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A position dependent force F→=y2xi^+x2yj^ N acts on a particle and displace it from x = 1 m,y = 2m to x = 2m, y = 1m. The work done by the force is

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a

103

b

10 j

c

20 j

d

Zero

answer is D.

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Detailed Solution

w=∫F→⋅dr→=∫(1,2)(2,1) y2xdx+x2ydy=12∫1,22.1 dx2y2=0

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