Two masses m₁ = 5 kg and m₂ = 10 kg , connected by an inextensible string over a frictionless pu

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4-2.Friction

hard

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 ........ $kg$

A$23.3$

B$43.3$

C$10.3 $

D$18.3 $

(JEE MAIN-2018)

Solution

$\begin{array}{l}
Given\,{m_1} = 5kg;{m_2} = 10kg;\mu = 0.15\\
For\,{m_1},\,{m_1}g – T = {m_1}a\\
\Rightarrow 50 – T = 5 \times a\\
and,\,T – 0.15\left( {m + 10} \right)g = \left( {10 + m} \right)a\\
For\,rest\,a = 0\\
or,\,50 = 0.15\left( {m + 10} \right)10\\
\Rightarrow \,5 = \frac{3}{{20}}\left( {m + 10} \right)\\
\frac{{100}}{3} = m + 10\,\,\,\therefore \,m = 23.3\,kg\,;
\end{array}$

Standard 11

Physics

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normal

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easy

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easy

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medium

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