Question

Two masses $m_1=5\text{kg}$ and $m_2=10\text{kg},$ connected through an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15 (${\mu }_k={\mu }_s=\mu )$ . The minimum weight $m$ that should be put on top of $m_2$ to stop the motion is:

• A. $18.3\text{kg}$
• B. $23.3\text{kg}$
• C. $43.3\text{kg}$
• D. $10.3\text{kg}$

Answer 1

The correct option is B $23.3\text{kg}$

Given : $m_1=5\text{kg};m_2=10\text{kg};\mu =0.15$


Applying the condition of equilibrium for both i.e ($m_2+m$) as a system & then block $m_1$ alone.

For mass $m_1$: $m_{1}g-T=m_{1}a=0$
(since block is not moving)
$\Rightarrow 50-T=0$
$\therefore T=50N--\left(1\right)$

For mass $m_2+m$:
$T-f_{max}=(10+m)a=0$

In limiting case,
$f_{max}=\mu N=\mu (m_2+m)g$
$\therefore f_{max}=0.15\times (10+m)g--\left(2\right)$

From Eq. (1), (2) and (4):
$50-(15+1.5m)=0$
$\therefore m=\frac{70}{3}=23.3\text{kg}$