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Question

# Two masses m1=5 kg, m2=10 kg connected by an inextensible massless string over a frictionless pulley, are released from rest as shown in the figure. The coefficient of static friction between the horizontal surface and block m2 is 0.15. The minimum mass m in kg that should be put on top of m2 to stop the motion is

A
23.33 kg
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B
43.3 kg
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C
10.3 kg
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D
18.3 kg
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Solution

## The correct option is A 23.33 kgfor minimum mass m kg placed on m2, we have to consider friction will act at its maximum value i.e fs=μsN Applying equilibrium condition for blocks: For block m1 in vertical direction, T=m1g ∴T=5×10=50 N For system of blocks (m+m2), In vertical direction, N=(m+m2)g Along horizontal direction: T=fs ⇒50=μsN=μs(m+m2)g ⇒50=(0.15×(m+10))×10=1.5m+15 ∴m=351.5=23.33 kg

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