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Consider a particle on which constant forces ${\vec{F}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k} N$ and ${\vec{F}}_{2} = 4 \hat{i} - 5 \hat{j} - 2 \hat{k} N$ act together resulting in a displacement from position ${\vec{r}}_{1} = 20 \hat{i} + 15 \hat{j} c m$ to . ${\vec{r}}_{2} = 5 \hat{k} c m$ The total work done on the particle is

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$- 0.50 J$

$+ 0.50 J$

$- 5.0 J$

$+ 5.0 J$

answer is A.

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Detailed Solution

$\vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2} = [5 \hat{i} - 3 \hat{j} + \hat{k}] N$ and $\vec{s} = {\vec{r}}_{2} - {\vec{r}}_{1} = [- 20 \hat{i} - 15 \hat{j} + 5 \hat{k}] c m$

$∴ W = \vec{F} . \vec{s} = [- 10 + 45 + 5] 10^{- 2}$

$= - 50 \times 10^{- 2} J$

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