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A force `F=-k(y 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)` |
Answer» For motion of the particle from (0,0)to(a,0), `vec(F)=-K(0hat(i)+ahat(j))rArrvec(F)=-Kaj` Displacement, `vec(r )=(ahat(i)+0hat(j))-(0hat(i)+0hat(j))=ahat(i)` so work done from (0,0)to(a,0) is given by `W=vec(F).vec(r )=-Kahat(j).ahat(i)=0` For motion (a,0) to (a,a) `vec(F)= -K(ahat(i)+hat(j))` and displacement, `vec(r )=(ahat(i)+ahat(j))-(ahat(i)+0hat(j))=ahat(j)` So work done from `(a,0)` to`(a,a)` `W=vec(F).vec(r)= -K(ahat(i)+ahat(j)).ahat(j)= -Ka^(2)` So total work done `= -Ka^(2)` |
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