Question

Two masses $m_1=5\text{kg},m_2=10\text{kg}$ connected by an inextensible massless string over a frictionless pulley, are released from rest as shown in the figure. The coefficient of static friction between the horizontal surface and block $m_2$ is $0.15$. The minimum mass $m$ in $\text{kg}$ that should be put on top of $m_2$ to stop the motion is:

• A. $23.33\text{kg}$
• B. $43.3\text{kg}$
• C. $10.3\text{kg}$
• D. $18.3\text{kg}$

Answer 1

The correct option is A $23.33\text{kg}$

For minimum mass $m\text{kg}$ to be placed on $m_2$, we have to consider that friction will act at its maximum value i.e $f_s={\mu }_{s}N$


Applying equilibrium condition for blocks:

For block $m_1$ in vertical direction,
$T=m_{1}g$
$\therefore T=5\times 10=50\text{N}$

For system of blocks $(m+m_2)$,
In vertical direction, $N=(m+m_2)g$

Along horizontal direction:
$T=f_s$
$\Rightarrow 50={\mu }_{s}N={\mu }_s(m+m_2)g$
$\Rightarrow 50=0.15\times (m+10)\times 10=1.5m+15$
$\therefore m=\frac{35}{1.5}=23.33\text{kg}$