Question

A block of mass $0.24kg$ is attached to a spring of force constant $4N/m$. The coefficient of friction between the block and the floor is $0.2$. 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 $0.18m$ and comes to rest for the first time. The initial velocity of the block in $m/s$ is $v={\left(\frac{N}{100}\right)}^{\frac{1}{2}}$ then, $N$ is __________. (Take $g=10m/s^2$ )

Answer 1

Answer: 126

Decrease in mechanical energy
$=$ Work done against friction
$\therefore \frac{1}{2}m{v}^2-\frac{1}{2}k{x}^2=\mu mgx$
$\therefore v=\sqrt{\frac{2\mu mgx+k{x}^2}{m}}$
$\therefore v=\sqrt{\frac{2\times 0.2\times 0.24\times 10\times (0.18)+4(0.18{)}^2}{0.24}}$
$\therefore v=\sqrt{\frac{63}{50}}$
$\therefore v=\sqrt{\frac{126}{100}}m/s$
$\Rightarrow N=126$