Question

A $1kg$ block collides with a horizontal weightless spring of force constant $2.75N/m$ as shown in figure. The block compresses the spring $4m$ from the free position. If the coefficient of kinetic friction between the block and the horizontal surface is $0.25$, find the speed(in $m/s)$ of the block at the instant of collision in $m/s$. $\left(g=10m/s^2\right)$

Answer 1

Answer: 8.00

WET
$\therefore W$ by friction $+W$ by Spring $=0-\frac{1}{2}m{V}^2$
$-\mu mgx-\frac{K}{2}{x}^2=-\frac{1}{2}m{V}^2$
$0.25\times 1\times \frac{10\times 4}{100}+\frac{2.75}{2}\times \frac{16}{1000}=\frac{1}{2}\times 1\times V^2$
$10+2.75\times 8=\frac{V^2}{2}$
$32=\frac{V^2}{2}$
$V=8m/s$