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प्रश्न
A force F = (10 + 0.5 x) N acts on a particle in the x-direction. The work done by the force in displacing the particle from x = 0 to x = 2 metre is ______.
विकल्प
42 J
63 J
21 J
31.5 J
MCQ
रिक्त स्थान भरें
उत्तर
A force F = (10 + 0.5 x) N acts on a particle in the x-direction. The work done by the force in displacing the particle from x = 0 to x = 2 metre is 21 J.
Explanation:
`W=int_0^2Fdx=int_0^2(10+0.5x)dx`
`=[10x+0.5x^2/2]_0^2`
= 21 J
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