A body of mass m is launched up on a rough inclined plane making an angle of 30 degrees with the hor

Q: Why can we use $s = \frac{1}{2} a_A T_A^2$ for the ascent when the initial velocity isn't zero?

Great question! While the block starts with an initial velocity $u$ at the bottom, it comes to rest at the top ($v=0$). Because the acceleration is constant, the motion is perfectly symmetric in time if we reverse it. So, the time it takes to go from $u$ to $0$ with deceleration $a_A$ is exactly the same as the time it would take to go from $0$ to $u$ with acceleration $a_A$. Therefore, we can mathematically treat the ascent distance as if it started from rest at the top.

Q: Why does the direction of friction change?

Kinetic friction is a reactive force that always opposes the relative sliding motion between two surfaces. When the block moves up, friction points down the incline. When the block slides down, friction points up the incline to oppose the descent.

Q: Why is the time of descent longer than the time of ascent?

During ascent, both gravity and friction work together to slow the block down, resulting in a large deceleration. During descent, gravity pulls the block down, but friction fights against gravity, resulting in a smaller net acceleration. A smaller acceleration means it takes more time to cover the same distance from rest.

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

Yes, you can use the Work-Energy Theorem to find the velocities and distances, but since the problem explicitly gives a relationship between the *times* of ascent and descent, using kinematics (equations of motion) is the most direct and elegant approach.

Q: What would happen if the incline was perfectly smooth?

If the incline was perfectly smooth, the coefficient of friction $\mu$ would be zero. In that case, the acceleration during ascent ($g \sin\theta$) would be exactly equal to the acceleration during descent ($g \sin\theta$). Consequently, the time of ascent would be exactly equal to the time of descent.

Question

A body of mass $m$ is launched up on a rough inclined plane making an angle of $30^\circ$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$. If the time of ascent is half of the time of descent. The value of $x$ is .......

Sigma Percentile · Accepted Answer

### The Setup: A Journey Up and Down

Imagine a block launched up a rough inclined plane. It travels a certain distance, comes to a momentary halt at its highest point, and then slides back down to where it started. The distance covered during the upward journey (ascent) is exactly equal to the distance covered during the downward journey (descent).

However, the times taken for these two halves of the journey are not the same. The problem states that the time of ascent is exactly half the time of descent. Why does this happen? The answer lies in the asymmetric nature of friction.

### The Physics of Ascent and Descent

Let's break down the forces acting on the block. When the block is moving **up** the incline, gravity pulls it downwards along the plane with a force of $mg \sin	heta$. Friction, which always opposes relative motion, also acts downwards along the plane with a force of $\mu mg \cos	heta$. Because both forces are acting in the same direction (down the incline), they work together to decelerate the block rapidly. The net acceleration during ascent is:

$$a_A = g \sin	heta + \mu g \cos	heta$$

Now, consider the block sliding **down** the incline. Gravity still pulls it downwards with $mg \sin	heta$. But friction, true to its nature, flips its direction to oppose the downward motion, now acting upwards along the plane. The net acceleration during descent is the difference between these two opposing forces:

$$a_D = g \sin	heta - \mu g \cos	heta$$

Because $a_A > a_D$, the block decelerates quickly on the way up, taking less time. On the way down, it accelerates more slowly, taking more time to cover the same distance.

### The Mathematical Bridge

We can link the distance, acceleration, and time using the second equation of motion, $s = ut + \frac{1}{2}at^2$.

For the descent, the block starts from rest, so the distance is simply $s = \frac{1}{2} a_D T_D^2$.

For the ascent, if we imagine the motion in reverse (starting from rest at the top and accelerating downwards at $a_A$), the distance is $s = \frac{1}{2} a_A T_A^2$.

Since the distance $s$ is the same for both journeys, we can equate them:

$$\frac{1}{2} a_A T_A^2 = \frac{1}{2} a_D T_D^2$$

Rearranging this gives us a beautiful relationship between the accelerations and the times:

$$\frac{a_A}{a_D} = \left(\frac{T_D}{T_A}ight)^2$$

### The Algebraic Climax

We are given that the time of ascent is half the time of descent, meaning $T_A = \frac{1}{2} T_D$, or $\frac{T_D}{T_A} = 2$. Squaring this ratio gives us $4$.

Now, let's substitute our expressions for $a_A$ and $a_D$ into the ratio equation. Notice that the acceleration due to gravity, $g$, cancels out completely:

$$\frac{\sin 30^\circ + \mu \cos 30^\circ}{\sin 30^\circ - \mu \cos 30^\circ} = 4$$

Substituting the standard trigonometric values $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$:

$$\frac{\frac{1}{2} + \mu \frac{\sqrt{3}}{2}}{\frac{1}{2} - \mu \frac{\sqrt{3}}{2}} = 4$$

Multiplying the numerator and denominator by $2$ simplifies the fraction:

$$\frac{1 + \sqrt{3}\mu}{1 - \sqrt{3}\mu} = 4$$

Now, we cross-multiply and solve for $\mu$:

$$1 + \sqrt{3}\mu = 4(1 - \sqrt{3}\mu)$$

$$1 + \sqrt{3}\mu = 4 - 4\sqrt{3}\mu$$

Bringing the $\mu$ terms to one side:

$$5\sqrt{3}\mu = 3$$

$$\mu = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$$

### The Final Reveal

The problem states that the coefficient of friction is given by the expression $\frac{\sqrt{x}}{5}$. By directly comparing our calculated value with this expression:

$$\frac{\sqrt{3}}{5} = \frac{\sqrt{x}}{5}$$

It is crystal clear that **$x = 3$**.

This problem is a fantastic demonstration of how friction breaks the time symmetry of motion on an incline. If the plane were perfectly smooth ($\mu = 0$), the time of ascent would exactly equal the time of descent. The presence of friction ensures the downward journey is always a more leisurely ride!

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Motion}

\text{Equating Distances}

\text{Acceleration during Ascent}

\text{Acceleration during Descent}

\text{Substituting Values}

\text{Using the Time Ratio}

\text{Solving for } \mu

\text{Finding } x

\text{Conclusion}

The Sigma Insight: Static and Kinetic Friction

The Setup

The Physics of Ascent and Descent

The Mathematical Bridge

The Algebraic Climax

The Final Reveal

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