Question

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement \[x\] is proportional to.
A. \[{x^2}\]
B. \[{e^x}\]
C. \[x\]
D. \[{\log _e}x\]

Answer 1

Hint:Use the formula for acceleration and velocity of the particle. Derive relation for acceleration of particle using the given information. Substitute the value in the form of change in velocity in this relation. Also substitute the value of change in time in this equation. Integrate this equation between the limits of velocity and displacement and derive the relation between the change or loss in kinetic energy of the particle and displacement of particle.

Formulae used:
The acceleration \[a\] of an object is
\[a = \dfrac{{dv}}{{dt}}\] …… (1)
Here, \[dv\] is the change in velocity of the particle and \[dt\] is the time interval.
The velocity \[v\] of an object is
\[v = \dfrac{{dx}}{{dt}}\] …… (2)
Here, \[dx\] is a change in displacement of the particle and \[dt\] is the time interval.

Complete step by step answer:
We have given that the retardation \[ - a\] of the particle is proportional to displacement \[x\] of the particle.
\[ - a \propto x\]
\[ \Rightarrow a = - kx\]
Substitute \[\dfrac{{dv}}{{dt}}\] for \[a\] in the above equation.
\[ \Rightarrow \dfrac{{dv}}{{dt}} = - kx\] …… (3)
Rearrange equation (2) for \[dt\].
\[dt = \dfrac{{dx}}{v}\]
Substitute for \[dt\] in equation (3).
\[ \Rightarrow \dfrac{{dv}}{{\dfrac{{dx}}{v}}} = - kx\]
\[ \Rightarrow vdv = - kxdx\]

Let us integrate both sides of the above equation.
\[ \Rightarrow \int_{{v_1}}^{{v_2}} {vdv} = \int_0^x { - kxdx} \]
\[ \Rightarrow \left[ {\dfrac{{{v^2}}}{2}} \right]_{{v_1}}^{{v_2}} = - k\left[ {\dfrac{{{x^2}}}{2}} \right]_0^x\]
\[ \Rightarrow \dfrac{{v_2^2}}{2} - \dfrac{{v_1^2}}{2} = - k\dfrac{{{x^2}}}{2} - 0\]
\[ \Rightarrow \dfrac{{v_2^2}}{2} - \dfrac{{v_1^2}}{2} = - k\dfrac{{{x^2}}}{2}\]

Let us multiply on both sides of the above equation by mass \[m\].
\[ \Rightarrow m\left( {\dfrac{{v_2^2}}{2} - \dfrac{{v_1^2}}{2}} \right) = - \dfrac{{km{x^2}}}{2}\]
We know that the left side of the above equation represents change in kinetic energy \[\Delta K\] of the particle.
\[ \Rightarrow \Delta K = - \dfrac{{km{x^2}}}{2}\]
From the above equation, we can write
\[ \therefore - \Delta K \propto {x^2}\]
Therefore, the loss in kinetic energy is proportional to \[{x^2}\].

Hence, the correct option is A.

Note:The students may think that how we have determined the change in kinetic energy of the particle is the loss in kinetic energy of the particle and not the increase in kinetic energy of the particle. But in the final relation between the change in kinetic energy and displacement, we can see that there is a negative sign before the change in kinetic energy of the particle which indicates that the kinetic energy of the particles is decreases and hence, there is loss in kinetic energy of the particle.