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A force F=(10+0.50x) acts on a particle ...

A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter. Find the work done by this force during a displacement from x=0, ox=2.0m

A

` 25 J`

B

`29 J`

C

` 21 J`

D

` 18 J`

Text Solution

Verified by Experts

The correct Answer is:
C
Work done for the displacement of particle `x=0` to `x=2 ` unit is
`W=int_(0)^(2)Fdx=int_(0)^(2)(10+0.5x)dx`
`=[10xx+0.5xx(x^(2))/(2)]_(0)^(2)=21J`
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