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(B) The net force acting on the particle is the vector sum of all individual forces:$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4$$\v

Vedclass · Accepted Answer

in $Y-Z$ plane

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(B) The net force acting on the particle is the vector sum of all individual forces:$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4$$\vec{F}_{net} = (5 + 2 - 6 - 1)\hat{i} + (-5 + 8 + 4 - 3)\hat{j} + (5 + 6 - 7 - 2)\hat{k}$$\vec{F}_{net} = 0\hat{i} + 4\hat{j} + 2\hat{k}$Since the particle starts from rest at the origin,its acceleration $\vec{a} = \frac{\vec{F}_{net}}{m}$ will be in the direction of the net force.Because the $x$-component of the force is $0$,the particle will not move along the $x$-axis.The force vector lies in the $Y-Z$ plane because it has non-zero components only in the $y$ and $z$ directions $(F_y = 4, F_z = 2)$.Therefore,the particle will move in the $Y-Z$ plane.

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