The arrangement shown in figure is at rest. An ideal spring of natural length `l_0` having spring constant `k=220Nm^-1`, is connected to block A. Blocks A and B are connected by an ideal string passing through a friction less pulley. The mass of each block A and B is equal to `m=2kg` when the spring was in natural length, the whole system is given an acceleration `veca` as shown. If coefficient of friction of both surfaces is `mu=0.25`, then find the maximum extension in `10(m)` of the spring. `(g=10ms^-2)`.

Explanation

Case I : `a=(4hati+4hatj)m//s^(2)`, `a_(x)=4m//s^(2)`, `a_(y)=4m//s^(2)`
`FBD` of blocks `A` and `B`
`N_(A)=m(g+a_(y))=28N`
`T=ma_(x)+muN_(A)=8+28mu`…..(`1`)
`N_(B)=ma_(x)=8N`
`T+muN_(B)=m(g+a_(y))`
`T+8mu=28`…..(`2`)
From Eqs.(`1`) and (`2`), `mu_(min)=5//9`
Case II : Let extension of spring `=x`, `mu'=(5//9)xx(9//35)=1//7`
From work energy theorem :
`mgx-mu'N_(A)x-mu'N_(B)x-ma_(x)x+ma_(y)x=1//2kx^(2)`
`20x-(1//7)x xx (28x)-(1//7)x xx (8x)=25x^(2)`
`x=0.6m`