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Step-by-Step Solution
Step 1: Understand the Given Condition
A particle moves along a straight line with a retardation (negative acceleration) proportional to its displacement. Mathematically, we write:
$ a = -kx $
where $k$ is a positive constant and $x$ is the displacement from some reference point.
Step 2: Express Acceleration in Terms of Velocity and Displacement
We use the relation for acceleration in one-dimensional motion:
$ a = \frac{dv}{dt} = v \frac{dv}{dx}. $
Substituting $ a = -kx $ into this gives:
$ v \frac{dv}{dx} = -kx.
$
Step 3: Integrate Both Sides
Rewrite the equation and integrate:
$ \int_{v_1}^{v_2} v \, dv = -k \int_{0}^{x} x \, dx.
$
Computing the integrals yields:
$ \left[ \frac{v^2}{2} \right]_{v_1}^{v_2} = -k \left[ \frac{x^2}{2} \right]_{0}^{x}.
$
So,
$ \frac{v_2^2 - v_1^2}{2} = - \frac{k x^2}{2}.
$
Step 4: Relate to Kinetic Energy
The kinetic energy $K$ is given by:
$ K = \frac{1}{2} m v^2.
$
Thus, the change in kinetic energy for velocities changing from $v_1$ to $v_2$ is:
$ \Delta K = \frac{1}{2} m \left(v_2^2 - v_1^2\right).
$
From our earlier result, $v_2^2 - v_1^2 = - k x^2.$ So,
$ \Delta K = \frac{1}{2} m \left(- k x^2\right) = - \frac{1}{2} m k x^2.
$
Step 5: infer the Proportionality
The negative sign indicates the kinetic energy is decreasing with increasing $x$. Hence, the loss in kinetic energy (magnitude) is proportional to $x^2$. In other words,
$ \text{Loss in } K \propto x^2.
Answer
The loss of kinetic energy for any displacement $x$ is proportional to $x^2$.
