Question

A $0.5Kg$ block slides from the point $A$ (see figure) on a horizontal track with an initial speed of $3m/s$ towards a weightless horizontal spring of length $1m$ and force constant $2N/m$. The part $AB$ of the track is frictionless and the part $BC$ has the coefficients of static and kinetic friction as $0.22$ and $0.2$ respectively. if the distance $AB$ and $BD$ are $2m$ and $2.14m$ respectively, find the total distance through which the block moves before it comes to rest completely. (Take $g=10m/s^2$)

Answer 1

As the block travels from A to B, there is no retardation. But for surface B to D, there is retardation which is calculated as

$f={\mu }_{k}mg=0.2\times 0.5\times 10=1.0N$

Retardation acceleration$=\frac{f}{m}=\frac{1.0}{0.5}=2.0m/s^2$

Now, $u=3m/s,a=2.0m/s^2,s=2.14m$

$\Rightarrow v^2-u^2=2as$

$\Rightarrow v^2=-2\left(2.2\right)\left(2.14\right)+\left(9\right)=9-8.56$

$=\sqrt{0}.44=0.66m/s$

$\Rightarrow$ The initial velocity at spring $=u=0.66m/s$

$v=0m/s$

and as the spring provides retardation acceleration; $a_r=\frac{-kx}{m}$, where

$x$:- Distance traveled by box to compress spring.

$K$:- Spring constant.

$a_r=\frac{-\left(2\right)x}{0.5}=-4xm/s^2$

Now, for spring $u=0.66m/s$

$v=0m/s$

$a=-4xm/s^2$

Now, $v^2=u^2+2as$

$\Rightarrow 0=(0.66{)}^2+(2\left)\right(-4x\left)\right(x)$

$\Rightarrow x^2=\frac{(0.66{)}^2}{8}=0.23m$

Now, Total distance covered$=2m+2.14+0.23=4.37m$