Question

The blocks of mass $m_1=1kg$ and $m_2=2kg$ at rest are connected by a spring of spring constant $k=2N/m$. The coeffcient of friction between blocks and horizontal surface is $\mu =\frac{1}{2}.$ Now the block of mass $m_1$ is imparted a velocity $u$ towards block of mass $m_2$ as shown, then the largest value of $u$ (in $m/s$) such that the block of mass $m_2$ never moves is
(Take $g=10m{s}^{-2}$)

Answer 1

The limiting frictional force on $m_2$ is $\mu m_{2}g=\frac{1}{2}\left(2\right)\times 10=10N$.
If the spring force exceeds this value, then $m_2$ will move.
Thus, maximum spring force is $10N$
$k{x}_{max}=10\Rightarrow x_{max}=\frac{10}{2}=5m(\because k=2N/m)$
The maximum compression in the spring is $5m$.
$m_1$ should stop after comperssing the spring by $5m$
Applying work energy theorem for the motion of $m_1$
$K_i+U_i+W_f=K_f+U_f\frac{1}{2}{m}_1{u}^2+0-\mu m_{1}g{x}_{max}=0+\frac{1}{2}\left(k\right)(x_{max}{)}^2\frac{1}{2}{m}_1{u}^2+0-\mu m_{1}g{x}_{max}=0+\frac{1}{2}(2\left)\right(5{)}^2{u}^2=100u=10m{s}^{-1}$