The blocks ov mass `m_(1)kg` 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=(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))`

Answer

10

Explanation

For the block of mass not to move, the maximum compression in the spring should be such that
`kx_(0) = mu m_2 g" " …(i)`
Applying work energy theorem to block of mass `m_1` we get

`1/2 m_1u^2 = 1/2 kx_(0)^2 + mu m_1 gx_(0)" "…(ii)`
From equations (i) and (ii) we get `1/2 m_1 u^2 = 1/2 (mu^2 m_2^2 g^2)/(K) + (mu^2 m_1m_2g^2)/(K)`
Putting the appropriate value we get `u = 10 m//s`.