A small block starts slipping down from a point B on an inclined plane AB, which is making an angle

Q: Why can't we use Newton's laws and kinematics?

You absolutely can! But it requires calculating acceleration for two different sections and matching final/initial velocities. The Work-Energy theorem does it in one single, elegant line.

Q: Why is the work done by the normal force zero?

The normal force is always perpendicular to the displacement of the block. Since $\cos(90^\circ) = 0$, the work done is zero.

Q: Does the mass of the block matter?

No! As you can see in the final equation, the mass $m$ cancels out completely. A feather and a truck would require the same coefficient of friction to stop.

Q: What if the entire plane was rough?

If the entire plane was rough, the work done by friction would be $-\mu mg \cos\theta \cdot AB$. The block would stop much earlier, or not even start moving if $\mu$ is large enough!

Q: Why is the work done by friction negative?

Friction always acts in the direction opposite to the relative motion. The angle between the friction force and displacement is $180^\circ$, and $\cos(180^\circ) = -1$.

Question

A small block starts slipping down from a point $B$ on an inclined plane $AB$, which is making an angle $	heta$ with the horizontal section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$, the coefficient of friction is given by $\mu = k 	an	heta$. Then, the value of $k$ is.........

Sigma Percentile · Accepted Answer

### Visualizing the Journey

Imagine you are standing at the top of a slide. The first part of the slide is perfectly smooth, but the second part is covered in rough sandpaper. This is exactly what our block experiences!

The block starts from rest at point $B$ on an inclined plane. It glides effortlessly down the smooth section $BC$, picking up speed. But as soon as it hits point $C$, it enters the rough section $CA$. The kinetic friction kicks in, slowing the block down until it comes to a complete halt exactly at the bottom, point $A$.

### The Magic of the Work-Energy Theorem

You might be tempted to use Newton's laws and kinematics here. You could calculate the acceleration on the smooth part, find the velocity at $C$, and then calculate the deceleration on the rough part to find where it stops. But there is a much more elegant way: **The Work-Energy Theorem**.

The Work-Energy Theorem states that the net work done on an object is equal to its change in kinetic energy:
$$W_{	ext{net}} = \Delta K$$

Since the block starts from rest ($K_i = 0$) and ends at rest ($K_f = 0$), the total change in kinetic energy is zero. Therefore, the net work done on the block must be exactly zero!

### Breaking Down the Work Done

Let's look at the forces doing work on our block:
1. **Gravity:** The component of gravity along the incline is $mg \sin	heta$. It pulls the block down the entire length of the incline, $AB$. So, the work done by gravity is positive: $W_{	ext{gravity}} = mg \sin	heta \cdot AB$.
2. **Friction:** Friction only acts on the rough section $CA$. It opposes the motion, so it does negative work. The frictional force is $f_k = \mu N = \mu mg \cos	heta$. The work done is: $W_{	ext{friction}} = -\mu mg \cos	heta \cdot AC$.
3. **Normal Force:** The normal force is perpendicular to the motion, so it does zero work.

### The Final Equation

We are given a crucial piece of geometric information: the smooth section $BC$ is twice as long as the rough section $AC$ ($BC = 2AC$). This means the total length of the incline $AB$ is:
$$AB = AC + BC = AC + 2AC = 3AC$$

Now, let's plug everything into our Work-Energy equation:
$$W_{	ext{gravity}} + W_{	ext{friction}} = 0$$
$$mg \sin	heta \cdot (3AC) - \mu mg \cos	heta \cdot (AC) = 0$$

Notice how beautifully the mass $m$, gravity $g$, and the length $AC$ cancel out from both sides! This tells us that the result is independent of the block's mass or the actual length of the incline.

Rearranging the terms, we get:
$$3 \sin	heta = \mu \cos	heta$$
$$\mu = 3 \frac{\sin	heta}{\cos	heta} = 3 	an	heta$$

The problem states that $\mu = k 	an	heta$. By comparing our result with this given expression, we can immediately see that:
**$k = 3$**

And there we have it! A seemingly complex two-part motion problem solved in just a few lines of elegant physics.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$W or k - E n er g y T h eor e m$

$I d e n t i f y in g W or k D o n e$

$R e l a t in g L e n g t h s$

$A ppl y in g t h e T h eor e m$

$S o l v in g f or μ$

$F in d in g k$

$T h e W a y F or w a r d$

The Sigma Insight: Work Done by Forces

Visualizing the Journey

The Magic of the Work-Energy Theorem

Breaking Down the Work Done

The Final Equation

Similar Questions

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A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $\frac{g}{2}$ upwards as shown in figure. Work done by normal reaction on block in time $t$ is

Comprehension Passage

Enter Numerical Value:

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; } Identifying Work Done

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

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

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

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

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