A particle is acted upon by constant forces 4 | vedclass

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(D) The total force $\vec{F}$ is the sum of the individual forces: $\vec{F} = (4\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} + \hat{j} - \hat{k}) = 7\hat

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) The total force $\vec{F}$ is the sum of the individual forces: $\vec{F} = (4\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} + \hat{j} - \hat{k}) = 7\hat{i} + 2\hat{j} - 4\hat{k}$. The displacement vector $\vec{d}$ is the difference between the final position vector and the initial position vector: $\vec{d} = (5\hat{i} + 4\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 4\hat{i} + 2\hat{j} - 2\hat{k}$. The work done $W$ is the dot product of the total force and the displacement: $W = \vec{F} \cdot \vec{d} = (7\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (4\hat{i} + 2\hat{j} - 2\hat{k})$. $W = (7 	imes 4) + (2 	imes 2) + (-4 	imes -2) = 28 + 4 + 8 = 40$ units.

$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.

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