Question

The blocks of mass $m_1=1kg$ and $m_2=2kg$ are connected by an ideal spring, rest on a rough horizontal surface. The spring is unstressed. The spring constant of spring is $K=2N/m$. The coefficient of friction between blocks and horizontal surface is $\mu =\frac{1}{2}.$ Now the left block is imparted a velocity $u$ towards right as shown. The largest value of $u$ (in $m/s$) such that the block of mass $m_2$ never moves is (Take $g=10m/s^2)$:

• A. 10
• B. 20
• C. 5
• D. 2

Answer 1

The correct option is A 10

The blocks of mass $m_1=1$ kg

For the block of mass $m_2,$ not to move, the maximum
compression in the spring $x_o$ should be such that

$k{x}_0=\mu m_{2}g....\left(1\right)$

Applying work energy theorem to block of mass $m_1$ we get

$\frac{1}{2}{m}_1{u}^2=\frac{1}{2}k{x}_0^2+\mu m_{1}g{x}_0....\left(2\right)$

From equation (1) and (2) we get

$\frac{1}{2}{m}_1{u}^2=\frac{1}{2}\frac{{\mu }^2{m}_2^2{g}^2}{K}+\frac{{\mu }^2{m}_1{m}_2{g}^2}{K}$

putting the appropriate value we get u$=10m/s.$