JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Work, Power & Energy

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.

Answer: 40

Solution

The displacement is

d = r_{f} - r_{i} = (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}

. The work done by the constant force is the scalar product

W = F \cdot d = (5) (3) + (2) (- 5) + (7) (5) = 15 - 10 + 35 = 40 J .

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

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