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 un-stretched. 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 = N/10. Then N is

Answer 1

$\frac{1}{2}m{V}^2=\mu mgx+\frac{1}{2}k{x}^2$

$\frac{1}{2}\left(0.18\right)V^2=\left(0.1\right)\left(0.18\right)\left(10\right)\left(0.06\right)+\frac{1}{2}2(36\times {10}^{-4})$

$0.09V^2=108\times {10}^{-4}+36\times {10}^{-4}$

$0.09V^2={10}^{-4}\left[144\right]$

$0.3V={10}^{-2}\left(12\right)$

$V=0.4m{s}^{-1}=\frac{N}{10}$

$\Rightarrow N=4$