Question

In the given figure, $m_1=200g,$$m_2=600g,$${\mu }_1=0.3,$${\mu }_2=0.8$$andK=200N/m.$The minimum force that must be applied on $m_1$ to just displace $m_2$ is (in N). (Take g = 10 m/ $s^2$)

Answer 1

Ans: 3 N
$K{x}_0=f_{12}$, $K{x}_0={\mu }_2{m}_{2}g$.
By work energy theorem for $m_1$ and spring $F{x}_0-{\mu }_1{m}_{1}g{x}_0-\frac{1}{2}K{x}_0^2=0$ $F={\mu }_1{m}_{1}g+\frac{K{x}_0}{2}=({\mu }_1{m}_1+\frac{{\mu }_2{m}_2}{2})g=3N$