A uniform force of (3hat i + hat j) text{ N} | vedclass

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(A) Given force $\vec{F} = (3\hat{i} + \hat{j}) 	ext{ N}$.Initial position $\vec{r}_1 = (2\hat{i} + \hat{k}) 	ext{ m}$.Final position $\vec{r}_2 = (

Vedclass · Accepted Answer

$9$

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(A) Given force $\vec{F} = (3\hat{i} + \hat{j}) 	ext{ N}$.Initial position $\vec{r}_1 = (2\hat{i} + \hat{k}) 	ext{ m}$.Final position $\vec{r}_2 = (4\hat{i} + 3\hat{j} - \hat{k}) 	ext{ m}$.Displacement $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + \hat{k}) = (2\hat{i} + 3\hat{j} - 2\hat{k}) 	ext{ m}$.Work done $W = \vec{F} \cdot \vec{d} = (3\hat{i} + \hat{j}) \cdot (2\hat{i} + 3\hat{j} - 2\hat{k})$.$W = (3 	imes 2) + (1 	imes 3) + (0 	imes -2) = 6 + 3 = 9 	ext{ J}$.

$A$ uniform force of $(3\hat i + \hat j) \text{ N}$ acts on a particle of mass $2 \text{ kg}$. The particle is displaced from position $(2\hat i + \hat k) \text{ m}$ to position $(4\hat i + 3\hat j - \hat k) \text{ m}$. The work done by the force on the particle is ............. $J$.

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