Problem 26 - University Physics with Modern Physics

Question

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

Step-by-Step Solution

Answer

The speed is approximately 17.15 m/s, independent of the incline's angle.

Step 1: Analyze Components of Gravity

Start by considering the gravitational force acting on the mass. This force can be split into two components relative to the incline: one parallel to the incline (\(mg \sin \alpha\)) and one perpendicular to the incline (\(mg \cos \alpha\)). Since the incline is frictionless, only the parallel component does work on the mass.

Step 2: Calculate Work Done by Gravity

The work done by the parallel component of gravity is given by the formula \(W = Fd\cos\theta\), where \(F\) is the force parallel to the incline, \(d\) is the displacement along the plane, and \(\theta\) is the angle between the force and displacement. Here, \(\theta = 0\), so \(W_{\parallel} = mg \sin \alpha \times \frac{h}{\sin \alpha} = mgh\).

Step 3: Establish Work-Energy Theorem Relation

According to the work-energy theorem, the work done by the force is equal to the change in kinetic energy of the mass. Since the mass starts from rest, \(W = \frac{1}{2}mv^2\). From our previous step, we know \(W = mgh\). Therefore, \(\frac{1}{2}mv^2 = mgh\), leading to \(v^2 = 2gh\).

Step 4: Determine Speed at Bottom of Incline

Using the derived equation \(v^2 = 2gh\), solve for the speed \(v = \sqrt{2gh}\). Notice that this expression for \(v\) does not depend on \(\alpha\), hence proving that the speed at the bottom is independent of the incline's angle.

Step 5: Apply Formula to Specific Height

Given a height of 15.0 m, substitute into the expression for speed: \(v = \sqrt{2g \cdot 15.0}\). Using the approximate value of \(g \approx 9.81 \text{ m/s}^2\), \(v = \sqrt{2 \times 9.81 \times 15.0}\). Compute this to find \(v \approx 17.15 \text{ m/s}\).

Key Concepts

Work-Energy Theorem

The Work-Energy Theorem is a fundamental principle in physics that describes how the work done on an object relates to its change in energy. It states that the work done by forces acting on a body is equal to the change in its kinetic energy.
This concept is particularly useful when dealing with inclined planes because it allows us to connect the work done by gravitational forces to the change in velocity of an object moving along the plane.
In the given problem, the work done by gravity as a mass slides down an inclined plane results in a change in kinetic energy, which can be expressed as:

• The work done by the gravitational force: \( W = mgh \)
• The change in kinetic energy: \( rac{1}{2}mv^2 - 0 = mgh \)

By setting these equations equal, we derive the final velocity, \( v = \sqrt{2gh} \), demonstrating that the work done results in this increase in speed independent of the angle, \( \alpha \), of the incline.

Gravitational Force Components

Understanding the components of gravitational force on an inclined plane is crucial to solving problems involving such scenarios. Gravity acts downwards, but it can be broken into two components when considering an inclined surface.

• Parallel to the incline: \( mg \sin \alpha \)
• Perpendicular to the incline: \( mg \cos \alpha \)

On a frictionless incline, only the parallel component contributes to the work done on the object. This is because the perpendicular component is balanced by the normal force of the plane.
The work done by the parallel component results in the movement of the object along the plane, contributing to its kinetic energy, ultimately leading to the calculated speed at the bottom: \( v = \sqrt{2gh} \). This demonstrates how gravitational force is split and used when an object moves down an incline.

Kinematic Equations

Kinematic equations describe the motion of objects, taking into account initial velocity, acceleration, displacement, and time. These equations are particularly handy for calculating various attributes of motion when forces such as gravity act upon an object.
However, in the context of inclined planes, kinematic equations often provide a secondary role, due to the emphasis on energy principles.
For example, consider how we use:

• The equation \( v^2 = u^2 + 2as \), where \( u \) is the initial velocity and \( a \) is acceleration.

By replacing acceleration with the gravitational component, \( g \sin \alpha \), we'd solve for velocity, reinforcing the findings from the work-energy theorem.
However, the elegance of the work-energy approach bypasses these steps, providing speed directly through energy considerations, thus simplifying calculations on inclines.

Inclined Plane Mechanics

Inclined Plane Mechanics examines the behavior of objects on sloped surfaces, exploring how forces like gravity influence motion. These mechanics highlight the significance of angles, surfaces, and force components in determining how objects move along an incline.
This analysis often focuses on:

• Decomposing gravitational forces into parallel and perpendicular components.
• Applying energy principles to relate work done to kinetic energy changes.

In frictionless scenarios, the mechanics simplify, as there is no resistance to account for. Here, the mass of an object sliding down is driven entirely by gravity's pull, showing how the incline's angle, \( \alpha \), does not affect the final speed.
As demonstrated, inclined plane mechanics blend forces and motion, offering rich insights into how objects behave on these surfaces, making it a key topic in physics.