Question

A force $F=-k(x\hat{i}+y\hat{j})$ here k is constant acts on a particle moving is the x-y plane.Starting from the origin,the particle is taken to (a,a) and then to $(a/\sqrt{2},0)$.The total work done by the force F on the particle is

• A. $-2k{a}^2$
• B. $2k{a}^2$
• C. $3k{a}^2/4$
• D. $-\left(k{a}^2/4\right)$

Answer 1

The correct option is C $3k{a}^2/4$

Work done, $W=\int F.dx=-k\int F_{x}dx+F_{y}dy=\frac{-k}{2}{}_{(a,a)}^{(\frac{a}{\sqrt{2}},0)}(x^2+y^2)=\frac{-k}{2}[(\frac{a^2}{2}+0)-(a^2+a^2)]=\frac{-k}{2}[\frac{a^2}{2}-2a^2]=\frac{-k}{2}\left[\frac{-3a^2}{2}\right]=\frac{3}{4}{a}^{2}k$