91Ó°ÊÓ

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 it 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 friction- less hill, starting from rest 15.0 m above the bottom.

Step by step solution

01

Analyzing the Components of Gravitational Force

When a mass slides down an incline, gravity acts vertically downward. We can decompose the gravitational force into two components: one parallel to the incline and one perpendicular to the incline.- The parallel component of the gravitational force is given by \( F_{\parallel} = mg \sin \alpha \).- The perpendicular component is \( F_{\perp} = mg \cos \alpha \).

02

Calculating the Work Done by Each Force Component

The work done by a force is the product of the force component along the direction of displacement and the displacement itself.- Work done by the parallel component, \( W_{\parallel} = mg \sin \alpha \times s \), where \( s \) is the distance along the incline.- For a smooth incline, the work done by the perpendicular component is zero because it does not cause displacement in its direction. Thus, \( W_{\perp} = 0 \).

03

Relating Incline Displacement to Initial Height

Using trigonometry, relate the distance \( s \) along the incline to the height \( h \): \[ s = \frac{h}{\sin \alpha} \].Substitute this into the work done by the parallel component: \[ W_{\parallel} = mg \sin \alpha \left( \frac{h}{\sin \alpha} \right) = mgh \].

04

Confirming Work Done using Energy Consideration

Total work done is \( W_{\text{total}} = mgh \), which equals the potential energy change if the object had fallen vertically. This shows the path taken didn't affect the work as gravity is a conservative force.

05

Applying the Work-Energy Theorem

According to the work-energy theorem: \[ \Delta KE = W_{\text{total}} = \frac{1}{2}mv^2 - 0 \], where initial kinetic energy is zero (starting from rest).\[ \Rightarrow v^2 = 2gh \rightarrow v = \sqrt{2gh} \].The speed at the bottom is independent of \( \alpha \), as work only depends on height \( h \).

06

Calculating the Speed from a Given Height

Substitute \( h = 15.0 \, \text{m} \) into the formula: \[ v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \, \text{m/s}^2 \cdot 15.0 \, \text{m}} \approx 17.15 \, \text{m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force Components

When a mass moves along an inclined plane, gravity pulls it straight towards the center of the Earth. However, not all of this gravitational force directly influences the motion of the mass down the incline. To better understand what happens, we decompose this force into two components:

• Parallel Component: This part pulls the object down the slope and is given by the formula \( F_{\parallel} = mg \sin \alpha \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \alpha \) is the angle of the incline.
• Perpendicular Component: This part presses the object into the surface of the incline and does not contribute to the movement down the slope. It is determined by \( F_{\perp} = mg \cos \alpha \).

The parallel component is what actually causes the object to slide down the incline, while the perpendicular component does no work since it does not cause the object to displace in its own direction. Understanding these components is crucial to analyzing how objects accelerate on inclined planes.

Inclined Plane Dynamics

The dynamics of an object sliding down an inclined plane can be fully understood by considering the forces in play. A crucial point here is recognizing how the gravitational force is applied and how this translates into motion.When you look at an object on an inclined plane, you will note that the angle of the incline, \(\alpha\), affects the gravitational force's parallel and perpendicular components. While one might initially think the angle should also impact the acceleration of the mass, it turns out that it doesn’t.This is demonstrated using work done calculations:

• The parallel force component over the incline does work equivalent to \( mgh \). This result matches what would occur if the object fell vertically through the height \( h \). Thus, both methods of descent lead to the same speed at the bottom, since work depends solely on the vertical height and not the path taken.
• This outcome is rooted in the simple relationship \( s = \frac{h}{\sin \alpha} \), which shows how inclined path length \( s \) relates to vertical height \( h \). Substituting this into the work formula, the dependence on \( \alpha \) cancels out.

Inclined plane dynamics illustrate how a body accelerates under the influence of gravitational force components and confirms that certain physical outcomes, like energy use, remain constant regardless of the path taken by the object.

Conservative Forces

Gravitational force is classified as a conservative force. This means that the work it does on a body only depends on the starting and ending positions, not the path taken. A conservative force like gravity can convert potential energy to kinetic energy or vice versa, without any loss of total mechanical energy.For practical purposes, consider the descent of the mass on an inclined plane:

• The total work converted from potential energy (mgh) to kinetic energy is path-independent, as highlighted by the work-energy theorem: \[ \Delta KE = W_{\text{total}} = \frac{1}{2}mv^2 \].
• Since the change in kinetic energy equals the work done by gravity, and this is independent of the angle \( \alpha \), it shows that the speed at the bottom only depends on the height \( h \).

By understanding conservative forces, students can better grasp how energy is conserved in various physical processes. This understanding is essential for further studies in physics, as it forms a foundation for analyzing motion and energy transformations across different systems.