A block of mass 2kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8N is applied on the block parallel to the floor, the force of friction between the block and floor is (take g=10ms−2)
A
2.8N
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B
8N
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C
2N
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D
zero
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Solution
The correct option is A2.8N fmax=μmg=0.4×2×10=8N Since the applied force is less than fmax so block will remain at rest. Force of friction will be equal to the applied force which is 2.8N.