A) \[-2K{{a}^{2}}\]
B) \[2K{{a}^{2}}\]
C) \[-K{{a}^{2}}\]
D) \[K{{a}^{2}}\]
Correct Answer: C
Solution :
While moving from (0,0) to (a,0) Along positive x-axis, y = 0 \[\therefore \ \vec{F}=-kx\hat{j}\] i.e. force is in negative y-direction while displacement is in positive x-direction. \\[{{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 \[\vec{F}=-k(y\hat{i}+a\hat{J})\] The first component of force,\[-ky\hat{i}\]will not contribute any work because this component is along negative x-direction \[(-\hat{i})\] while displacement is in positive y-direction (a,0) to (a,a). The second component of force i.e. \[-ka\hat{j}\]will perform negative work \\[{{W}_{2}}=(-ka\hat{j})\ (a\hat{j})\]= \[(-ka)\ (a)\ =-k{{a}^{2}}\] So net work done on the particle \[W={{W}_{1}}+{{W}_{2}}\] = \[0+(-k{{a}^{2}})=-k{{a}^{2}}\]
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