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  4. Two masses m 1=10 kg and m 2=20 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/f8887363c171918b432f5add2ab37ed7-.png /> The coefficient of friction of horizontal surface is 0.30. The minimum weight m that should be put on top of m2 to stop the motion is kg

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Q. Two masses $m _{1}=10\, kg$ and $m _{2}=20\, kg$, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure.
image
The coefficient of friction of horizontal surface is $0.30$. The minimum weight $m$ that should be put on top of $m_{2}$ to stop the motion is___________ $kg$

Laws of Motion

Solution:

image
At equilibrium, $F = T$
$\mu\left( m _{2}+ m \right) g = m _{1} g$
$\therefore \mu(20+ m ) g =10\, g$
$\therefore 20+ m =\frac{10}{0.30}$
$\therefore 20+ m =\frac{100}{3}$
$\therefore m =\frac{40}{3}=13.33\, kg$