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A block of mass 2 kg is lying on a floor. The coefficient of kinetic friction is 0.4 If a force of 2 5 N is applied onthe block as shown in the figure, then the magnitude of frictional force will be(a) 7.5 N(c) 2.5 N(UPSEAT 2002(b) 10 N(d) 5 N |
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Answer» given that , 》mass of block = 2kg 》force applied = 2.5 》coefficient of static friction, µs= 0.4 ☆Maximum static frictional force, fmax = µs× m × g = 0.4 × 2 × 9.8 => fmax= 7.84 N Since, the applied force is less than the maximum static frictional force, the frictional force on the block is equal to the applied force = 2.5 N. ☆it is due to the fact that static friction is a self adjusting force. |
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