/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A crate of mass \(M\) starts fro... [FREE SOLUTION] | 91Ó°ÊÓ

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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: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with \(y\) positive upward. (b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (c) Why did the normal force not enter into your solution?

Short Answer

Expert verified
The speed at the bottom is \( v = \sqrt{2g\sin(\alpha) d} \). Normal force doesn't work on the crate, thus doesn't impact energy calculations.

Step by step solution

01

Define Variables and Understand the Problem

We need to find the speed of a crate at the bottom of a frictionless ramp. We are given:- Mass of crate: \( M \)- Incline angle: \( \alpha \)- Distance traveled down the ramp: \( d \)We will solve it using energy principles by considering two different potential energy baselines.
02

Use Energy Conservation for Scenario (a)

Let's assume potential energy zero at the bottom of the ramp (height \( y = 0 \) at the bottom). At the top of the ramp, the crate has gravitational potential energy but no kinetic energy.- Potential Energy at Top: \( U_i = Mg\sin(\alpha) d \)- Kinetic Energy at Top: \( K_i = 0 \)At the bottom, all potential energy is converted to kinetic energy.- Kinetic Energy at Bottom: \( K_f = \frac{1}{2} M v^2 \)- Potential Energy at Bottom: \( U_f = 0 \)Apply conservation of energy:\[ Mg\sin(\alpha) d = \frac{1}{2} M v^2 \]Solve for speed, \( v \):\[ v = \sqrt{2g\sin(\alpha) d} \]
03

Use Energy Conservation for Scenario (b)

Now, assume potential energy zero level at the top of the ramp.- Potential Energy at Top: \( U_i = 0 \)- Kinetic Energy at Top: \( K_i = 0 \)At the bottom, the crate is entirely converted to kinetic energy, losing potential energy.- Kinetic Energy at Bottom: \( K_f = \frac{1}{2} M v^2 \)- Potential Energy at Bottom: \( U_f = -Mg\sin(\alpha) d \)Apply conservation of energy:\[ 0 = \frac{1}{2} M v^2 - Mg\sin(\alpha) d \]Re-arrange and solve for speed, \( v \):\[ v = \sqrt{2g\sin(\alpha) d} \]
04

Explanation for the Absence of Normal Force

The normal force does not do work as it is perpendicular to the direction of motion. Since the ramp is frictionless, all the force and energy changes are due to gravity in the direction of the ramp. Thus, normal force does not enter into the energy calculations.

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Key Concepts

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

Kinetic Energy
When you hear about kinetic energy, think about movement! This type of energy is all about objects that are moving. The faster something is moving and the more mass it has, the more kinetic energy it has. This is why kinetic energy is often given by the formula:
  • \( K = \frac{1}{2} mv^2 \)
Here, \( m \) stands for mass, and \( v \) stands for velocity. So, kinetic energy grows with both mass and speed! It's like a snowball effect: the more you have, the more it builds up.
When something rolls down a hill, it's a classic example of potential energy becoming kinetic energy. By the time the object reaches the bottom, like our crate on the ramp, it has mostly converted its potential energy into kinetic energy. Understanding this transformation helps us see how energy shifts from one type to another, which is crucial in many physics problems.
Potential Energy
Potential energy is stored energy that depends on the position of an object. Imagine it as energy waiting in the wings to become kinetic energy when the conditions are right. The formula for gravitational potential energy when an object is above ground level is:
  • \( U = mgh \)
In this formula, \( h \) is the height above the ground, \( m \) is the mass, and \( g \) is the gravitational acceleration (approximately 9.81 m/s² on Earth). So, the higher up something is, the more potential energy it has.
When you set the potential energy zero at the top of the ramp, the crate begins with zero potential energy but gains kinetic energy as it descends. Conversely, setting zero at the bottom means the crate has potential energy initially, which converts as it moves down the ramp. Regardless, the try runs on the conservation principle. This ensures total energy remains constant but changes form.
Frictionless Inclined Plane
An inclined plane without friction offers the perfect setting to study pure energy transformations. The absence of friction means no energy is lost to heat or deformation, making calculations neat and tidy.
Imagine a smooth slide; the motion is controlled entirely by gravity without other forces interfering. In our exercise, the normal force doesn't contribute to any energy changes because it acts perpendicular to the motion. Without friction, the only forces we need to consider are gravity pulling the crate downhill and the reaction of the normal force. Both must adhere to energy conservation.
  • The smooth interaction lets the potential energy at the top completely convert into kinetic energy at the bottom.
  • It showcases the conservation of energy and how it manages to work in its unaltered form.
This makes frictionless inclined planes excellent for grasping fundamental physics principles.

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Most popular questions from this chapter

The Great Sandini is a \(60-\mathrm{kg}\) circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 \(\mathrm{N} / \mathrm{m}\) that he will compress with a force of 4400 \(\mathrm{N}\) . The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40 \(\mathrm{N}\) during the 4.0 \(\mathrm{m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 \(\mathrm{m}\) above his initial rest position?

A particle with mass \(m\) is acted on by a conservative force and moves along a path given by \(x=x_{0} \cos \omega_{0} t\) and \(y=y_{0} \sin \omega_{0} t\) where \(x_{0}, y_{0},\) and \(\omega_{0}\) are constants. (a) Find the components of the force that acts on the particle. (b) Find the potential energy of the particle as a function of \(x\) and \(y .\) Take \(U=0\) when \(x=0\) and \(y=0 .\left(\text { c) Find the total energy of the particle when (i) } x=x_{0 .}\right.\) \(y=0\) and \(\left(\text { ii) } x=0, y=y_{0}\right.\)

A hydroelectric dam holds back a lake of surface area \(3.0 \times 10^{6} \mathrm{m}^{2}\) that has vertical sides below the water level. The water level in the lake is 150 \(\mathrm{m}\) above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted into electrical energy with 90\(\%\) efficiency. (a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000 \(\mathrm{kg} / \mathrm{m}^{3} .(\mathrm{b})\) What volume of water must pass through the dam to produce 1000 kilo-watt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

A block with mass \(m\) is attached to an ideal spring that has force constant \(k .\) (a) The block moves from \(x_{1}\) to \(x_{2},\) where \(x_{2}>x_{1} .\) How much work does the spring force do during this displacement? (b) The block moves from \(x_{1}\) to \(x_{2}\) and then from \(x_{2}\) to \(x_{1}\) . How much work does the spring force do during the displacement from \(x_{2}\) to \(x_{1} ?\) What is the total work done by the spring during the entire \(x_{1} \rightarrow x_{2} \rightarrow x_{1}\) displacement? Explain why you got the answer you did. (c) The block moves from \(x_{1}\) to \(x_{3},\) where \(x_{3}>x_{2}\) . How much work does the spring force do during this displacement? The block then moves from \(x_{3}\) to \(x_{2}\) . How much work does the spring force do during this displacement? What is the total work done by the spring force during the \(x_{1} \rightarrow x_{3} \rightarrow x_{2}\) displacement? Compare your answer to the answer in part (a), where the starting and ending points are the same but the path is different.

A force of 800 \(\mathrm{N}\) stretches a certain spring a distance of 0.200 \(\mathrm{m} .\) (a) What is the potential energy of the spring when it is stretched 0.200 \(\mathrm{m} ?(\mathrm{b})\) What is its potential energy when it is compressed 5.00 \(\mathrm{cm} ?\)

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