A particle moved from position $$r1$$ → $$=$$ $$3i^$$ $$+2j^-6k^$$ to position $$r2$$→ $$=$$ $$14i^+13j^$$ $$+$$ $$9$$ $$k^$$ under the action of a force (4i^+j^+3k^) newtons. Find the work done.

displacement : s→ $$=$$ $$r2$$→ $$-r1$$→. Now find W $$=$$ F→ .s→ s→$$=$$ $$14i^+13j^+9k^-3i^-2j^+6k^$$ $$=11i^+11j^+15k^W$$ $$=$$ $$44+11+45$$ $$=$$ $$100$$ J

Work done by Diff type of forces

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Question

A particle moved from position $\vec{r_{1}}$ = $3 \hat{i} + 2 \hat{j}$ $- 6 \hat{k}$ to position $\vec{r_{2}}$ = 14 $\hat{i}$ + $13 \hat{j} + 9 \hat{k}$ under the action of a force ( $4 \hat{i} + \hat{j} + 3 \hat{k}$ ) newtons. Find the work done.

Easy

Solution

displacement : $\vec{s}$ = $\vec{r_{2}} - \vec{r_{1}}$ . Now find W = $\vec{F} . \vec{s}$
$\vec{s}$ = 14 $\hat{i}$ +13 $\hat{j}$ +9 $\hat{k}$ -3 $\hat{i}$ -2 $\hat{j}$ +6 $\hat{k}$ =11 $\hat{i}$ +11 $\hat{j}$ +15 $\hat{k}$
W = 44+11+45 = 100 J

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