/*! 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 ball rolls up a ramp, turns ar... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball rolls up a ramp, turns around, and comes back down. When does it have the greatest gravitational energy? The greatest kinetic energy? [Based on a problem by Serway and Faughn.]

Short Answer

Expert verified
Greatest gravitational energy: highest point; greatest kinetic energy: lowest point.

Step by step solution

01

Understanding Gravitational and Kinetic Energy

Gravitational potential energy is highest when an object is at its highest point because potential energy depends on height in the gravitational field. Kinetic energy, on the other hand, reaches its maximum when the object is moving at its fastest speed.
02

Determining the Maximum Gravitational Energy

The ball's maximum gravitational potential energy occurs at the highest point on the ramp. At this position, the height is greatest, and all the kinetic energy has been converted into potential energy momentarily. This happens right before it begins to roll back down.
03

Identifying the Maximum Kinetic Energy

The ball's maximum kinetic energy occurs when it is moving fastest. This is at the bottom of the ramp after it has rolled back down, as all potential energy at the top has been converted back into kinetic energy due to gravitational acceleration.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy represents the energy stored due to an object's position in a gravitational field. It's the energy that has the potential to be converted into other forms, such as kinetic energy. Gravitational potential energy depends on two main factors:
  • The object's mass
  • The height of the object above a reference point
Mathematically, this energy is given by the formula: \[ PE_{gravity} = mgh \]where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) near the Earth's surface), and \( h \) is the height above the ground.
For example, when the ball in our scenario reaches the highest point on the ramp, all the kinetic energy is converted into gravitational potential energy. At that moment, the height \( h \) is maximized, resulting in the maximum gravitational potential energy. Understanding this concept helps to clarify how energy transformations occur during the ball's motion.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's an indication of how much work an object can do as it moves. When the ball is rolling up and down the ramp, its kinetic energy changes continuously depending on its velocity.
The formula for kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. When analyzing the motion of the ball, we find that its kinetic energy is greatest when the ball moves at its fastest speed. This occurs at the lowest point of the ramp, once the ball has descended from the highest point. At this point, gravitational potential energy has been almost entirely converted into kinetic energy.
This energy transformation explains why the ball has maximum speed at the bottom of the ramp, showcasing the intricate relationship between height, motion, and energy.
Conservation of Energy
The conservation of energy principle is a fundamental concept in physics stating that energy in a closed system is constant. It cannot be created nor destroyed, only transformed from one form to another. This principle is perfectly illustrated by the motion of the ball on the ramp.
As the ball climbs the ramp, its kinetic energy decreases while gravitational potential energy increases because it's rising higher into the gravitational field. When the ball reaches the top, kinetic energy is minimized, and gravitational potential energy is maximized. As the ball rolls back down, the process reverses — gravitational potential energy is converted back into kinetic energy, causing the ball to speed up.
Understanding conservation of energy helps clarify why the ball never loses its combined kinetic and potential energy, assuming no other forces such as friction come into play. It highlights how energy is redistributed within mechanical systems based on position and speed.

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Most popular questions from this chapter

A grasshopper with a mass of \(110 \mathrm{mg}\) falls from rest from a height of \(310 \mathrm{~cm}\). On the way down, it dissipates \(1,1 \mathrm{~mJ}\) of heat due to air resistance. At what speed, in \(\mathrm{m} / \mathrm{s}\), does it hit the ground? (solution in the pdf version of the book)

The following table gives the amount of energy required in order to heat, melt, or boil a gram of water. $$\begin{array}{ll}\hline \text { heat } 1 \mathrm{~g} \text { of ice by } 1^{\circ} \mathrm{C} & 2.05 \mathrm{~J} \\ \text { melt } 1 \mathrm{~g} \text { of ice } & 333 \mathrm{~J} \\ \text { heat } 1 \mathrm{~g} \text { of liquid by } 1^{\circ} \mathrm{C} & 4.19 \mathrm{~J} \\ \text { boil } 1 \mathrm{~g} \text { of water } & 2500 \mathrm{~J} \\ \text { heat } 1 \mathrm{~g} \text { of steam by } 1^{\circ} \mathrm{C} & 2.01 \mathrm{~J} \\ \hline\end{array}$$ (a) How much energy is required in order to convert \(1.00 \mathrm{~g}\) of ice at \(-20^{\circ} \mathrm{C}\) into steam at \(137^{\circ} \mathrm{C}\) ? (answer check available at lightandmatter.com) (b) What is the minimum amount of hot water that could melt \(1.00 \mathrm{~g}\) of ice? (answer check available at lightandmatter.com)

On page 83, I used the chain rule to prove that the acceleration of a free- falling object is given by \(a--g .\) In this problem, you'll use a different technique to prove the same thing. Assume that the acceleration is a constant, \(a\), and then integrate to find \(v\) and \(y\), including appropriate constants of integration. Plug your expressions for \(v\) and \(y\) into the equation for the total energy, and show that \(a=-g\) is the only value that results in constant energy.

A ball falls from a height \(h\). Without air resistance, the time it takes to reach the floor is \(\sqrt{2 h / g}\). A numerical version of this calculation was given in program time2 on page 92. Now suppose that air resistance is not negligible. For a smooth sphere of radius \(r\), moving at speed \(v\) through air of density \(\rho\), the amount of energy \(d Q\) dissipated as heat as the ball falls through a height \(d y\) is given (ignoring signs) by \(d Q=(\pi / 4) \rho v^{2} r^{2} d y\). Modify the program to incorporate this effect, and find the resulting change in the fall time in the case of a \(21 \mathrm{~g}\) ball of radius \(1.0 \mathrm{~cm}\), falling from a height of \(1.0 \mathrm{~m}\). The density of air at sea level is about \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\). Turn in a printout of both your program and its output. Answer: \(0.34 \mathrm{~ms}\).

An idealized pendulum consists of a pointlike mass \(m\) on the end of a massless, rigid rod of length \(L\). Its amplitude, \(\theta\), is the angle the rod makes with the vertical when the pendulum is at the end of its swing. Write a numerical simulation to determine the period of the pendulum for any combination of \(m, L\), and \(\theta\). Examine the effect of changing each variable while manipulating the others.

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