The system is released from rest and the block of mass 1kg is found to have a speed of 0.3m/s after it has descended through a distance of 1m. The value of coefficient of kinetic friction is
A
0.24
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B
0.36
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C
0.12
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D
0.48
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Solution
The correct option is C0.12
As 1kg block is lowered by 1m, pulley attached to it also gets lowered by 1m. This releases 2m string length which is provided by displacement of 4kg block by 2m rightwards. (x)4kg=2(x)1kg ⇒(v)4kg=2(v)1kg ∴(v)4kg=0.6m/s Speed of 4kg block will be 0.6m/s after descending through h=1m.
Applying Work Energy Theorem on system of blocks, Wext=ΔKE...(i) where Wext=Wf+Wgravity Wf=−μNd=−μ(4×10×2)=−80μJ Wgravity=+mgh=+1×10×1=10J Substituting in Eq.(i), Wfriction+Wgravity=KEf−KEi ⇒−80μ+10=[12×4×(0.6)2+12×1×(0.3)2]−(0+0) ⇒−80μ+10=1825+9200 ⇒−80μ=153200−10 ⇒−80μ=−1847200 ∴μ=0.115≃0.12 Coefficient of kinetic friction between the block and table is approximately 0.12.