Question

A $1kg$ body slides from the point $A$ (see figure), along a horizontal track with an initial speed of $6m{s}^{-1}$, towards a weightless horizontal spring of length $2m$ and force constant $4N{m}^{-1}$. The part $AB$ of the track is frictionless and the part $BC$ has the coefficients of static and dynamic friction as $0.22$ and $0.2$, respectively. Find the total distance through which the block moves, before it comes to rest completely . ($g=10m{s}^{-2}$), If $AB=4m$ and $BD=4.28m$

Answer 1

Refer above image 1 and 2.

At $B$

vel.of block $=6m/s$

(because surface of frictionless)

vel. at $D$ $a=\frac{f_k}{m}=2m/s^2$

$V_D^2-V_B^2=2as$

$\Rightarrow V_D^2-6^2=-2\times 2\times 4.28$

$V_D^2=18.88$

$V_D=4.35m/s$

Let us assume spring compression is $x$

Refer image 3.

From work energy theorem

(from $D$ to $E$)

$W_s+W_{f_k}=\Delta KE$

$-\frac{1}{2}k{x}^2+2x=0-\frac{1}{2}\times 1\times v_V^2$

$\Rightarrow \frac{kx}{2}+2c=\frac{1}{2}\times 18.88$

$x=1.72$

At $E$

Refer image .4

here $kx>f_{max}$

Hence $f_k^{\prime }$ acts $f_2=2N$

Applying work-energy theorem (from E-F)

total work-done

$=\Delta KE$

$\Rightarrow W{D}_s+W{D}_{f_k}=\Delta KE$

$\Rightarrow \frac{1}{2}k{x}^2-2(x+y)=0$

$\Rightarrow \frac{1}{2}\times 4\times {1.72}^2-2(1.72+y)$

$=0$

$\Rightarrow 5.92-3.44=2y$

$y=1.24$

Total Distance $=AB+BD+x+x+y$

$=4+4.28+1.72+1.72+1.24$

$=12.96m$