Question

A block of mass $m = 1$ kg. moving on a horizontal surface with speed ${u_t} = 2m{s^{ - 1}}$, enters a rough patch ranging from $x = 0.10m$ to $x = 2.01m$. The retarding force F, on the block in this range is inversely proportional to x over this range.
${F_r} = \dfrac{{ - K}}{x}$ for $0.1 < x < 2.01m$
$ = 0$ for $x < 0.1m$ and $x < 2.01m$. Where $K = 0.5J$. What is the final kinetic energy and speed ${v_f}$ of the block as it crosses this patch?

Answer 1

Hint: As the force is retarding in nature, so its velocity decreases. By mathematical calculation, final kinetic energy can be found.
1. Final Kinetic energy \[ = \dfrac{1}{2}m{v_f}^2\]
2. Initial Kinetic energy $ = \dfrac{1}{2}m{v_i}^2$
3. Force, $F = ma = mv\dfrac{{dv}}{{dx}}$
Where ${v_f}$ is the final velocity
${v_i}$ is the initial velocity
m is mass
x is distance

Complete step by step answer:
In the question, it is given that the retarding force acting on block is
${F_r} = \dfrac{{ - K}}{x}$ For $0.1 < x < 2.01m$
$ = 0$ For $x < 0.1$ and $x < 2.01m$
So, where K is constant x is the range.
Now, as we know that
$
  F = ma \\
   = m\dfrac{{dv}}{{dt}} = m\dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dx}} \\
   = m\dfrac{{dx}}{{dt}}\dfrac{{dv}}{{dx}} = mv\dfrac{{dv}}{{dx}} \\
 $
So, $F = mv\dfrac{{dv}}{{dx}}$
Where, m is mass, v is velocity, t is time and x is distance.
Now, equating both forces, we have
$mv\dfrac{{dv}}{{dx}} = \dfrac{{ - K}}{x}$ For $0.1 < x < 2.01$
$\left( {\dfrac{{ - m}}{K}} \right)vdv = \dfrac{1}{x}dx$
Integrating both sides as for $x = 0.1 \to {v_i}$ and $x = 2.01 \to {v_f}$
So, \[\left( {\dfrac{{ - m}}{L}} \right)\int\limits_{{v_i}}^{{v_f}} {vdv = \int\limits_{0.1}^{2.01} {\dfrac{1}{x}dx} } \]
\[
   \Rightarrow \left( {\dfrac{{ - m}}{K}\left[ {\dfrac{{{v^2}}}{2}} \right]_{{v_i}}^{{v_f}} = \left[ {\ln x} \right]_{0.1}^{2.01}} \right) \\
   \Rightarrow \dfrac{{ - m}}{K}\left[ {\dfrac{{{v_f}^2}}{2} - \dfrac{{{v_i}^2}}{2}} \right] = \left[ {\ln {\text{ 2}}{\text{.01 - ln 0}}{\text{.1}}} \right] \\
   \Rightarrow \dfrac{{m{v_f}^2}}{2} - \dfrac{{m{v_i}^2}}{2} = - K\left[ {\ln {\text{ }}\left( {2.01} \right) - \ln \left( {0.1} \right)} \right] \\
   \Rightarrow {K_f} = - K\left[ {\ln {\text{ 2}}{\text{.01}} - \ln 0.1} \right] + \dfrac{{m{v_i}^2}}{2} \\
   = \left( { - 0.5} \right)\left[ {0.69 - \left( { - 2.30} \right)} \right] + \dfrac{{1 \times {2^2}}}{2} \\
  {K_f} = \left( { - 0.5} \right)\left( {2.99} \right) + 2 \\
  {K_f} = 0.505{\text{ Joule}} \\
  {{\text{K}}_f} \simeq 0.50{\text{ Joule}} \\ \]

Where ${K_f}$ is final kinetic energy $ = \dfrac{1}{2}m{v_f}^2$ and ${K_i}$ is initial kinetic energy$ = \dfrac{1}{2}m{v_i}^2$.

Note:
Basic formulas of integration are used remember that
$\int {xdx = \dfrac{{{x^2}}}{2}} $ and $\int {\dfrac{1}{x}dx = \ln x} $.