Let's now work through an example using conservation of energy. Let's say a small, rubber ball with a mass of 1.00 kg is launched straight up in the air from ground level by a toy compressed-air cannon. Assume that the cannon is small, so we'll consider the initial height of the ball to be zero when it leaves the cannon. The initial speed of the ball when it leaves the cannon is 11.5 m/s. Our goal is to find the maximum height the ball reaches before it begins to fall. You may have solved a similar type of problem when studying kinematics. But now, let's apply conservation of energy to this problem. If we assume, reasonably, that air resistance is minimal, then our ball–Earth system is a closed system. That means the total energy of the system is constant, or \( E_f = E_i \), the final total energy is equal to the initial total energy. Since the initial potential energy and final kinetic energy are zero, our equation now becomes \( E_{P,f} = E_{K,i} \). We can then substitute the formula for kinetic energy, \( E_K = \frac{1}{2}mv^2 \), and the formula for gravitational potential energy, \( E_p = mgh \), \( mgh_f = \frac{1}{2}mv_i^2 \). Now it's just a matter of doing the algebra, solving for the final height \( h_f \), and substituting values to find \( h_f \). Notice that the mass \( m \) divides out of both sides of the equation, so the value of the mass is not needed to find the final height. Calculate the maximum height of the ball in meters. \( h_f = 6.17 \) ✗ Double-check your algebra when solving for the final speed. Remember to cancel out the mass on both sides of the equation. Remember that the value of \( g \) is 9.80 m/s², and remember that the speed is squared.
Let's now work through an example using conservation of energy. Let's say a small, rubber ball with a mass of 1.00 kg is launched straight up in the air from ground level by a toy compressed-air cannon. Assume that the cannon is small, so we'll consider the initial height of the ball to be zero when it leaves the cannon. The initial speed of the ball when it leaves the cannon is 11.5 m/s. Our goal is to find the maximum height the ball reaches before it begins to fall. You may have solved a similar type of problem when studying kinematics. But now, let's apply conservation of energy to this problem. If we assume, reasonably, that air resistance is minimal, then our ball–Earth system is a closed system. That means the total energy of the system is constant, or \( E_f = E_i \), the final total energy is equal to the initial total energy. Since the initial potential energy and final kinetic energy are zero, our equation now becomes \( E_{P,f} = E_{K,i} \). We can then substitute the formula for kinetic energy, \( E_K = \frac{1}{2}mv^2 \), and the formula for gravitational potential energy, \( E_p = mgh \), \( mgh_f = \frac{1}{2}mv_i^2 \). Now it's just a matter of doing the algebra, solving for the final height \( h_f \), and substituting values to find \( h_f \). Notice that the mass \( m \) divides out of both sides of the equation, so the value of the mass is not needed to find the final height. Calculate the maximum height of the ball in meters. \( h_f = 6.17 \) ✗ Double-check your algebra when solving for the final speed. Remember to cancel out the mass on both sides of the equation. Remember that the value of \( g \) is 9.80 m/s², and remember that the speed is squared.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:Let's now work through an example using conservation of energy.
Let's say a small, rubber ball with a mass of 1.00 kg is launched straight up in the air from ground level by a toy compressed-air cannon. Assume that the cannon is small, so we'll consider the initial height of the ball to be zero when it leaves the cannon. The initial speed of the ball when it leaves the cannon is 11.5 m/s.
Our goal is to find the maximum height the ball reaches before it begins to fall.
You may have solved a similar type of problem when studying kinematics. But now, let's apply conservation of energy to this problem.
If we assume, reasonably, that air resistance is minimal, then our ball–Earth system is a closed system. That means the total energy of the system is constant, or
\( E_f = E_i \),
the final total energy is equal to the initial total energy.
Since the initial potential energy and final kinetic energy are zero, our equation now becomes
\( E_{P,f} = E_{K,i} \).
We can then substitute the formula for kinetic energy, \( E_K = \frac{1}{2}mv^2 \), and the formula for gravitational potential energy, \( E_p = mgh \),
\( mgh_f = \frac{1}{2}mv_i^2 \).
Now it's just a matter of doing the algebra, solving for the final height \( h_f \), and substituting values to find \( h_f \). Notice that the mass \( m \) divides out of both sides of the equation, so the value of the mass is not needed to find the final height.
Calculate the maximum height of the ball in meters.
\( h_f = 6.17 \)
✗ Double-check your algebra when solving for the final speed. Remember to cancel out the mass on both sides of the equation. Remember that the value of \( g \) is 9.80 m/s², and remember that the speed is squared.
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