A force $\mathbf{F} = - K(y\mathbf{i} + x\mathbf{j})$ (where $K$ is a positive constant) acts on a particle moving in the $xy$-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 particles is

Question

Dr. Abubacker · Accepted Answer

(c) While moving from (0,0) to ($a,0$)
Along positive $x$-axis, $y = 0\ 	herefore\overrightarrow{F} = - kx\widehat{j}$
$i.e$., force is in negative $y$- direction while displacement is in positive $x$-direction
$	herefore W_{1} = 0$
Because force is perpendicular to displacement
Then particle moves from ($a,0$) to ($a.a$) along a line parallel to $y$-axis $(x = + a)$ during this $\overrightarrow{F} = - k(y\widehat{i} + a\widehat{j})$
The first component of force, $- ky\widehat{i}$ will not contribute any work because this component is along negative $x$-direction $( - \widehat{i})$ while displacement is in positive $y$-direction $(a,0)$ to $(a,a)$. The second component of force $i.e.$ $- ka\widehat{j}$ will perform negative work
$	herefore W_{2} = \left( - ka\widehat{j} ight)\left( a\widehat{j} ight) = ( - ka)(a) = - ka^{2}$
So net work done on the particle $W = W_{1} + W_{2}$
$= 0 + \left( - ka^{2} ight) = - ka^{2}$

Explanation

Reviewed by Dr. Abubacker

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