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A particle is made to move from the origin in three spells of equal distances, first along the $x-$ axis, second parallel to $y-$ axis and third parallel to $z-$ axis. One of the forces acting on it is has constant magnitude of $50\,N$ and always acts along the direction of motion. Work done by this force in the three spells of motion are equal and total work done in all the three spells is $300\,J$. The final coordinates of the particle will be
$(2,\, 2,\, 2)\,m$
$(4,\, 4,\, 4)\,m$
$(6,\, 6,\, 6)\,m$
$(10,\, 10,\, 10)\,m$
Solution
$\because \quad \mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}$
$\Rightarrow \quad(S)=\frac{W}{F}=\frac{300}{50}=6$
$\Rightarrow \quad S=|x|+|y|+|z|$
$\& \because|x|=|y|=|z|$
$\Rightarrow$ Final condinate of point is $(2,2,2)$
