Question

A block of mass $0.18\, kg$ is attached to a spring of force constant $2 \,N / m$. The coefficient of friction between the block and the floor is $0.1$. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of $0.06\, m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v=\frac{N}{10}$. Then $N$ is

Answer 1

Decrease in mechanical energy $=$ Work done against friction
$\therefore \frac{1}{2} m v^{2}-\frac{1}{2} k x^{2}=\mu m g x$
or $v=\sqrt{\frac{2 \mu m g x+k x^{2}}{m}}$
Substituting the values, we get
$v=0.4 \,m / s =\left(\frac{4}{10}\right) \,m / s$