = mgh
` 1000 xx 9.8 xx 20 = 196 xx 10^(3) J `
Work done to impart velocity to the body = `1/2 mv^(2)`
` 1/2 xx 10^(3) xx 16 =8 xx 10^(3)J`
Work done against frictional force ` 500 xx 20 = 10 xx 10^(3) J`
Total work done = ` 214 xx 10^(3) J`
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