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A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with \(y\) positive upward, and (b) at the top of the ramp with y positive upward. (c) Why didn't the normal force enter into your solution?

Short Answer

Expert verified
The speed at the bottom is the same, \(v = \sqrt{2gd \sin(\alpha)}\), in both cases since the normal force does no work.

Step by step solution

01

Define Energy Conservation Principle

The principle of conservation of mechanical energy states that the total mechanical energy (potential energy + kinetic energy) in a system remains constant if only conservative forces are doing work. In this problem, because the ramp is frictionless, the only forces acting are conservative (gravity).
02

Solve for Speed Using Potential Energy Zero at Bottom (a)

With potential energy zero at the bottom, initially, the crate has maximum potential energy and zero kinetic energy. Initial potential energy is given by: \[ U_i = Mgd \sin(\alpha) \]At the bottom, potential energy is zero, and all energy is converted to kinetic energy: \[ K_f = \frac{1}{2}Mv^2 \]Using the conservation of energy, \( U_i = K_f \), solve for \( v \): \[ Mgd \sin(\alpha) = \frac{1}{2}Mv^2 \] \[ v = \sqrt{2gd \sin(\alpha)} \]
03

Solve for Speed Using Potential Energy Zero at Top (b)

When potential energy is zero at the top, initially, the crate has zero potential energy and zero kinetic energy (since it starts from rest). At the bottom, potential energy is \(-Mgd \sin(\alpha)\) and kinetic energy is \(\frac{1}{2} Mv^2\). Applying energy conservation: \[ 0 = -Mgd \sin(\alpha) + \frac{1}{2}Mv^2 \] Solving for \( v \): \[ \frac{1}{2} Mv^2 = Mgd \sin(\alpha) \] \[ v = \sqrt{2gd \sin(\alpha)} \]
04

Explanation of the Normal Force's Role (c)

The normal force acts perpendicular to the surface of the ramp. Since it is perpendicular to the direction of motion, it does no work on the crate. Only the component of gravitational force parallel to the ramp has an effect on the crate's motion, hence the normal force does not enter into the energy conservation equations.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. When an object moves, it possesses kinetic energy, which depends on its mass and speed. The formula for kinetic energy is given by:
\[ K = \frac{1}{2} mv^2 \]In this formula, \( K \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity. You can see that kinetic energy increases with both the mass of the object and the square of its speed. This means that even small increases in speed can lead to large increases in kinetic energy.
In the context of the crate sliding down the inclined plane, it starts with no kinetic energy at the top (since it starts from rest). As it moves down due to the force of gravity, its speed increases, thereby increasing its kinetic energy. By the time it reaches the bottom of the ramp, all its potential energy has been converted into kinetic energy, leading to the speed we solved for.
Potential Energy
Potential energy is stored energy, often associated with an object's position in a force field, like gravity. The most common type of potential energy we encounter is gravitational potential energy. It is calculated using the formula:
\[ U = mgh \]Here, \( U \) is the potential energy, \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height above the zero potential energy level. Potential energy decreases as the object moves towards the zero level, and increases when the object is raised.
In our scenario, the crate begins with potential energy at the top of the ramp depending on its height above the bottom. If we choose the bottom of the ramp to be our zero level (Part a), the initial potential energy is \( Mgd \sin(\alpha) \), which is completely converted to kinetic energy at the bottom. Meanwhile, if the top is chosen as the zero level (Part b), the potential energy at the bottom becomes negative, yet it results in the same speed because of conservation laws.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, used to assist in raising or lowering objects with less effort. This angle, denoted as \(\alpha\), is crucial as it affects the component of gravitational force acting parallel to the surface.
Inclined planes are simple machines that make it easier to move objects upwards by spreading the required force over a longer distance. When a crate moves down an inclined plane, its path opens a competition between gravitational pull and the angle of descent.
The force down the slope due solely to gravity can be found using:
  • Parallel Force: \( Mg \sin(\alpha) \)
  • Perpendicular Force (Normal): \( Mg \cos(\alpha) \), which doesn't affect speed
The normal force counteracts the perpendicular gravitational force, doing no work, thus not affecting the kinetic energy calculation.
Gravity
Gravity is the force that pulls objects towards the center of the Earth. It gives weight to physical objects and causes them to fall when dropped. The force of gravity is what provides the potential energy to the crate at the top of the inclined plane.
In physics, gravity is responsible for the acceleration of any object in free fall, including the component of force that acts parallel to the inclined plane. In this exercise, it is the force that helps convert the crate's potential energy at the top of the ramp into kinetic energy as it moves down.
This force is given by:
  • Gravitational Force: \( Mg \)
  • Gravitational Acceleration: \( g \approx 9.81 \text{ m/s}^2 \)
The effect of gravity is maximized when there are no other forces, like friction or air resistance, acting against it, as is the case on our frictionless ramp. This makes gravity the sole driver of the crate's motion down the plane, showcasing the elegance of conservation of energy in a pure form.

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Most popular questions from this chapter

A spring of negligible mass has force constant \(k =\) 1600 N/m. (a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.20-kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.

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