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Question

# A force →F=−k(y^i+x^j) (where k is a positive constant) acts on a particle moving in the x−y plane. Starting from the origin, the particle is taken along the positive x−axis to the point (a,0) and then parallel to the y−axis to point (a,a). Then, total work done by the force is

A
2ka2
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B
2ka2
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C
ka2
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D
ka2
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Solution

## The correct option is C −ka2Given, →F=−k(y^i+x^j)=−ky^i−kx^j Applying the fundamental equation for work done as the force is variable, W=∫→F.→dr [→dr=dx^i+dy^j] ⇒W=∫(−ky^i−kx^j).(dx^i+dy^j) ⇒W=∫−kydx−kxdy ...(1) For displacement along path OA, y is constant and x is variable. ⇒y=0, dy=0 x goes from 0 to a From Eq.(1), WOA=∫0−0=0 ...(2) For displacement along path AB, x is constant and y is variable. ⇒x=a, dx=0 Limit of y goes from 0 to a. ∴From Eq.(1), WAB=∫0−ka dy WAB=−kaa∫0dy WAB=−ka[y]a0=−ka2 ....(3) Therefore, total work done by the force is given as, W=WOA+WAB ∴W=−ka2

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