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A force F=Kx^(2) acts on a particle at a...

A force `F=Kx^(2)` acts on a particle at an angle of `60°` with the x–axis. the work done in displacing the particle from `x_(1)` to `x_(2)` will be –

A

`k/6 (x_(2)^(2) - x_(1)^(2))`

B

`(k)/(6) (x_(2)^(3) - x_(1)^(3))`

C

`k(x_(2)-x_(1))`

D

`k(x_(2)^(3) - x_(1)^(3))`.

Text Solution

Verified by Experts

The correct Answer is:
B
We know work done `=dW= overset(to)(F).doverset(to)(x)= Fdx cos theta`
`therefore W= int_(x_1)^(x_2) F. dx .cos60^@ = (1)/(2) [ int_(x_1)^(x_2) kx^(2).dx]`
`rArr W= (k)/(2) int_(x_1)^(x_2) x^(2) dx=(k)/(2) [ (x^3)/(3)]_(x_1)^(x_2) = (k)/(6) (x_(2)^(3) - x_(1)^(3))`.
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