A small particle moves to position 5i^−2j^+k^ from its initial position 2i^+3j^-4k^ under the action of force 5i^+2j^+7k^ N. The value of work done will be______J.

Displacement=5i^−2j^+k^-2i^+3j^-4k^ =3i^−5j^+5k^Work done =F→·r→ = 40 unit

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A small particle moves to position $5 \hat{i} - 2 \hat{j} + \hat{k}$ from its initial position $2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ under the action of force $5 \hat{i} + 2 \hat{j} + 7 \hat{k} N$ . The value of work done will be______J.

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answer is 40.

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Detailed Solution

Displacement
$= (5 \hat{i} - 2 \hat{j} + \hat{k}) - (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 3 \hat{i} - 5 \hat{j} + 5 \hat{k}$
Work done $= \vec{F} \cdot \vec{r}$ = 40 unit

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