1.

A 1000 kg elevator rises from rest in the basement to the fourthfloor,a distance of 20 m. Asit passes the fourth floorits speedis4 m s^(-1) . There is a constantfrictionalforceof 500 N. The workdoneby the liftingmechanism is

Answer»

`196 XX 10^(3)`J
` 204 xx 10^(3) J`
` 214 xx 10^(3) J `
`203 xx 10^(5) J`

Solution :Work doneagaint gravitational FORCE = mgh
` 1000 xx 9.8 xx 20 = 196 xx 10^(3) J `
Work done to impartvelocityto the body = `1/2 mv^(2)`
` 1/2 xx 10^(3) xx 16 =8 xx 10^(3)J`
WORKDONE against frictional force ` 500 xx 20= 10 xx10^(3) J`
Totalworkdone = ` 214 xx 10^(3) J`


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