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A block of mass 4 kg is kept on a rough horizontal surface. The coefficient of static friction is 0.8. If a force of 19 N is applied on the block parallel to the floor, then the force of friction between the block and the floor is (a) \( 32 N \quad \) (b) \( 18 N \) (c) \( 19 N \) |
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Answer» The correct option is (C) 19 N Given, Mass of block (m)=4 kg Coefficient of static friction (μ)=0.8 Applied force (F) = 19 N Now, limiting friction (fs)max=μsN=μsmg = 0.8×4×10 = 32 N Since, applied force is less than the limiting friction. So, force of friction, f = 19 N |
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