Motion of a Body on an Inclined Plane: Comprehensive NEET Physics Notes

1. Motion on an Inclined Plane

1.1 Introduction to Motion on an Inclined Plane

When a body is placed on an inclined plane, it experiences a gravitational force that can be resolved into two components:

1. Parallel to the plane (causing the body to move downwards along the incline).
2. Perpendicular to the plane (balanced by the normal reaction force from the surface).

For a body of mass $m$ placed on a plane inclined at an angle $\theta$ to the horizontal, the forces acting on the body are:

• The weight of the body: $mg$, acting vertically downward.
• The component of the weight parallel to the inclined plane: $mg\sin \theta$.
• The component of the weight perpendicular to the inclined plane: $mg\cos \theta$.
• The normal reaction force from the surface: $N=mg\cos \theta$.
• The frictional force (if any), opposing the motion: maximum static friction force $f_s={\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction.

1.2 Equations of Motion on an Inclined Plane

The motion of the body on the inclined plane depends on whether friction is present.

Without Friction: If there is no friction, the only force causing motion along the plane is the component of gravity parallel to the plane, $mg\sin \theta$. The net force along the incline is:

$F_{\text{net}}=mg\sin \theta$

According to Newton's second law, the acceleration of the body is:

$a=\frac{F_{\text{net}}}{m}=g\sin \theta$

Thus, the acceleration of a body on a frictionless inclined plane is directly proportional to the sine of the angle of inclination.

With Friction: If friction is present, the net force acting along the inclined plane is reduced by the frictional force. The equation becomes:

$F_{\text{net}}=mg\sin \theta -f_k$

Where $f_k={\mu }_{k}N={\mu }_{k}mg\cos \theta$ (for kinetic friction). Therefore, the net force is:

$F_{\text{net}}=mg\sin \theta -{\mu }_{k}mg\cos \theta$

The acceleration of the body is:

$a=g(\sin \theta -{\mu }_k\cos \theta )$

1.3 Conditions for Motion to Start

For the body to start moving on an inclined plane, the force parallel to the incline must overcome static friction. The condition is:

$mg\sin \theta >{\mu }_{s}mg\cos \theta$

This leads to the critical angle at which motion begins:

$\tan {\theta }_{\text{min}}={\mu }_s$

At this angle, the static friction is overcome, and the body starts sliding down the incline.

1.4 Example of Motion on an Inclined Plane

Example 1: A block of mass 5 kg is placed on a smooth inclined plane with an angle of $30^{\circ }$. The acceleration of the block down the incline can be calculated as:

$a=g\sin 30^{\circ }=9.8\times 0.5=4.9{\text{m/s}}^2$

Thus, the block accelerates down the incline at $4.9{\text{m/s}}^2$ in the absence of friction.

Did You Know?

Galileo's study of inclined planes was crucial in developing the concept of inertia, which laid the foundation for Newton's laws of motion.

NEET Tip:

When dealing with inclined plane problems, always resolve the forces into components parallel and perpendicular to the plane for clarity. This will help you apply Newton's laws more effectively during the NEET exam.

Common Misconception:

A common mistake is assuming that friction always acts in the opposite direction of motion. In reality, static friction opposes the impending motion, while kinetic friction opposes relative motion between surfaces.

Quick Recap:

• The weight of the body is divided into components parallel and perpendicular to the inclined plane.
• Acceleration without friction is given by: $a=g\sin \theta$.
• Acceleration with friction is given by: $a=g(\sin \theta -{\mu }_k\cos \theta )$.
• A body begins to slide when $\tan \theta >{\mu }_s$.

NEET Problem-Solving Strategy:

For NEET questions involving inclined planes:

1. Break down the gravitational force into its components.
2. Apply Newton’s second law in both directions (parallel and perpendicular to the incline).
3. If friction is involved, include it in your calculations.
4. Be aware of the condition for motion to start, especially when dealing with static friction.

Practice Questions:

1. A block of mass 10 kg is placed on a plane inclined at an angle of $45^{\circ }$. Assuming no friction, calculate the acceleration of the block.
2. A block is sliding down an inclined plane with an angle of $30^{\circ }$ and a coefficient of kinetic friction of 0.2. What is its acceleration?
3. A block just starts to slide down a plane inclined at $20^{\circ }$. What is the coefficient of static friction?
4. A block of mass 7 kg is placed on a rough inclined plane with ${\mu }_k=0.3$ and $\theta =40^{\circ }$. What is the acceleration of the block as it slides down?
5. Calculate the minimum force required to push a 5 kg block up a plane inclined at $25^{\circ }$ if the coefficient of friction is 0.4.

Solutions:

1. The acceleration is: $a=g\sin 45^{\circ }=9.8\times 0.707=6.93{\text{m/s}}^2$
2. The acceleration is: $a=g(\sin 30^{\circ }-{\mu }_k\cos 30^{\circ })=9.8\times (0.5-0.2\times 0.866)=3.2{\text{m/s}}^2$
3. The coefficient of static friction is: ${\mu }_s=\tan 20^{\circ }=0.364$
4. The acceleration is: $a=g(\sin 40^{\circ }-0.3\cos 40^{\circ })=4.35{\text{m/s}}^2$
5. The minimum force required is: $F=mg(\sin \theta +{\mu }_s\cos \theta )=5\times 9.8\times (0.422+0.4\times 0.906)=36.83\text{N}$

Final Recommendations (Based on Improvements):

1. Visual Aids: Include free-body diagrams for various cases (with and without friction) to illustrate the forces more clearly. For example, show how the forces are resolved into components along and perpendicular to the plane.
2. Mnemonics for Memorability: Introduce simple mnemonics to help students remember the key formulas. For example, "Sin Slide, Cos Hold" can remind students that $mg\sin \theta$causes sliding, and