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. An impulse is given to the block as shown in the figure. The block slides a distance of $0.06m$ 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

Answer: 4

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