A force of 100 N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the surface and the block is μ = 13. The frictional force acting on the block is

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15 N downwards

25 N upwards

20 N downwards

30 N upwards

answer is C.

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Detailed Solution

N' = 100 cos 300 = 100.32 = 503NNet driving force = (100 × 12-30) = 20 N (upward)flimiting13(503) = 50 N∴ friction = 20 N (downwards)

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