When a body slides down from rest along a smooth inclined plane making an angle of 30 degrees with t

Q: Why is the distance 's' the same in both cases?

The problem statement explicitly says 'through the same distance'. This is the critical constraint that allows us to equate the two kinematic equations.

Q: Why do we use kinetic friction and not static friction?

The problem states that the body 'slides down'. Since there is relative motion between the block and the inclined plane, we must use the coefficient of kinetic friction.

Q: What is the physical significance of alpha being greater than 1?

Because the rough plane has friction, the net downward acceleration is smaller than on the smooth plane. A smaller acceleration means it will take a longer time to cover the same distance. Hence, the new time $\alpha T$ must be greater than $T$, which implies $\alpha > 1$.

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

Yes, you can. However, since the problem gives information about 'time' rather than 'final velocity', using Newton's laws and kinematic equations ($s = ut + \frac{1}{2}at^2$) is much more direct and avoids unnecessary intermediate variables.

Q: How did we get the general formula $\mu = \tan\theta (1 - 1/\alpha^2)$?

If you replace $30^\circ$ with $\theta$ in the derivation, the smooth acceleration is $g \sin\theta$ and the rough acceleration is $g(\sin\theta - \mu \cos\theta)$. Equating distances gives $\frac{1}{2}(g \sin\theta)T^2 = \frac{1}{2}g(\sin\theta - \mu \cos\theta)(\alpha T)^2$. Canceling terms leads directly to $\mu = \tan\theta (1 - 1/\alpha^2)$.

Question

When a body slides down from rest along a smooth inclined plane making an angle of $30^\circ$ with the horizontal, it takes time $T$. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time $\alpha T$, where $\alpha$ is a constant greater than 1. The coefficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}} \left( \frac{\alpha^2 - 1}{\alpha^2} \right)$, where $x$ is ......... .

Sigma Percentile · Accepted Answer

This problem is a beautiful exploration of kinematics combined with Newton's Laws of Motion. It asks us to compare the time taken by a block to slide down a smooth inclined plane versus a rough inclined plane. Let's break it down step-by-step.

### Analyzing the Setup
Imagine a block of mass $m$ resting on an inclined plane at an angle of $30^\circ$. Gravity pulls it straight down with a force $mg$. We can resolve this gravitational force into two perpendicular components: $mg \cos 30^\circ$ acting perpendicular to the plane (into the surface), and $mg \sin 30^\circ$ acting parallel to the plane (down the slope). The plane pushes back with a normal force $N$, which perfectly balances the perpendicular component, so $N = mg \cos 30^\circ$.

### The Smooth Plane Scenario
First, let's consider the case where the plane is perfectly smooth. Since there is no friction, the only force driving the block down the incline is the parallel component of gravity.

Using Newton's Second Law, we can write:
$$ma_1 = mg \sin 30^\circ$$

The mass $m$ cancels out, giving us the acceleration $a_1$:
$$a_1 = g \sin 30^\circ = \frac{g}{2}$$

The block starts from rest ($u = 0$) and travels a distance $s$ in time $T$. Using the second equation of motion, $s = ut + \frac{1}{2}at^2$, we get:
$$s = \frac{1}{2} a_1 T^2 = \frac{1}{2} \left(\frac{g}{2}ight) T^2 = \frac{g}{4} T^2$$

Let's hold onto this equation. It represents the distance covered in terms of the time $T$.

### The Rough Plane Scenario
Now, let's reset and look at the rough plane. As the block slides down, kinetic friction opposes its motion, acting upwards along the incline. This frictional force is given by $f_k = \mu N = \mu mg \cos 30^\circ$.

The net force pulling the block down is now reduced by this friction:
$$ma_2 = mg \sin 30^\circ - \mu mg \cos 30^\circ$$

Dividing by mass, we find the new acceleration $a_2$:
$$a_2 = g \sin 30^\circ - \mu g \cos 30^\circ = g \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} ight)$$

Because the acceleration is smaller, the block takes a longer time, $\alpha T$, to cover the exact same distance $s$. Plugging this into the distance formula:
$$s = \frac{1}{2} a_2 (\alpha T)^2 = \frac{1}{2} g \left( \frac{1 - \sqrt{3}\mu}{2} ight) \alpha^2 T^2 = \frac{g}{4} (1 - \sqrt{3}\mu) \alpha^2 T^2$$

### Equating and Solving
The problem states that the distance $s$ is the same in both scenarios. So, we can equate our two expressions for $s$:
$$\frac{g}{4} T^2 = \frac{g}{4} (1 - \sqrt{3}\mu) \alpha^2 T^2$$

Notice how beautifully the $\frac{g}{4}$ and $T^2$ terms cancel out on both sides, leaving us with a very simple relation:
$$1 = (1 - \sqrt{3}\mu) \alpha^2$$

Now, it's just simple algebra to isolate $\mu$. We divide by $\alpha^2$:
$$\frac{1}{\alpha^2} = 1 - \sqrt{3}\mu$$

Rearranging the terms to get $\sqrt{3}\mu$ on one side:
$$\sqrt{3}\mu = 1 - \frac{1}{\alpha^2} = \frac{\alpha^2 - 1}{\alpha^2}$$

Finally, dividing by $\sqrt{3}$ gives us the expression for the coefficient of kinetic friction:
$$\mu = \frac{1}{\sqrt{3}} \left( \frac{\alpha^2 - 1}{\alpha^2} ight)$$

### Final Calculation
The problem gave us the format of the friction coefficient as $\frac{1}{\sqrt{x}} \left( \frac{\alpha^2 - 1}{\alpha^2} ight)$. Comparing our derived expression with the given one, it is crystal clear that $x$ must be exactly equal to **3**.

**Pro Tip:** If the angle of inclination was a general angle $	heta$ instead of $30^\circ$, the formula generalizes to $\mu = 	an	heta \left( 1 - \frac{1}{\alpha^2} ight)$. Memorizing this can be a huge time-saver in competitive exams!

Enter Numerical Value:

Visualized Solution

$Free Body Diagram$

$Acceleration on Smooth Plane$

$Distance Covered (Smooth)$

$Introducing Friction$

$Acceleration on Rough Plane$

$Distance Covered (Rough)$

$Equating the Distances$

$Solving for Coefficient of Friction μ$

$Comparing with Given Expression$

$The Way Forward$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup Imagine a block of mass $m$ resting on an inclined plane at an angle of $3 0^{\circ}$

The Smooth Plane Scenario First, let's consider the case where the plane is perfectly smooth

The Rough Plane Scenario Now, let's reset and look at the rough plane

Equating and Solving The problem states that the distance $s$ is the same in both scenarios

Final Calculation The problem gave us the format of the friction coefficient as $\frac{1}{x} (\frac{α ^{2} - 1}{α ^{2}})$

Similar Questions

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A block of mass $m$ slides along a floor, while a force of magnitude $F$ is applied to it at an angle $θ$ as shown in figure. The coefficient of kinetic friction is $μ_{k}$ . Then, the block's acceleration $a$ is given by ( $g$ is acceleration due to gravity)

List-I

List-II

A block of mass $2 kg$ rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$ . The frictional force on the block is

Enter Numerical Value:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance Covered (Smooth)

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance Covered (Rough)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Coefficient of Friction μ

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

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