Question:
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}}, $ the work done against friction is
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}}, $ the work done against friction is
Updated On: Aug 1, 2022
- 200 J
- 100 J
- zero
- 1000 J
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The Correct Option is B
Solution and Explanation
Net work done in sliding a body up to a height h on inclined plane = Work done against gravitational force + Work done against frictional force $ \Rightarrow $ $ W={{W}_{g}}+{{W}_{g}} $ ?(i) but $ W=300\,J $ $ {{W}_{g}}=mgh=2\times 10\times 10=200\,J $ Putting in E (i) we get $ 300=200+{{W}_{f}} $ $ {{W}_{f}}=300-200=100\,\text{J} $
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Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.