A force →F=2x^j acts on a particle. The particle starts from point (0,0) and moves to the point (3,9) following the path y=x2. The total work done by the force →F on the particle is
A
20J
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B
36J
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C
−26J
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D
27J
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Solution
The correct option is B36J Given variable force →F=2x^j
Displacement vector →ds=dy^j
workdone (W)=∫→F.→ds=∫30(2x^j).(dy^j) W=∫302xdy
As the path followed is y=x2⇒dy=2xdx ⇒W=∫30(2x)(2xdx)⇒W=∫304x2dx ⇒W=43x3|30=36J