Assertion A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops

Q: Why is the final energy in Case II not zero?

While the kinetic energy becomes zero when the block stops, it has gained height. Therefore, it possesses gravitational potential energy ($mgh$) at its highest point.

Q: Doesn't the work done by friction depend on the distance traveled?

Yes, it does! The block travels a shorter distance on the incline because both gravity and friction oppose its motion. This results in less work done by friction, which perfectly matches our energy calculation showing a smaller decrease in mechanical energy.

Q: Why doesn't the coefficient of friction change when we tilt the surface?

The coefficient of friction ($\mu$) is a measure of how two specific surfaces interact at a microscopic level. It depends on the materials (like wood on wood, or rubber on concrete), not on how hard they are pressed together or the angle of the surface.

Q: But the frictional force changes on an incline, right?

Exactly! The *frictional force* ($f = \mu N$) changes because the normal force ($N = mg \cos\theta$) decreases on an incline. However, the *coefficient* of friction ($\mu$) itself remains constant.

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

Yes, you can! You could find the acceleration in both cases, calculate the stopping distance, and then find the work done by friction ($f \cdot d$). However, the energy method is much faster and more elegant for this specific question.

Question

**Assertion** A block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^{\circ}$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation.
**Reason** The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

Options:

(A) (a) If Assertion is true, Reason is true; Reason is the correct explanation for Assertion

(B) (b) If Assertion is true, Reason is true; Reason is not a correct explanation for Assertion

(C) (c) If Assertion is true; Reason is false

(D) (d) If Assertion is false; Reason is true

Sigma Percentile · Accepted Answer

### The Tale of Two Surfaces: A Journey of Energy and Friction

Have you ever wondered why it's harder to push a heavy box up a ramp than across a flat floor? It seems intuitive, right? Gravity is fighting against you. But what happens to the *energy* of an object when it slides and eventually stops? This classic JEE problem takes us on a fascinating journey comparing a block sliding on a flat surface versus an inclined plane, testing our deep understanding of the Work-Energy Theorem and the true nature of friction.

### Analyzing the Setup

Imagine a block of mass $m$ sliding across a rough horizontal floor with an initial velocity $v$. We know from experience that it will eventually grind to a halt. Why? Because kinetic friction is doing negative work, draining the block's kinetic energy and converting it into heat.

Now, imagine taking that exact same block, with the exact same initial velocity $v$, and launching it up a ramp tilted at $30^{\circ}$. Again, it stops. But here is the million-dollar question: in which scenario did the block lose more mechanical energy?

### The Master Equation: Mechanical Energy

To answer this, we need to look at the mechanical energy ($E$), which is the sum of kinetic energy ($K$) and potential energy ($U$). 
$$E = K + U$$

The decrease in mechanical energy ($\Delta E$) is simply the initial energy minus the final energy. According to the Work-Energy Theorem, this decrease is exactly equal to the magnitude of the work done by non-conservative forces—in this case, friction.

### Case I: The Horizontal Slide

Let's break down the first scenario. The block starts with a kinetic energy of $\frac{1}{2}mv^2$. Since it's moving on a flat surface, its height doesn't change, so we can set its initial and final gravitational potential energy to zero.

When the block finally stops, its velocity is zero, meaning its final kinetic energy is also zero. 
Therefore, the total decrease in mechanical energy is:
$$\Delta E_1 = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$$

All of its initial kinetic energy was eaten up by friction.

### Case II: The Uphill Battle

Now, let's look at the inclined plane. The block starts with the same initial kinetic energy, $\frac{1}{2}mv^2$. However, as it slides up the ramp, it gains height. Let's call the maximum height it reaches $h$.

When it stops, its kinetic energy is zero, but it now possesses gravitational potential energy equal to $mgh$. 
So, what is the decrease in mechanical energy this time?
$$\Delta E_2 = 	ext{Initial Energy} - 	ext{Final Energy}$$
$$\Delta E_2 = \frac{1}{2}mv^2 - mgh$$

### The Grand Comparison

Now we compare the two energy losses. 
In Case I, the loss was $\frac{1}{2}mv^2$. 
In Case II, the loss is $\frac{1}{2}mv^2$ *minus* a positive quantity ($mgh$).

Mathematically, it is crystal clear:
$$\Delta E_2 < \Delta E_1$$

The decrease in mechanical energy is indeed smaller in the second situation! Why does this happen physically? On the incline, gravity is helping friction slow the block down. Because gravity is doing negative work, the block travels a much shorter distance before stopping. A shorter distance means friction has less opportunity to do work, resulting in a smaller loss of mechanical energy. Thus, the **Assertion is absolutely true**.

### The Nature of Friction

Finally, let's examine the Reason statement. It claims that the coefficient of friction ($\mu$) decreases as the angle of inclination increases.

This is a classic trap! The coefficient of friction is a fundamental property of the two materials in contact (like wood on concrete, or steel on ice). It is a measure of their microscopic roughness and molecular bonding. It does **not** care whether the surface is flat, tilted, or upside down.

What *does* change on an incline is the normal force ($N = mg \cos	heta$), which in turn changes the actual frictional *force* ($f = \mu N$). But the coefficient $\mu$ itself remains stubbornly constant. Therefore, the **Reason is false**.

By mastering the distinction between the frictional force and the coefficient of friction, and by trusting the infallible logic of energy conservation, we can confidently conquer problems like this!

Select Answer:

Visualized Solution

Visualizing the Two Cases

Mechanical Energy

Case I: Horizontal Surface

Case II: Inclined Surface

Comparing $Δ E_{1} and Δ E_{2}$

Analyzing the Reason

Conclusion on Reason

The Sigma Insight: Conservation of Mechanical Energy

The Tale of Two Surfaces

Analyzing the Setup

The Master Equation

Case I

Case II

The Grand Comparison

The Nature of Friction

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