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A block of mass 4kg is at rest on a rough horizontal surface. Coefficient of static friction μs=0.4 and Coefficient of kinetic friction μk=0.3. A horizontal force F = 4t N Where t is time in second, is applied on the block. Then the frictional force acting on the block at t = 5s is

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answer is 3.

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Detailed Solution

Maximum static frictional force (fs)max=0.4 x 4 x 10 N =16 NAt t = 5s, applied force F = 4 x 5N = 20 NSince, F>(fs)max , the block will sideThe frictional force acting on the block =fk=0.3 x 4 x 10 N=12 N

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