हिंदी

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 ______. -

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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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