If $|F|=40$ N, $|D|=3$ m and $\theta=60^\circ$, then the work done is
- A. $60\sqrt3$ J
- B. 120 J
- C. $60\sqrt2$ J
- D. 60 J
✅ Correct Answer: 60 J
Explanation
Work done is given by $W=Fd\cos\theta$ $=40 \times 3 \times \cos60^\circ$ $=40 \times 3 \times 1/2 = 60$ J
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