Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible… | JEE Main 2018

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Friction Basics

Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is:

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Answer: (c)

To stop the motion, tension must equal the hanging weight: $T = m_1 g = 50\text{ N}$ .
Static friction on $(m_2 + m)$ must hold this: $\mu(m_2 + m)g \ge 50$ .
$0.15(10 + m)(10) \ge 50 \Rightarrow 10 + m \ge 33.3 \Rightarrow m \ge 23.3\text{ kg}$
The calculation gives $23.3\text{ kg}$ ; the official key is (c) $27.3\text{ kg}$ .

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