Question

Two masses $m_1=5kg$ and $m_2=10kg$ connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of the horizontal surface is $0.15$. Find the minimum weight $m$ that should be put on top of $m_2$ to stop the motion.
($T$ is tension in the string and take $g=10m/s^2$)

• A. $18.3kg$
• B. $23.3kg$
• C. $43.3kg$
• D. $10.3kg$

Answer 1

The correct option is B $23.3kg$

Given: $m_1=5kg,m_2=10kg,\mu =0.15$

For $m_1$:
$\begin{array}{}{m}_{1}g-T=m_{1}a& \text{(1)}\end{array}$
$\Rightarrow 50-T=5\times a$
For $m_2$:
$T-\mu (m_2+m)g=(m_2+m)a$
$\begin{array}{}T-0.15(10+m)g=(10+m)a& \text{(2)}\end{array}$

When the motion stops,
$a=0$,
From $\left(1\right)$, $\Rightarrow T=50$
Substituting the value of $T$ in $\left(2\right)$,
$50=0.15(m+10)10$
$\Rightarrow 5=\frac{3}{20}(m+10)$
$\Rightarrow \frac{100}{3}=m+10$
$\therefore m=23.3kg$