A spring block system is placed on a rough horizontal surface having coefficient of friciton `mu`. The spring is given initial elongation `3mumg//k` (where m=mass of block and k=spring constant) and the block is released from rest. For the subsequent motion, find

a. Initial acceleration of block
b. Maximum compression in spring
c. Maximum speed of the block

Explanation

a. From Newton's second law, we get
`ma=3mumg-mumgimpliesa=2mug`
b. From work energy principle
`W_(spring)+W_(spring)=0`
`1/2k(x_0^2-x_1^2)-mumg(x_1+x_0)=0`
`x_0-x_1=(2mumg)/(k)impliesx_1=(mumg)/(k)`
c. Speed will be maximum where net force is zero
`mumg=kx`
`implies x=(mumg)/(k)` (extension)
Now from work-energy principle, we have
`1/2k((mumg)/(k))^2(3^2-1^2)-(2mumg)/(k)(mumg)=1/2mv^2`
Solving we get, `v=2mugsqrt(m/k)`