A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The co

Q: Why is the change in kinetic energy zero?

The particle is released from rest at point $P$ (initial velocity is zero) and comes to rest at point $R$ (final velocity is zero). Therefore, both initial and final kinetic energies are zero, making the change $\Delta K = 0$.

Q: Why is the normal force different on the incline compared to the horizontal surface?

On a horizontal surface, gravity acts entirely perpendicular to the surface, so the normal force balances the full weight ($mg$). On an incline, gravity has two components. The normal force only balances the perpendicular component, which is $mg \cos\theta$.

Q: Can we solve this using Newton's laws and kinematics instead of energy?

Yes, absolutely! You could find the acceleration on the incline, calculate the velocity at the bottom, and then use that as the initial velocity for the horizontal deceleration phase. However, the Work-Energy Theorem is much faster and less prone to algebraic errors.

Q: Why did we use $h/\sin 30^\circ$ for the length of the incline?

If you look at the right-angled triangle formed by the incline, the height $h$ is the opposite side to the $30^\circ$ angle, and the length of the incline $L$ is the hypotenuse. From trigonometry, $\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{L}$, which rearranges to $L = \frac{h}{\sin 30^\circ}$.

Q: What if the particle didn't stop at $R$ but kept moving?

If the particle had a final velocity at $R$, then the final kinetic energy would not be zero. The Work-Energy equation would then be $W_{\text{friction}} = \frac{1}{2}mv_f^2 - mgh$. You would need more information to solve the problem in that case.

Question

A point particle of mass $m$, moves along the uniformly rough track $PQR$ as shown in the figure. The coefficient of friction between the particle and the rough track equals $\mu$. The particle is released, from rest, from the point $P$ and it comes to rest at a point $R$. The energies, lost by the ball, over the parts, $PQ$ and $QR$, of the track, are equal to each other, and no energy is lost when particle changes direction from $PQ$ to $QR$. The values of the coefficient of friction $\mu$ and the distance $x(= QR)$, are respectively close to

Options:

(A) 0.2 and 6.5 m

(B) 0.2 and 3.5 m

(C) 0.29 and 3.5 m

(D) 0.29 and 6.5 m

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Tracks

Imagine a particle perched at the top of an inclined plane, ready to embark on a journey. It starts from rest at point $P$, slides down the rough incline $PQ$, and then glides along a rough horizontal surface $QR$ until it finally exhausts all its energy and comes to a halt at $R$.

Our mission is to uncover two mysteries: the coefficient of friction $\mu$ (which is uniform across the entire track) and the distance $x$ it travels on the horizontal surface. The problem gifts us a beautiful constraint: the energy lost to friction on the incline is exactly equal to the energy lost on the horizontal surface.

### The Work-Energy Connection

To solve this elegantly, we turn to the **Work-Energy Theorem**. Since the particle starts from rest and ends at rest, its overall change in kinetic energy is zero ($\Delta K = 0$).

This implies that the total mechanical energy lost by the particle is simply its initial gravitational potential energy. At a height $h = 2	ext{ m}$, this energy is $mgh$. Where did this energy go? It was entirely consumed by the work done against friction along the path $P 	o Q 	o R$.

### Equating the Energy Losses

Let's break the journey into two parts. First, the inclined plane $PQ$. Using trigonometry, the length of the incline is $L = \frac{h}{\sin 30^\circ} = \frac{2}{1/2} = 4	ext{ m}$. As the particle slides down, the normal force is $mg \cos 30^\circ$, making the frictional force $f_1 = \mu mg \cos 30^\circ$. The energy lost here is the work done by friction:

$$W_1 = f_1 L = \mu mg \cos 30^\circ 	imes 4$$

Next, on the horizontal surface $QR$, the normal force is simply $mg$, so the frictional force is $f_2 = \mu mg$. The energy lost over distance $x$ is:

$$W_2 = f_2 x = \mu mg x$$

We are told that $W_1 = W_2$. Equating them yields a magical cancellation:

$$\mu mg \cos 30^\circ 	imes 4 = \mu mg x$$

$$x = 4 \cos 30^\circ = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.46	ext{ m}$$

### The Final Piece of the Puzzle

Now, we need to find $\mu$. We know the total energy lost is $W_1 + W_2$, which must equal the initial potential energy $mgh$. Since $W_1 = W_2$, we can write:

$$2 W_2 = mgh$$

Substituting our expression for $W_2$:

$$2 (\mu mg x) = mgh$$

$$2 \mu x = h$$

Plugging in the values $x = 2\sqrt{3}$ and $h = 2$:

$$2 \mu (2\sqrt{3}) = 2$$

$$\mu = \frac{1}{2\sqrt{3}} \approx \frac{1}{3.464} \approx 0.288$$

Rounding to two decimal places, we get $\mu \approx 0.29$ and $x \approx 3.5	ext{ m}$. This perfectly matches option (c). By leveraging the Work-Energy Theorem, we bypassed complex kinematic equations and arrived at the solution with pure elegance!

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Visualized Solution

$Visualizing the Setup$

$Work-Energy Theorem$

$Energy Lost on Incline P Q$

$Energy Lost on Horizontal QR$

$Equating the Energy Losses$

$Total Energy Conservation$

$Calculating μ$

$Final Conclusion$

The Sigma Insight: Work Done by Forces

The Setup

The Work-Energy Connection

Equating the Energy Losses

The Final Piece of the Puzzle

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Lost on Incline PQ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Lost on Horizontal QR

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

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

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

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