A block A of mass 5 Kg is placed on a rough table. The coefficient of static & kinetic friction between surfaces of block and table be 0.4 & 0.3 respectively. If the force F exerted on the block is 10 N (g = 10 m/s2) the force of friction between block and table is.
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Solution
Given:- mass=5 kg Coefficient of static friction=μs=0.4 Then limiting force of friction= fs=μsR =μsmg =0.4 X 5 X 9.8 =19.6 N Since:-F=10 N As applied force F < fs, therefore,block does not move further, as static friction is self adjusting.