At time t=0, particle starts moving along the x-axis. If its kinetic energy increases uniformly with

Q: Why is power constant?

Power is the rate of change of kinetic energy ($P = \frac{dK}{dt}$). Since kinetic energy increases uniformly (linearly) with time ($K \propto t$), its derivative with respect to time is a constant.

Q: Can we solve this using equations of motion?

No, the standard equations of motion ($v = u + at$, etc.) are only valid for constant acceleration. Here, the force (and thus acceleration) is changing with time.

Q: Why did we use $P = Fv$?

The work-energy theorem states that the work done by the net force equals the change in kinetic energy. Differentiating both sides with respect to time gives Power = $\frac{dK}{dt}$. And mechanically, Power is also defined as the dot product of Force and velocity.

Q: What does 'increases uniformly' mean?

It means the quantity increases at a constant rate. Mathematically, it implies a linear relationship, so the quantity is directly proportional to time.

Q: Is the force increasing or decreasing?

Since $F \propto \frac{1}{\sqrt{t}}$, as time $t$ increases, the denominator increases, which means the net force acting on the particle is decreasing over time.

Question

At time $t=0$, particle starts moving along the x-axis. If its kinetic energy increases uniformly with time $t$, the net force acting on it must be proportional to

Options:

(A) $\sqrt{t}$

(B) constant

(C) $t$

(D) $1/\sqrt{t}$

Sigma Percentile · Accepted Answer

### The Core Concept: Decoding "Uniform Increase"

When a physics problem states that a quantity "increases uniformly," it is giving you a massive hint. It means the rate of increase is constant, which translates to a linear relationship.

In our case, the kinetic energy $K$ increases uniformly with time $t$. Mathematically, we can express this as $K \propto t$, or $K = ct$, where $c$ is a positive constant.

This simple linear equation is the key that unlocks the entire problem. From here, we can take two different paths to find the force: the Power perspective and the Kinematics route.

### The Power Perspective

Let's think about power. Power $P$ is defined as the rate at which work is done, which is also the rate of change of kinetic energy.

So, $P = \frac{dK}{dt}$. Since we know $K = ct$, taking the derivative with respect to time gives us $P = c$. This means the power delivered to the particle is **constant**.

Now, recall the mechanical definition of power: $P = Fv$. Since power is constant, the product of force and velocity must also be constant.

This implies that $F \propto \frac{1}{v}$. To find how force depends on time, we need to know how velocity depends on time.

We know $K = \frac{1}{2}mv^2$. Equating this to $ct$, we get $\frac{1}{2}mv^2 = ct$, which means $v^2 \propto t$, or $v \propto \sqrt{t}$.

Substituting this back into our force proportionality, we get $F \propto \frac{1}{\sqrt{t}}$.

### The Kinematics Route

Alternatively, we can use pure kinematics. We already established that $v \propto \sqrt{t}$. Let's write this as $v = k t^{1/2}$, where $k$ is another constant.

To find the force, we need the acceleration. Acceleration $a$ is the rate of change of velocity, $a = \frac{dv}{dt}$.

Differentiating our velocity expression with respect to time, we get $a = k \cdot \frac{1}{2} t^{-1/2}$.

This shows that $a \propto t^{-1/2}$, or $a \propto \frac{1}{\sqrt{t}}$.

According to Newton's Second Law, $F = ma$. Since mass is constant, the force is directly proportional to the acceleration.

Therefore, $F \propto \frac{1}{\sqrt{t}}$.

### Final Thoughts

Both methods beautifully converge to the same result. The net force acting on the particle is inversely proportional to the square root of time.

This is a classic JEE problem that tests your ability to seamlessly connect energy, power, and kinematics. Always remember to translate phrases like "increases uniformly" into precise mathematical statements!

Animated Solution for Physics - Work, Energy, and Power: At time $t = 0$ , particle starts moving along the x-axis. If its kinetic energy increases uniformly with time $t$ , the net force acting on it must be proportional to

Select Answer:

Visualized Solution

$Given Condition$

$K and v Relation$

$Power and Force$

$Force Proportionality$

$Alternate Method: Kinematics$

$Conclusion$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Core Concept

The Power Perspective

The Kinematics Route

Final Thoughts

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K and v Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power and Force

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alternate Method: Kinematics

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