A force of (4 hat{i}+2 hat{j}+hat{k}) text{ N | vedclass

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(B) Given force $\vec{F} = (4 \hat{i} + 2 \hat{j} + \hat{k}) 	ext{ N}$.Initial position $\vec{r_1} = (2 \hat{i} + 2 \hat{j} + \hat{k}) 	ext{ m}$.Fin

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Given force $\vec{F} = (4 \hat{i} + 2 \hat{j} + \hat{k}) 	ext{ N}$.Initial position $\vec{r_1} = (2 \hat{i} + 2 \hat{j} + \hat{k}) 	ext{ m}$.Final position $\vec{r_2} = (4 \hat{i} + 3 \hat{j} + 2 \hat{k}) 	ext{ m}$.Displacement $\vec{S} = \vec{r_2} - \vec{r_1} = (4-2) \hat{i} + (3-2) \hat{j} + (2-1) \hat{k} = (2 \hat{i} + \hat{j} + \hat{k}) 	ext{ m}$.Work done $W = \vec{F} \cdot \vec{S}$.$W = (4 \hat{i} + 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + \hat{j} + \hat{k})$.$W = (4 	imes 2) + (2 	imes 1) + (1 	imes 1) = 8 + 2 + 1 = 11 	ext{ J}$.

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