Two masses $m_1 = 5\ kg$ and $m_2 = 10\ kg$, 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 $m_2$ to stop the motion is $...... kg$
JEE MAIN 2018, Difficult
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Given $m_1=5 \ kg ; m_2=10 \ kg ; \mu=0.15$
For $m_1, m_1 g-T=m_1 a$
$\Rightarrow 50-T=5 \times a$
and, $T-0.15(m+10) g=(10+m) a$
For rest $a=0$
or, $50=0.15(m+10) 10$
$\Rightarrow 5=\frac{3}{20}(m+10)$
$\frac{100}{3}=m+10 \therefore m=23.3 \ kg ;$
For $m_1, m_1 g-T=m_1 a$
$\Rightarrow 50-T=5 \times a$
and, $T-0.15(m+10) g=(10+m) a$
For rest $a=0$
or, $50=0.15(m+10) 10$
$\Rightarrow 5=\frac{3}{20}(m+10)$
$\frac{100}{3}=m+10 \therefore m=23.3 \ kg ;$
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