The arrangement shown in the diagram is moving in space with an acceleration `a=4(hati+hatj)m//s^(2)`. An ideal spring of natural length `l_(0)` having spring constant `k=50 N//m`, is connected to block A. Blocks `A` and `B` are connected by an ideal string passing through a frictionless pulley. Mass of each block `A` and `B` is equal to `m=2kg`.
(`a`) Calculate the minimum value of coefficient of friction so that spring remains in its natural length. (`b`) If `mu` is reduced to `(9//35)^(th)` of the original value, calculate the maximum extension of the spring.

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`