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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

Initially, both the blocks are at rest on horizontal surface as shown in the figure. Find the minimum value of force F in N so that sliding starts between the blocks g = 10 m s - 2

Correct answer

80

Step-by-step solution

F - 10 μ g - 20 μ g = 10 a .......(i) μ 10 g = 10 a ..................(ii) From (i) and (ii) F = 40 μ g F = 80   N

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