The blocks ov 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=(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

`1-m//s`

Explanation

For the block of mass `m_(2)`, not to move, the maximum compression in the spring `x_(0)` should be such that
`kx_(0)=mu m_(2)g …..(1)`
Applying work energy theorem to block of mass `m_(1)` we get
`(1)/(2)m_(1)u^(2)=(1)/(2)(mu^(2)m_(2)^(2)g^(2))/(K)+(mu^(2)+m_(1)m_(2)g^(2))/(K)` putting the appropriate value we get `u=10m//s`.