Question

A $0.5kg$ block slides from the points $A$ 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 a coefficient of static and kinetic friction as $0.22$ and $0.2$ respectively. Find the total distance through which the block moves before it comes to rest completely. $(g=10m/s^2)$. the answer is X.3, find X ??

Answer 1

Initially, the block moves with constant speed $3m/s$ on a smooth surface.

Then for a distance $2.14m,$ it is subjected to friction provided by rough surface.

friction force $=\mu N=\mu mg=0.2\times 0.5\times 9.8=0.98N$

Retardation experienced by the block $=$ force/mass $0.98/0.5=1.96m/s^2$

speed after travelling $2.14m$ rough path:- $v^2=9-2\times 1.96\times 2.14$,

we get $v=0.78m/s$

after travelling 2.14 m rough path, the block is stopped by spring. Let the spring has compressed by a distance x.

Then the stopping force provided by the spring is $kx,$ where $k$ is force constant of the spring.

since the block is stopped, we get the distance x moved from$,x=\frac{u^2}{2\times a_r}$ ..............$\left(1\right)$

where $u$ is initial velocity and $a_r$ is the retardation provided by the spring$.$

Retardation is given by$,a_r=\frac{kx}{m}=\frac{2x}{0.5}=4x$

hence we rewrite $e{q}^n\left(1\right)$ to get $x:x=\frac{u^2}{\left(2\times 4x\right)}$

$0r,x^2=\frac{u^2}{8}=\frac{\left(0.78\times 0.78\right)}{8}=0.07605$

hence $x=0.28m=28cm$