InterviewSolution
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InterviewSolution
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A force `F=-k(^hati + x^hatj)` (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 the point `(a,a)`. The total work done by the force F on the particle is (a) `-2ka^(2)` , (b) `2ka^(2) , (c)`-ka^(2)` , (d) `ka^(2)` |
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Answer» Correct Answer - A::B::C `dW=F.dr`, where `dr=dxhat(i) + dyhat(j) + dzhat(k)` and `F=k(yhati + xhatj)` `:. dW=-k(ydx + xdy)=-kd(xy)` `:. W=-int_(0,0)^(a,a)dW =-kint_(0,0)^(a,a)d(xy)` `=-k[xy]_(0,0)^(a,a)`, `W=-ka^(a)` `:.` The correct option is `(c)`. |
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