A block starts moving up an inclined plane of inclination 30 degrees with an initial velocity of v0.

Q: Why does the direction of friction change between the upward and downward journeys?

Kinetic friction is a resistive force that always acts in the direction opposite to the relative motion between the surfaces. When the block moves up, friction points down. When the block slides down, friction points up.

Q: Why is the final velocity for the upward journey taken as zero?

The block is moving against gravity and friction, which causes it to decelerate. It will continue to move up until its velocity drops exactly to zero. That point marks the maximum distance $s$ it travels up the incline.

Q: Can we solve this problem using the Work-Energy Theorem instead of kinematics?

Absolutely! The total work done by friction over the entire round trip is $-2f_k s$. This must equal the total change in kinetic energy: $\frac{1}{2}m(\frac{v_0}{2})^2 - \frac{1}{2}mv_0^2$. You will arrive at the exact same result, and it's a fantastic way to verify your answer.

Q: Why did we ignore the mass of the block?

If you look closely at the acceleration equations, every force term ($mg \sin\theta$ and $\mu_k mg \cos\theta$) contains the mass $m$. When we write $F_{net} = ma$, the mass $m$ cancels out entirely. The motion is independent of the block's mass.

Q: What would happen if the static friction was very high?

If the coefficient of static friction $\mu_s$ was greater than or equal to $\tan 30^\circ$, the block would travel up, stop, and then just stay there! It wouldn't slide back down because static friction would be strong enough to balance the downward pull of gravity.

Question

A block starts moving up an inclined plane of inclination $30^\circ$ with an initial velocity of $v_0$. It comes back to its initial position with velocity $\frac{v_0}{2}$. The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{I}{1000}$, the nearest integer to $I$ is .......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the bottom of a rough inclined plane, and you give a block a sharp push upwards with an initial velocity $v_0$. The block shoots up, but it's fighting a tough battle. Not only is gravity pulling it back down, but the rough surface is also scraping against it, generating kinetic friction that opposes its motion.

Eventually, the block runs out of steam, stops for a microscopic fraction of a second at its highest point, and then surrenders to gravity, sliding back down to where it started. However, when it returns, its velocity is only $\frac{v_0}{2}$. It has lost a significant amount of kinetic energy to friction. Our mission is to find the coefficient of kinetic friction, $\mu_k$, that caused this energy loss.

### The Upward Journey: Fighting a Double Headwind

Let's break the motion into two distinct phases. During the upward journey, the block experiences a severe retardation. Why? Because both the component of gravity along the incline ($mg \sin 30^\circ$) and the force of kinetic friction ($f_k = \mu_k mg \cos 30^\circ$) are pointing downwards, directly opposing the velocity.

We can write the magnitude of this retardation, let's call it $a_1$, as:
$$a_1 = g \sin 30^\circ + \mu_k g \cos 30^\circ$$

Using the third equation of kinematics, $v^2 - u^2 = 2as$, we can find the distance $s$ it travels before coming to rest ($v=0$):
$$0 - v_0^2 = -2a_1 s \implies s = \frac{v_0^2}{2a_1}$$

### The Downward Journey: A Slight Tailwind

Now, the block starts its descent from rest. Gravity is still pulling it down the incline ($mg \sin 30^\circ$), but friction, being the ultimate contrarian, flips its direction! It now points *up* the incline, trying to slow the block's fall.

The net acceleration downwards, $a_2$, is the difference between these forces:
$$a_2 = g \sin 30^\circ - \mu_k g \cos 30^\circ$$

The block travels the exact same distance $s$ back to the start, reaching a final velocity of $\frac{v_0}{2}$. Applying the kinematics equation again:
$$\left(\frac{v_0}{2}\right)^2 - 0 = 2a_2 s \implies s = \frac{v_0^2}{8a_2}$$

### The Grand Equating

Since the distance $s$ is identical for both trips, we can set our two expressions equal to each other:
$$\frac{v_0^2}{2a_1} = \frac{v_0^2}{8a_2}$$

Notice how beautifully the $v_0^2$ terms cancel out! This tells us that the result is independent of how hard you initially pushed the block. Cross-multiplying gives us a remarkably simple relationship between the accelerations:
$$4a_2 = a_1$$

### Final Calculation

Now, we substitute our expressions for $a_1$ and $a_2$ into this relation. We can also divide out the acceleration due to gravity, $g$, from every term:
$$4(\sin 30^\circ - \mu_k \cos 30^\circ) = \sin 30^\circ + \mu_k \cos 30^\circ$$

Plugging in the standard trigonometric values ($\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$):
$$4\left(\frac{1}{2} - \mu_k \frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \mu_k \frac{\sqrt{3}}{2}$$

Expanding the left side:
$$2 - 2\sqrt{3}\mu_k = \frac{1}{2} + \mu_k \frac{\sqrt{3}}{2}$$

Bringing the constants to one side and the $\mu_k$ terms to the other:
$$2 - \frac{1}{2} = \mu_k \frac{\sqrt{3}}{2} + 2\sqrt{3}\mu_k$$
$$\frac{3}{2} = \frac{5\sqrt{3}}{2}\mu_k$$

Solving for $\mu_k$:
$$\mu_k = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$$

Using the approximation $\sqrt{3} \approx 1.732$, we get:
$$\mu_k = \frac{1.732}{5} = 0.3464$$

The problem asks for the nearest integer $I$ where $\mu_k \approx \frac{I}{1000}$. Multiplying our result by $1000$, we find that **$I = 346$**.

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Motion}

\text{Forces During Upward Motion}

\text{Distance Travelled Upward}

\text{Forces During Downward Motion}

\text{Distance Travelled Downward}

\text{Equating the Distances}

\text{Substituting Accelerations}

\text{Solving for } \mu_k

\text{Calculating } \mu_k

\text{Finding the Integer } I

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Upward Journey

The Downward Journey

The Grand Equating

Final Calculation

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