Question

Two masses $m_1=5\text{kg}\text{and}{m}_2=10\text{kg}$ connected by an inextensible string over a frictionless pulley are moving as shown in the figure. The coefficient of static friction of horizontal surface is 0.15. The minimum mass $m$ that should be put on top of $m_2$ to stop the motion is -

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

Answer 1

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


$m_1=5\text{kg},m_2=10\text{kg}$
As the system is at rest, $T=m_{1}g=50\text{N}$
$T-0.15(m+10)g=0$
$50=0.15(m+10)\times 10$
$5=\frac{3}{20}(m+10)$
$\frac{100}{3}=m+10$
$m=23.3\text{kg}$