A particle moves in a straight line with retardation proportional to its displacement. Its loss of k

Q: Why do we use $v \frac{dv}{dx}$ instead of $\frac{dv}{dt}$ for acceleration?

We use $v \frac{dv}{dx}$ because the problem gives us a relationship between acceleration and displacement ($x$), not time ($t$). Using the chain rule allows us to eliminate time and directly integrate with respect to position.

Q: Why is there a negative sign in $a = -kx$?

The problem states that the particle experiences 'retardation', which means the acceleration is acting in the opposite direction of the velocity to slow the particle down. The negative sign mathematically represents this opposing direction.

Q: What does the constant $k$ represent physically?

The constant $k$ is a proportionality constant that dictates how rapidly the retardation increases with displacement. If this were a spring system, $k$ would be related to the spring constant divided by the mass ($k = \frac{K_{spring}}{m}$).

Q: Can we solve this using the Work-Energy Theorem directly?

Yes, absolutely! The force is $F = ma = -mkx$. The work done is the integral of force over displacement: $W = \int_0^x -mkx dx = -\frac{1}{2}mkx^2$. Since Work = Change in KE, the loss in KE is $\frac{1}{2}mkx^2$, which is proportional to $x^2$.

Q: Why is the loss in KE proportional to $x^2$ and not just $x$?

Because the retarding force itself is not constant; it increases linearly with $x$. When you calculate the work done (which equals the loss in KE), you are finding the area under a linear Force vs. Displacement graph, which forms a triangle. The area of that triangle depends on $x \cdot x$, resulting in an $x^2$ dependence.

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

Options:

(A) $x^2$

(B) $e^x$

(C) $x$

(D) $\log_e x$

Sigma Percentile · Accepted Answer

This problem is a classic example of how kinematics seamlessly blends into the concepts of work and energy. Let's break down the physics and the math behind it.

### Analyzing the Setup
We are told that a particle is moving in a straight line and experiencing a **retardation** that is proportional to its displacement. Retardation is simply negative acceleration. If we let the displacement be $x$, we can write the acceleration $a$ as:
$$a = -kx$$
Here, $k$ is a positive proportionality constant. The negative sign is crucial—it tells us that the force (and thus the acceleration) is acting in the opposite direction of the displacement, actively slowing the particle down.

### The Master Equation
To find the loss in kinetic energy, we need a relationship between velocity and displacement. The standard definition of acceleration is $a = \frac{dv}{dt}$. However, this involves time, which we don't care about here. Instead, we use the chain rule to express acceleration in terms of position:
$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\frac{dv}{dx}$$

Substituting this into our retardation equation gives us a beautiful differential equation:
$$v\frac{dv}{dx} = -kx$$

### Integration and Execution
Now, we separate the variables to prepare for integration. We bring all the $v$ terms to one side and the $x$ terms to the other:
$$v \, dv = -kx \, dx$$

Let's assume the particle starts at $x = 0$ with an initial velocity $v_0$, and at some arbitrary displacement $x$, its velocity is $v$. We integrate both sides with these limits:
$$\int_{v_0}^v v \, dv = -\int_0^x kx \, dx$$

Evaluating the integrals, we get:
$$\left[ \frac{v^2}{2} ight]_{v_0}^v = -k \left[ \frac{x^2}{2} ight]_0^x$$
$$\frac{v^2}{2} - \frac{v_0^2}{2} = -\frac{kx^2}{2}$$

### Revealing the Kinetic Energy
We are almost there! The expression looks suspiciously close to kinetic energy. Let's multiply the entire equation by the mass of the particle, $m$:
$$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mkx^2$$

The left side of the equation is exactly the final kinetic energy minus the initial kinetic energy, which is the change in kinetic energy ($\Delta K$):
$$\Delta K = -\frac{1}{2}mkx^2$$

The negative sign explicitly shows that the kinetic energy is decreasing. The **loss** in kinetic energy is the magnitude of this change:
$$	ext{Loss in KE} = \frac{1}{2}mkx^2$$

Since $\frac{1}{2}mk$ is a constant, we can conclude that the loss in kinetic energy is directly proportional to the square of the displacement:
$$	ext{Loss in KE} \propto x^2$$

### The Ninja Method: Work-Energy Theorem
Wait, could we have done this faster? Absolutely! The Work-Energy Theorem states that the net work done on a particle equals its change in kinetic energy ($W = \Delta K$).

The force acting on the particle is $F = ma = m(-kx) = -mkx$.

The work done by this force as the particle moves from $0$ to $x$ is:
$$W = \int_0^x F \, dx = \int_0^x (-mkx) \, dx = -mk \left[ \frac{x^2}{2} ight]_0^x = -\frac{1}{2}mkx^2$$

Since $W = \Delta K$, the change in kinetic energy is $-\frac{1}{2}mkx^2$. The loss in kinetic energy is simply the magnitude of this change, which is $\frac{1}{2}mkx^2$. Thus, the loss is proportional to $x^2$. Boom! Two lines of math and we arrive at the exact same destination.

Animated Solution for Physics - Work, Energy, and Power: 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

Select Answer:

Visualized Solution

Visualizing the Motion

Setting up the Differential Equation

Separating Variables

Integrating Both Sides

Evaluating the Integrals

Connecting to Kinetic Energy

Final Conclusion

The Sigma Insight: Work Done by Forces

Analyzing the Setup We are told that a particle is moving in a straight line and experiencing a retardation that is proportional to its displacement

The Master Equation To find the loss in kinetic energy, we need a relationship between velocity and displacement

Integration and Execution Now, we separate the variables to prepare for integration

Revealing the Kinetic Energy We are almost there! The expression looks suspiciously close to kinetic energy

The Ninja Method

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