Question

A block of mass $0.24\, kg$ is attached to a spring of force constant $4\, N / m$. 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 $0.18\, m$ 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 $g =10\, m / s ^{2}$ )

Answer 1

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