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Question

A force F=(2x^i+3y2^j) N acts on a body and displaces it from the origin (0,0) to (2 m,4 m).The work done (in J) by the force is

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Solution

Given, F=(2x^i+3y2^j) N
Position vector of initial position r1=(0^i+0^j)
Position vector of final position of the body is, r2=(2^i+4^j) m
Work done by a variable force is given by,
W=r2r1F.dr
W=r2r1(2x^i+3y2^j).(dx^i+dy^j)
[dr=dx^i+dy^j]
Putting the limits of (x,y) as (0,0) to (2,4)
W=(2,4)(0,0)2xdx+3y2dy ...(1)
W=[x2+y3](2,4)(0,0)
W=[(4+64)(0+0)]
W=68 J
Work done by the force is 68 J.

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