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. Numerical answer type. Work it out, then open the answer.

Answer: 40. s = (5-2)i + (-2-3)j + (1+4)k = 3i - 5j + 5k m W = F = 5(3) + 2(-5) + 7(5) = 15 - 10 + 35 Work done = 40 J.

A small particle moves to position 5i - 2j + k from its initial… | JEE Main 2023 (1 Feb, Shift 1)

Q11JEE Main 2023 · 1 FEB · SHIFT 1Easy

Work by a Constant Force

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 $5\hat{i} + 2\hat{j} + 7\hat{k}\text{ N}$ . The value of work done will be __________ J.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 40

$\vec{s} = (5-2)\hat{i} + (-2-3)\hat{j} + (1+4)\hat{k} = 3\hat{i} - 5\hat{j} + 5\hat{k}$ m
$W = \vec{F}\cdot\vec{s} = 5(3) + 2(-5) + 7(5) = 15 - 10 + 35$
Work done $= 40$ J.

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