Question

A particle starts from rest from the top of an inclined plane and again comes to rest on reaching the bottom most point of incline plane. If the coefficient of friction on some part of inclined plane is $3\tan \theta$ and rest portion is smooth then the ratio of the rough length of the inclined plane to smooth length would be: (' $\theta$ ' is the plane angle) (1) $3:1$ (2) $\frac{\sin \theta }{{\cos }^2\theta }$ (3) $1:2$ (4) $\sqrt{2}:1$

Short answer

1:2

Step-by-step solution

Understand the Problem

A particle starts from rest at the top of an inclined plane and comes to rest at the bottom. There is friction on a part of the plane, with a coefficient of friction equal to 3 $\tan \theta$, and another part of the plane is smooth. The task is to find the ratio of the rough length to the smooth length of the inclined plane.

Notes on Forces Involved

The forces involved are gravity, normal force, and friction. For the smooth part, there is no friction, while in the rough part, friction force contributes to stopping the particle. The angle $\theta$ of the inclined plane affects both the normal force and the frictional force.

Applying Kinetic Energy Concepts

The particle starts and ends with zero kinetic energy, so by the work-energy principle, the work done by friction must equate to the work done by gravity on the rough segment: $W_f=W_g$.

Express Work Done by Gravity

The work done by gravity over the entire length of the plane ($L$) is equal to $mgL\sin \theta$, where $m$ is the mass of the particle and $g$ is the acceleration due to gravity.

Express Work Done by Friction

The work done by friction in the rough section of length $L_r$ is $F_f\times L_r$. The friction force $F_f$ can be given by $(3\tan \theta )\times mg\cos \theta$. Simplifying, this gives us: $3mg\sin \theta$. Therefore, $W_f=(3mg\sin \theta )\times L_r$.

Equate Work Done by Gravity and Friction

Since $W_f=W_g$, we set $3mg\sin \theta \times L_r=mg\sin \theta \times L$, dividing through by $mg\sin \theta$, we get $3L_r=L$.

Find Ratio

Given the entire length is sum of rough and smooth parts, let $L=L_r+L_s$. Thus, $3L_r=L_r+L_s$, solving we find $L_s=2L_r$. Therefore, the ratio of the rough length to smooth length is $1:2$.

Key concepts

kinetic energy

Kinetic energy is the energy possessed by an object due to its motion. In this exercise, the particle starts and ends with zero kinetic energy. This is crucial because it guides how we use the work-energy principle to solve the problem. Kinetic energy is represented as: $KE=\frac{1}{2}m{v}^2$, where $m$ is the mass of the particle and $v$ is its velocity.
Since the particle starts from rest and comes to rest, its initial and final kinetic energies are zero. Any loss or gain in kinetic energy results from the work done by various forces.
In our scenario, the only forces doing work on the particle are gravity and friction. The work done by these forces alters the kinetic energy throughout the particle's motion, which is why understanding kinetic energy is vital to solving this exercise.

coefficient of friction

The coefficient of friction ($\mu$) is a measure of how much frictional force exists between two surfaces. It's represented by the ratio of the frictional force ($F_f$) to the normal force ($N$): $\mu =\frac{F_f}{N}$.
In this problem, the coefficient of friction on the rough segment is given as $3\tan \theta$. This tells us how strong the friction force is in relation to gravitational force on the inclined plane. Friction opposes the motion of the particle, and its role is to bring the particle to rest.
The normal force on the inclined plane is $mg\cos \theta$, where $m$ is the particle’s mass and $g$ is the acceleration due to gravity. Therefore, the frictional force on the rough segment can be expressed as: $F_f=\mu N=(3\tan \theta )\times (mg\cos \theta )$. This simplifies to $3mg\sin \theta$. Understanding the coefficient of friction allows us to quantify the work done by friction as the particle moves down the slope.

work-energy principle

The work-energy principle states that the work done by all forces acting on an object is equal to the change in its kinetic energy. Mathematically, this is expressed as: $W=\mathrm{\Delta }KE$.
In this exercise, we apply the work-energy principle to relate the work done by gravity and friction to the particle’s motion. Since the particle’s initial and final kinetic energies are zero, we have: $W_g+W_f=0$.
Work done by gravity ($W_g$) over the entire length of the incline is $mgL\sin \theta$. Work done by friction ($W_f$) on the rough segment is $-3mg\sin \theta \times L_r$ (negative because friction opposes motion). Equating these, we get:
$mgL\sin \theta =3mg\sin \theta \times L_r$.
By simplifying, we find the relationship between the rough and smooth lengths of the incline: $3L_r=L$.
Given the total length $L$ is the sum of the rough and smooth sections: $L=L_r+L_s$, simplifying further we get $3L_r=L_r+L_s$ which leads to $L_s=2L_r$. Therefore, the ratio of rough length to smooth length is $1:2$. This calculation leverages the work-energy principle effectively to solve for the ratio of the inclined plane's sections.