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(A) The work done by a constant force is given by $W = \vec{F} \cdot \vec{S}$. Since the force acts along the direction of motion,$W = F \cdot S$.Give

Vedclass · Accepted Answer

$(2, 2, 2) \ m$

Vedclass · Answer

(A) The work done by a constant force is given by $W = \vec{F} \cdot \vec{S}$. Since the force acts along the direction of motion,$W = F \cdot S$.Given the total work $W_{total} = 300 \ J$ and force $F = 50 \ N$,the total distance $S_{total}$ is $S_{total} = \frac{W_{total}}{F} = \frac{300}{50} = 6 \ m$.The motion occurs in three equal spells of distance $d$. Thus,$3d = 6 \ m$,which means $d = 2 \ m$.The particle moves $2 \ m$ along the $x$-axis,$2 \ m$ parallel to the $y$-axis,and $2 \ m$ parallel to the $z$-axis.Therefore,the final coordinates are $(2, 2, 2) \ m$.

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