A force F = (5i + 3j + 2k) N is applied over a particle which displaces it from its origin to the point r = (2i - j) m The work done on the particle in joules is (a) +10 (b) +7 (c) -7 (d) +13

Answer: (b). s = 2i - j m (from the origin). W = F = 5(2) + 3(-1) + 2(0) = 7 J So the answer is (b) +7.

A force F = (5i + 3j + 2k) N is applied over a particle which… | JEE Main 2004 (AIEEE)

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Work by a Constant Force

A force $\vec{F} = \left(5\vec{i} + 3\vec{j} + 2\vec{k}\right)\text{ N}$ is applied over a particle which displaces it from its origin to the point $\vec{r} = \left(2\vec{i} - \vec{j}\right)\text{ m}$ The work done on the particle in joules is

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Answer: (b)

$\vec{s} = 2\hat{i} - \hat{j}$ m (from the origin).
$W = \vec{F}\cdot\vec{s} = 5(2) + 3(-1) + 2(0) = 7$ J
So the answer is (b) $+7$ .

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