A particle of mass m is initially at rest at the origin. It is subjected to a force and starts movin

Q: Wait, why did we use the chain rule to differentiate kinetic energy?

Because kinetic energy $K$ is a function of velocity $v$, but we need its derivative with respect to time $t$. So, $\frac{d}{dt}(v^2) = \frac{d(v^2)}{dv} \cdot \frac{dv}{dt} = 2v \frac{dv}{dt}$.

Q: Why is a constant force in one dimension always conservative?

A conservative force is one where the work done in a closed loop is zero, or equivalently, the work done is path-independent. For a constant 1D force $F$, the work done from $x_1$ to $x_2$ is simply $F(x_2 - x_1)$, which only depends on the endpoints, making it conservative.

Q: Could we have used the Work-Energy theorem directly instead of setting up a differential equation?

Yes! The Work-Energy theorem states $W = \Delta K$. Power is the rate of doing work, $P = \frac{dW}{dt} = \frac{dK}{dt} = \gamma t$. Since $P = Fv$, we have $Fv = \gamma t$. However, to find how $v$ or $F$ individually depend on time, you still need to express $F$ as $m \frac{dv}{dt}$ and integrate, which leads to the exact same differential equation!

Q: Why did we take the positive square root when solving for velocity?

The particle starts from rest and is subjected to a force that increases its kinetic energy (since $\gamma t$ is positive for $t>0$). This means the speed must be increasing. We can define the direction of this initial motion as the positive X-axis, so $v$ is positive.

Q: What does the problem mean by '$\gamma$ is a positive constant of appropriate dimensions'?

In physics, equations must be dimensionally consistent. Since $\frac{dK}{dt}$ has dimensions of Power (Energy/Time, or $ML^2T^{-3}$), the term $\gamma t$ must also have dimensions of Power. Therefore, the dimensions of $\gamma$ must be Power/Time, or $ML^2T^{-4}$. The problem states this so you don't worry about unit mismatches.

Question

A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the $X$-axis. Its kinetic energy $K$ changes with time as $\frac{dK}{dt} = \gamma t$, where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

Options:

(A) The force applied on the particle is constant

(B) The speed of the particle is proportional to time

(C) The distance of the particle from the origin increases linearly with time

(D) The force is conservative

Sigma Percentile · Accepted Answer

### The Energy-Time Connection

Imagine a particle of mass $m$ sitting peacefully at the origin. Suddenly, a mysterious force awakens, pushing it along the $X$-axis. We aren't given the force directly; instead, we are given a fascinating clue about its energy: the rate at which its kinetic energy $K$ changes with time is directly proportional to time itself.

Mathematically, this is expressed as:
$$ \frac{dK}{dt} = \gamma t $$

This single equation holds the DNA of the particle's entire journey. To unlock it, we must translate kinetic energy into the language of kinematics. We know that kinetic energy is given by $K = \frac{1}{2}mv^2$.

Let's differentiate this with respect to time using the chain rule. The mass $m$ is constant, and the derivative of $v^2$ is $2v \frac{dv}{dt}$.

This gives us:
$$ \frac{dK}{dt} = \frac{1}{2}m \left(2v \frac{dv}{dt}\right) = mv \frac{dv}{dt} $$

### Unveiling the Kinematics

Now, we can bridge the given information with our kinematic derivative. By equating the two expressions for $\frac{dK}{dt}$, we form a powerful differential equation:
$$ mv \frac{dv}{dt} = \gamma t $$

To solve this, we separate the variables, grouping velocity terms on one side and time terms on the other:
$$ v \, dv = \frac{\gamma}{m} t \, dt $$

Since the particle starts from rest at the origin, its initial velocity is $0$ at $t = 0$. We integrate both sides from the start of the motion to an arbitrary time $t$ and velocity $v$:
$$ \int_0^v v \, dv = \int_0^t \frac{\gamma}{m} t \, dt $$

Evaluating these integrals is straightforward. The integral of $v$ is $\frac{v^2}{2}$, and the integral of $t$ is $\frac{t^2}{2}$.

This yields:
$$ \frac{v^2}{2} = \frac{\gamma}{m} \frac{t^2}{2} $$

The factors of $2$ cancel out beautifully. Taking the square root of both sides, we find the velocity as a function of time:
$$ v = \sqrt{\frac{\gamma}{m}} t $$

This result is profound! It tells us that the speed of the particle is directly proportional to time ($v \propto t$). Therefore, **statement (b) is absolutely true**.

### The Nature of the Force

With the velocity in hand, we can easily uncover the force driving this motion. According to Newton's Second Law, force is mass times acceleration ($F = ma$), and acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$).

Let's differentiate our velocity expression:
$$ a = \frac{d}{dt} \left( \sqrt{\frac{\gamma}{m}} t \right) = \sqrt{\frac{\gamma}{m}} $$

The acceleration is a constant! Now, we multiply by mass to find the force:
$$ F = m \sqrt{\frac{\gamma}{m}} = \sqrt{\gamma m} $$

Since $\gamma$ and $m$ are both constants, the force $F$ is also a constant. This confirms that **statement (a) is true**.

Furthermore, in one-dimensional motion, any constant force is inherently conservative. The work done by such a force depends only on the initial and final positions, not the path taken. Thus, **statement (d) is also true**.

### The Final Piece: Displacement

Finally, let's investigate the particle's displacement. We know that velocity is the rate of change of position ($v = \frac{ds}{dt}$).

Substituting our velocity expression, we get:
$$ \frac{ds}{dt} = \sqrt{\frac{\gamma}{m}} t $$

To find the displacement $s$, we integrate with respect to time from $t = 0$ to $t$:
$$ s = \int_0^t \sqrt{\frac{\gamma}{m}} t \, dt = \sqrt{\frac{\gamma}{m}} \frac{t^2}{2} $$

This equation reveals that the distance $s$ from the origin increases quadratically with time ($s \propto t^2$), not linearly. Therefore, **statement (c) is false**.

In conclusion, the correct statements that describe this particle's elegant motion are (a), (b), and (d).

Select Answer:

Visualized Solution

$\frac{d K}{d t} = γ t$

$\frac{d}{d t} (\frac{1}{2} m v^{2})$

$m v \frac{d v}{d t} = γ t$

$v d v = \frac{γ}{m} t d t$

$\int_{0}^{v} v d v = \int_{0}^{t} \frac{γ}{m} t d t$

$\frac{v ^{2}}{2} = \frac{γ}{m} \frac{t ^{2}}{2}$

$v = \frac{γ}{m} t$

$a = \frac{d v}{d t}$

$F = ma$

$F = constant$

$v = \frac{d s}{d t}$

$s = \int_{0}^{t} \frac{γ}{m} t d t$

$s \propto t^{2}$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Energy-Time Connection

Unveiling the Kinematics

The Nature of the Force

The Final Piece

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdK​=γt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtd​(21​mv2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mvdtdv​=γt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vdv=mγ​tdt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫0v​vdv=∫0t​mγ​tdt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2v2​=mγ​2t2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=mγ​​t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=dtdv​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=ma

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=dtds​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s=∫0t​mγ​​tdt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s∝t2