A force of magnitude 50, N acting along hat{i | vedclass

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(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector along the direction $\hat{i} + \hat{j} + \hat{k}$.Unit vector $

Vedclass · Accepted Answer

$-50 \sqrt{3}\, J$

Vedclass · Answer

(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector along the direction $\hat{i} + \hat{j} + \hat{k}$.Unit vector $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.So,$\vec{F} = 50 	imes \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{50}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\, N$.The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4-5)\hat{i} + (8-9)\hat{j} + (6-7)\hat{k} = -\hat{i} - \hat{j} - \hat{k}\, m$.The work done $W = \vec{F} \cdot \vec{d} = \left[ \frac{50}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) ight] \cdot (-\hat{i} - \hat{j} - \hat{k})$.$W = \frac{50}{\sqrt{3}} (-1 - 1 - 1) = \frac{50}{\sqrt{3}} (-3) = -50\sqrt{3}\, J$.

$A$ force of magnitude $50\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a point mass from $(5, 9, 7)$ to $(4, 8, 6)$. The work done by this force during this displacement is:

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