A 1text{ kg} block subjected to two simultane | vedclass

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(C) The resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2 = (2+3)\hat{i} + (3-1)\hat{j} + (4-2)\hat{k} = (5\hat{i} + 2\hat{j} + 2\hat{k})	ext{ N}$.The

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$35$

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(C) The resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2 = (2+3)\hat{i} + (3-1)\hat{j} + (4-2)\hat{k} = (5\hat{i} + 2\hat{j} + 2\hat{k})	ext{ N}$.The displacement vector $\vec{d}$ is along the direction $(3\hat{i} - 4\hat{j})$.The unit vector in this direction is $\hat{u} = \frac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + (-4)^2}} = \frac{3\hat{i} - 4\hat{j}}{5}$.Thus,the displacement vector is $\vec{d} = 25 \hat{u} = 25 	imes \frac{3\hat{i} - 4\hat{j}}{5} = 5(3\hat{i} - 4\hat{j}) = (15\hat{i} - 20\hat{j})	ext{ m}$.The work done $W$ is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d} = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j} + 0\hat{k})$.$W = (5 	imes 15) + (2 	imes -20) + (2 	imes 0) = 75 - 40 = 35	ext{ J}$.

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