A small block of mass `m=1kg` is attached with one end of the spring of force constant `K=110Nm^-1`. Other end of the spring is fixed to a rough plane having coefficient of friction `mu=0.2`. The spring is kept in its natural length by an inextensible thread ties between its ends as shown in figure. If the thread is burnt, calculate the elongation of the spring when the block attains static equilibrium position.

Answer

`64/11cm`

Explanation

When the string is burnt, the block starts moving down. Due to friction and spring force, the block slows down and stops. Applying work-energy theorem between points 1 and 2,
`W_(t otal)=Deltak`(for block)
`W_(gravity)+W_(f riction)+W_(spri ng)=K_(fi nal)-K_(i nitial)`

`mgsinthetax_1-mumgcos thetax_1-1/2kx_1^2=0-0`
`1xx10xx3/5x_1-0.2xx1xx10xx4/5x_1-(1)/(2)110xxx_1^2=0`
`implies x_1=0.08m` or `8cm`
As at position 2, we can find that there is net force on the block in upward direction, i.e.
`kx_1gtmumgcostheta+mgsintheta`
So the block will not be permanently at rest at position 2, and it will start moving up, let it stop at position (3). Now consider position 2 and 3, consider the (block+spring) a system.
Now work done by external force = change in total energy of the system.
Let at this position, there be a stretch `x_2` in the spring.

`W_(external)=DeltaE`
`W_(f riction)=DeltaE_(gravitational)+DeltaE_(spri ng)`
`-mumgcostheta(x_1-x_2)=mgy+[1/2kx_2^2-1/2kx_1^2]`
`-mumgcostheta(x_1-x_2)=mg(x_1-x_2)sin theta`
`-1/2k(x_1-x_2)(x_1+x_2)`
`-mumgcostheta=mgsintheta-1/2k(x_1+x_2)`
`1/2k(x_1+x_2)=mg(sintheta+mucostheta)`
`(x_1+x_2)=(2mg(sintheta+mucostheta))/(k)`
`x_2=(2mg(sintheta+mucostheta))/(k)-x_1`
`=(2xx1xx10(3/5+0.2xx4/5))/(110)-0.08=64/11cm`