Chapter 8 Review Using Key Terms Define Work and Power to Include There Formulas

Learning Objectives

By the end of this section, you volition exist able to:

• Codify the principle of conservation of mechanical energy, with or without the presence of non-conservative forces
• Use the conservation of mechanical energy to summate various properties of simple systems

In this section, nosotros elaborate and extend the result we derived in Potential Energy of a System, where we re-wrote the work-energy theorem in terms of the change in the kinetic and potential energies of a particle. This will lead u.s.a. to a discussion of the of import principle of the conservation of mechanical free energy. As yous keep to examine other topics in physics, in later chapters of this book, you will see how this conservation law is generalized to encompass other types of energy and free energy transfers. The final section of this chapter provides a preview.

The terms 'conserved quantity' and 'conservation law' have specific, scientific meanings in physics, which are different from the everyday meanings associated with the use of these words. (The aforementioned comment is as well true about the scientific and everyday uses of the word 'work.') In everyday usage, you could conserve water by not using information technology, or past using less of information technology, or past re-using it. Water is equanimous of molecules consisting of ii atoms of hydrogen and one of oxygen. Bring these atoms together to form a molecule and you create water; dissociate the atoms in such a molecule and you destroy water. Still, in scientific usage, a conserved quantity for a system stays abiding, changes past a definite corporeality that is transferred to other systems, and/or is converted into other forms of that quantity. A conserved quantity, in the scientific sense, can be transformed, only not strictly created or destroyed. Thus, in that location is no physical law of conservation of water.

Systems with a Single Particle or Object

We first consider a system with a single particle or object. Returning to our development of Equation 8.ii, retrieve that we first separated all the forces interim on a particle into conservative and non-bourgeois types, and wrote the work washed by each type of force as a separate term in the piece of work-energy theorem. We then replaced the work washed by the bourgeois forces by the modify in the potential free energy of the particle, combining information technology with the change in the particle's kinetic energy to get Equation 8.2. Now, nosotros write this equation without the middle step and define the sum of the kinetic and potential energies, $K+U=E;$ to be the mechanical energy of the particle.

Conservation of Energy

The mechanical free energy East of a particle stays abiding unless forces exterior the system or non-bourgeois forces do work on it, in which case, the modify in the mechanical energy is equal to the work done by the non-conservative forces:

$$W_{\text{nc},AB}=\text{Δ}{\left(K+U\right)}_{AB}=\text{Δ}{\mathrm{Eastward}}_{AB}.$$

8.12

This statement expresses the concept of energy conservation for a classical particle as long as there is no non-conservative work. Call back that a classical particle is just a signal mass, is nonrelativistic, and obeys Newton'southward laws of movement. In Relativity, we will see that conservation of energy even so applies to a not-classical particle, but for that to happen, we have to make a slight aligning to the definition of energy.

It is sometimes convenient to separate the case where the work done past non-conservative forces is zero, either because no such forces are causeless present, or, like the normal force, they exercise zero work when the motion is parallel to the surface. So

$$0=W_{\text{nc},AB}=\text{Δ}{\left(\mathrm{Chiliad}+U\right)}_{AB}=\text{Δ}{\mathrm{Due\; east}}_{AB}.$$

8.13

In this example, the conservation of mechanical energy can exist expressed as follows: The mechanical energy of a particle does not change if all the non-conservative forces that may deed on it do no piece of work. Agreement the concept of energy conservation is the important thing, not the particular equation you lot use to limited information technology.

Problem-Solving Strategy

Conservation of Energy

1. Identify the torso or bodies to be studied (the organisation). Often, in applications of the principle of mechanical energy conservation, we written report more than than one body at the same fourth dimension.
2. Place all forces acting on the body or bodies.
3. Determine whether each force that does piece of work is bourgeois. If a non-conservative force (e.k., friction) is doing work, then mechanical free energy is not conserved. The organization must then be analyzed with non-conservative piece of work, Equation 8.thirteen.
4. For every force that does work, choose a reference point and decide the potential energy part for the force. The reference points for the diverse potential energies practice not have to be at the same location.
5. Employ the principle of mechanical energy conservation past setting the sum of the kinetic energies and potential energies equal at every point of interest.

Case viii.seven

Simple Pendulum

A particle of mass chiliad is hung from the ceiling by a massless string of length 1.0 g, as shown in Figure eight.7. The particle is released from rest, when the angle between the string and the downward vertical direction is $30\text{°.}$ What is its speed when it reaches the lowest signal of its arc?

Figure viii.7 A particle hung from a string constitutes a simple pendulum. Information technology is shown when released from rest, forth with some distances used in analyzing the motion.

Strategy

Using our trouble-solving strategy, the first step is to define that we are interested in the particle-Earth organization. Second, only the gravitational force is acting on the particle, which is conservative (step 3). Nosotros fail air resistance in the trouble, and no work is washed by the string tension, which is perpendicular to the arc of the motion. Therefore, the mechanical free energy of the organisation is conserved, as represented by Equation 8.13, $0=\text{Δ}\left(\mathrm{Thou}+U\right)$ . Considering the particle starts from residuum, the increase in the kinetic free energy is just the kinetic energy at the lowest bespeak. This increase in kinetic energy equals the decrease in the gravitational potential free energy, which nosotros can calculate from the geometry. In pace 4, we choose a reference point for zero gravitational potential energy to be at the lowest vertical bespeak the particle achieves, which is mid-swing. Lastly, in step 5, we prepare the sum of energies at the highest point (initial) of the swing to the lowest point (final) of the swing to ultimately solve for the final speed.

Solution

We are neglecting non-conservative forces, so we write the energy conservation formula relating the particle at the highest betoken (initial) and the everyman bespeak in the swing (final) every bit

$$K_{\text{i}}+U_{\text{i}}={\mathrm{Thou}}_{\text{f}}+U_{\text{f}}.$$

Since the particle is released from rest, the initial kinetic free energy is cipher. At the everyman point, we ascertain the gravitational potential energy to be naught. Therefore our conservation of energy formula reduces to

$$\begin{array}{ccc}\hfill 0+kgh& =\hfill & \frac{1}{2}m{v}^2+0\hfill \\ \hfill \mathrm{five}& =\hfill & \sqrt{2\mathrm{thousand}h}.\hfill \end{array}$$

The vertical tiptop of the particle is not given direct in the trouble. This tin can be solved for past using trigonometry and two givens: the length of the pendulum and the angle through which the particle is vertically pulled up. Looking at the diagram, the vertical dashed line is the length of the pendulum string. The vertical height is labeled h. The other partial length of the vertical string tin be calculated with trigonometry. That slice is solved for by

$$\text{cos}\theta =x\text{/}L,\mathrm{ten}=L\text{cos}\theta .$$

Therefore, past looking at the 2 parts of the string, we can solve for the height h,

$$\begin{array}{ccc}\hfill x+h& =\hfill & L\hfill \\ \hfill L\text{cos}\theta +h& =\hfill & L\hfill \\ \hfill h& =\hfill & L-\mathrm{50}\text{cos}\theta =L(1-\text{cos}\theta ).\hfill \end{array}$$

We substitute this top into the previous expression solved for speed to summate our result:

$$v=\sqrt{2g\mathrm{Fifty}\left(1-\text{cos}\theta \right)}=\sqrt{\mathrm{ii}\left(9.8{\text{m/s}}^{\mathrm{ii}}\right)\left(\mathrm{one}\text{m}\right)\left(1-\text{cos}30\text{°}\right)}=\mathrm{ane.62}\text{m/s}.$$

Significance

Nosotros found the speed directly from the conservation of mechanical energy, without having to solve the differential equation for the move of a pendulum (meet Oscillations). We can arroyo this problem in terms of bar graphs of total energy. Initially, the particle has all potential energy, being at the highest betoken, and no kinetic energy. When the particle crosses the lowest point at the bottom of the swing, the energy moves from the potential energy column to the kinetic free energy cavalcade. Therefore, we tin can imagine a progression of this transfer as the particle moves betwixt its highest signal, lowest point of the swing, and back to the highest point (Figure 8.8). As the particle travels from the lowest betoken in the swing to the highest point on the far right mitt side of the diagram, the energy bars go in reverse society from (c) to (b) to (a).

Effigy viii.viii Bar graphs representing the total free energy (E), potential energy (U), and kinetic free energy (K) of the particle in different positions. (a) The total free energy of the system equals the potential free energy and the kinetic energy is zero, which is found at the highest point the particle reaches. (b) The particle is midway between the highest and lowest point, and so the kinetic energy plus potential energy bar graphs equal the full energy. (c) The particle is at the lowest point of the swing, so the kinetic energy bar graph is the highest and equal to the total energy of the system.

Check Your Understanding 8.vii

How high above the bottom of its arc is the particle in the unproblematic pendulum to a higher place, when its speed is $0.81\text{thou}\text{/}\text{s}?$

Example viii.8

Air Resistance on a Falling Object

A helicopter is hovering at an distance of $1\text{km}$ when a panel from its underside breaks loose and plummets to the basis (Figure 8.9). The mass of the panel is $\mathrm{fifteen}\text{kg},$ and it hits the ground with a speed of $45\text{k}\text{/}\text{s}$ . How much mechanical energy was dissipated past air resistance during the panel's descent?

Figure eight.9 A helicopter loses a panel that falls until information technology reaches last velocity of 45 m/south. How much did air resistance contribute to the dissipation of energy in this problem?

Strategy

Step i: Here just one body is being investigated.

Step two: Gravitational force is acting on the panel, also as air resistance, which is stated in the problem.

Step 3: Gravitational force is bourgeois; yet, the non-conservative strength of air resistance does negative work on the falling panel, and so we tin use the conservation of mechanical energy, in the grade expressed by Equation eight.12, to notice the energy dissipated. This free energy is the magnitude of the piece of work:

$$\text{Δ}{E}_{\text{diss}}=\left|W_{\text{nc,if}}\right|=\left|\text{Δ}{\left(K+U\right)}_{\text{if}}\right|.$$

Step 4: The initial kinetic free energy, at $y_{\text{i}}=1\text{km},$ is goose egg. We set the gravitational potential energy to zero at ground level out of convenience.

Step five: The non-conservative work is ready equal to the energies to solve for the work dissipated by air resistance.

Solution

The mechanical energy prodigal by air resistance is the algebraic sum of the gain in the kinetic energy and loss in potential free energy. Therefore the calculation of this energy is

$$\begin{array}{cc}\hfill \text{Δ}{E}_{\text{diss}}& =\left|K_{\text{f}}-K_{\text{i}}+U_{\text{f}}-U_{\text{i}}\right|\hfill \\ & =\left|\frac{1}{2}\left(15\text{kg}\right){\left(45\text{m/s}\right)}^2-0+0-\left(15\text{kg}\right)\left(9.8{\text{m/southward}}^2\right)\left(1000\text{thousand}\right)\right|=130\text{kJ}.\hfill \end{array}$$

Significance

About of the initial mechanical free energy of the panel $\left(U_{\text{i}}\right)$ , 147 kJ, was lost to air resistance. Detect that we were able to calculate the energy dissipated without knowing what the force of air resistance was, only that information technology was dissipative.

Cheque Your Understanding 8.eight

You lot probably recall that, neglecting air resistance, if you throw a projectile straight up, the fourth dimension it takes to accomplish its maximum top equals the time it takes to fall from the maximum peak back to the starting height. Suppose you cannot neglect air resistance, equally in Instance eight.viii. Is the time the projectile takes to become upward (a) greater than, (b) less than, or (c) equal to the time it takes to come back downwardly? Explain.

In these examples, we were able to use conservation of energy to calculate the speed of a particle just at detail points in its motion. Only the method of analyzing particle motion, starting from energy conservation, is more powerful than that. More than advanced treatments of the theory of mechanics allow yous to calculate the total fourth dimension dependence of a particle'south move, for a given potential energy. In fact, it is ofttimes the case that a meliorate model for particle motion is provided by the form of its kinetic and potential energies, rather than an equation for force acting on information technology. (This is especially true for the quantum mechanical description of particles like electrons or atoms.)

We tin can illustrate some of the simplest features of this energy-based arroyo past because a particle in one-dimensional movement, with potential energy U(x) and no non-bourgeois interactions nowadays. Equation 8.12 and the definition of velocity require

$$\begin{array}{ccc}\hfill K& =\hfill & \frac{1}{2}m{v}^2=E-U\left(x\right)\hfill \\ \hfill v& =\hfill & \frac{dx}{dt}=\sqrt{\frac{2\left(\mathrm{East}-U\left(x\right)\right)}{m}}.\hfill \end{array}$$

Separate the variables x and t and integrate, from an initial fourth dimension $t=0$ to an arbitrary time, to get

$$t={\displaystyle \underset{0}{\overset{t}{\int }}dt=}{\displaystyle \underset{{\mathrm{10}}_0}{\overset{\mathrm{10}}{\int }}\frac{d\mathrm{10}}{\sqrt{\mathrm{two}\left[E-U\left(x\right)\right]\text{/}m}}}.$$

8.14

If y'all can do the integral in Equation eight.14, then yous tin solve for x as a function of t.

Example viii.nine

Abiding Acceleration

Use the potential energy $U\left(\mathrm{10}\right)=\text{−}E\left(\mathrm{ten}\text{/}{x}_0\right),$ for $E>0,$ in Equation 8.14 to find the position x of a particle as a office of time t.

Strategy

Since we know how the potential energy changes as a function of x, we tin can substitute for $U\left(x\right)$ in Equation 8.14, integrate, then solve for 10. This results in an expression of x as a office of time with constants of free energy E, mass yard, and the initial position $x_0.$

Solution

Following the kickoff ii suggested steps in the in a higher place strategy,

$$t={\displaystyle \underset{x_0}{\overset{x}{\int }}\frac{dx}{\sqrt{\left(2E\text{/}m{x}_0\right)\left(x_0-x\right)}}}=\frac{\mathrm{one}}{\sqrt{\left(2\mathrm{East}\text{/}m{x}_0\right)}}{\left|\mathrm{-2}\sqrt{\left({\mathrm{ten}}_0-x\right)}\right|}_{x_0}^x=-\frac{2\sqrt{\left(x_0-\mathrm{10}\right)}}{\sqrt{\left(2E\text{/}\mathrm{1000}{\mathrm{10}}_0\right)}}.$$

Solving for the position, we obtain $\mathrm{ten}\left(t\right)=x_0-\frac{\mathrm{ane}}{2}\left(E\text{/}m{x}_0\right)t^{\mathrm{ii}}$ .

Significance

The position every bit a function of fourth dimension, for this potential, represents one-dimensional movement with abiding dispatch, $a=\left(E\text{/}m{x}_0\right),$ starting at residual from position $x_0.$ This is not so surprising, since this is a potential free energy for a abiding force, $F=\text{−}dU\text{/}d\mathrm{ten}=E\text{/}{x}_0,$ and $a=F\text{/}\mathrm{yard}.$

Check Your Understanding 8.9

What potential energy $U\left(x\right)$ tin can you lot substitute in Equation eight.13 that will consequence in motion with abiding velocity of 2 m/s for a particle of mass 1 kg and mechanical energy i J?

Nosotros volition look at some other more physically appropriate example of the use of Equation 8.thirteen after nosotros have explored some further implications that can be drawn from the functional form of a particle'due south potential free energy.

Systems with Several Particles or Objects

Systems generally consist of more than one particle or object. However, the conservation of mechanical energy, in one of the forms in Equation viii.12 or Equation 8.13, is a key law of physics and applies to any organization. You lot just have to include the kinetic and potential energies of all the particles, and the piece of work washed by all the not-bourgeois forces acting on them. Until you larn more about the dynamics of systems composed of many particles, in Linear Momentum and Collisions, Fixed-Centrality Rotation, and Angular Momentum, information technology is better to postpone discussing the awarding of energy conservation to and so.

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Source: https://openstax.org/books/university-physics-volume-1/pages/8-3-conservation-of-energy

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