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A block of mass m =2kg is placed in equilibrium on moving planck acceleration with acceleration a=4m/s if coefficient of friction between, is 0.2 the friction force acting on block id |
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Answer» Answer: 3.924 N Explanation: Given data in the question : MASS of the block, m = 2 kg ACCELERATION of the plank = 4 m/s² coefficient of friction between, μ = 0.2 It is ALSO given that the block is in equilibrium with the plank. Now, the frictional force is given as: f = μN where, N is the normal reaction N = mg g is the acceleration due to the gravity thus, N = 2 × 9.81 = 19.62 Newton hence, the friction force acting on the block is f = 0.2 × 19.62 = 3.924 N |
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