For the figure shown below, the coefficient of friction between 2kg and 4kg block is 0.1 and between 4kg and ground is 0.2. The graph of friction force acting on 4kg block by the ground as a function of time is
A
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B
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C
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D
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Solution
The correct option is A
f1max=μ1N1=0.1×2g=2N f2max=μ2N2=0.2×6g=12N
For 0<F<2N: f1=F i.e Tension will be zero and f2=f1 ∴f2=0→2N (rightward)
For 2N<F<4N: F=f1max+T i.e Tension is 0<T<2N. f2=f1max−T=2−T ∴f2=2→0N (rightward)
For 4N<F<16N: F=f1max+T i.e tension T will be 2N<T<14N f2=T−f1max=T−2 ∴f2=0→12N (leftward)
For F>16N: Lower block will start moving since T>f1max+f2max. Hence friction will be kinetic (constant) and act leftward.