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Question

# A block of mass 2 kg is placed on the floor. The coefficient of ststic friction is 0.4. A force F of 3 N is applied on the block as shown in figure. The force of friction between the block and the floor is (Take g=10ms−2)

A
3 N
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B
8 N
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C
4 N
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D
6 N
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Solution

## The correct option is C 3 NGiven,mass of block, m=2kgacceleration due to gravity, g=10m/s2Coefficient of friction, μ=0.4External applied force on the block, F=3N Limiting friction value, in this case, can be given as:Fmax=μmgFmax=0.4×2×10Fmax=8NBut as we know the friction is a self-adjusting force when the applied external force used to be less or equal to the limiting friction value.And, here F=3N<Fmax=8NHence, the friction force experienced by the block, in this case, will be, f=F=3N [Applied force]So, the correct option is (A)

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