Question

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

Answer 1

$m_1$= 5 Kg and $m_2=$10 kg

As the system is at rest, T= 50N

$T-0.15(m+10)g=(10+m)a$

a=0, for rest

$50=0.15(m+10)\times 10$

$5=\frac{3}{20}(m+10)$

$\frac{100}{3}$=m+10

m=23.3 kg