A uniform force of (3i^+j^)newton$$ acts on a particle of mass $$2$$ kg. Hence the particle is displaced from position $$(2i^+k^) metre to position (4i^+3j^-k^) metre. The work done by the force on the particle is

Applied force F→ $$=$$ $$3i^+j^Displacement$$ S → $$=$$ $$r2$$→$$-r1$$→ $$=$$ $$2i^+3j^-2k^Hence$$ work done W $$=$$ F→.S→ $$=$$ $$6+3+0$$ $$=$$ $$9$$ J

Work done by Diff type of forces

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Question

A uniform force of $(3 \hat{i} + \hat{j})$ newton acts on a particle of mass 2 kg. Hence the particle is displaced from position $(2 \hat{i} + \hat{k})$ metre to position $(4 \hat{i} + 3 \hat{j} - \hat{k})$ metre. The work done by the force on the particle is

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Solution

$Applied force \vec{F} = 3 \hat{i} + \hat{j}$
Displacement $\vec{S} = \vec{r_{2}} - \vec{r_{1}} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k}$
Hence work done W = $\vec{F} . \vec{S} = 6 + 3 + 0 = 9 J$

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