xy- સમતલ માં ગતિ કરતાં કણ પર બળ F = - K(yi + | vedclass

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Question

(c)While moving from $(0,0)$ to $(a,0)$
Along positive $x-$axis, $y = 0$ $	herefore \;\vec F = - kx\hat j$
i.e. force is in negative $y-$direction

Vedclass · Accepted Answer

$ - K{a^2}$

Vedclass · Answer

(c)While moving from $(0,0)$ to $(a,0)$
Along positive $x-$axis, $y = 0$ $	herefore \;\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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