A force F = 20 + 10 y acts on a particle...

A force `F = 20 + 10 y` acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle form `y = 0` to `y = 1` m is

A

`20 J`

B

`30 J `

C

`5 J`

D

`25 J`

Text Solution

Verified by Experts

The correct Answer is:

D

Given : `F = 20 + 10 y`
work done, `W = int F.dy`
`= int_0^1 (20 + 10 y) dy = [20 y + 10/(2 y^2)]_(0)^(1) = 20 + 10/2 = 25J`.

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