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NEET2008PhysicsWork, Power and EnergyActual

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m . Taking g=10 ~m / s ^2 , work done against friction is

Options

  1. A200 J
  2. B100 J
  3. Czero
  4. D1000 J

Correct answer

B. 100 J

Step-by-step solution

Net work done in sliding a body up to a height h on inclined plane array r = Work done against gravitational force + Work done against frictional force array W=W_g+W_f ...(i) but W=300 ~J W_g=m g h=2 10 10=200 ~J Putting in Eq. (i), we get gathered 300=200+W_f W_f=300-200=100 ~J gathered

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