Question

A block of mass $0.18kg$ is attached to a spring of force-constant $2N/m$. The coefficient of friction between the block and the floor is $0.1$. Initially the block is at rest and the spring is unstretched. The block slides to a distance of $0.06m$ and stops for the first time when an impulse is given to the block as shown in the figure. Then the initial velocity of the block (in $m/s$) is

Answer 1

Let speed $v$ after impulse
Using Work energy theorem
$\therefore$ Loss in KE = gain in PE + Work done against friction

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2+\mu mgx$
$\frac{1}{2}\left(0.18\right)v^2=\frac{1}{2}\left(2\right)(0.06{)}^2+(0.1\left)\right(0.18\left)\right(10\left)\right(0.06)$
$0.09v^2=0.0036+0.0108$
$v^2=0.16$
$v=0.4m/s$