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99 Percentile Qs Bank for JEE MainPhysicsWork, Power and Energy

A 1 ~kg box placed at the origin starts sliding along x -axis under the action of a force F =F i . Its acceleration as a function of x is given by a(x)= . x where =5 ~s ⁻² . The work done by F in moving the box from x =2 ~cm to x =5 ~cm in joule is

Options

  1. A52.5 10⁻⁴
  2. B105.5 10⁻⁴
  3. C17.0 10⁻⁴
  4. D34.0 10⁻⁴

Correct answer

A. 52.5 10⁻⁴

Step-by-step solution

Work done = Fdx = _ 0.02 ^ 0.05 ~m ( x ) dx aligned & = 1 5 2 [ x ^2 ]_ 0.02 ^ 0.05 & = 5 2 [25 10⁻⁴-4 10⁻⁴ ] & = 5 2 21 10⁻⁴ & =52.5 10⁻⁴ Joule. aligned

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