The concept of energy is one of the most important topics in science and engineering. In everyday life, we think of energy in terms of fuel for transportation and heating, electricity for lights and appliances, and foods for consumption. However, these ideas do not really define energy. They merely tell us that fuels are needed to do a job and that those fuels provide us with something we call energy.

Energy is present in the Universe in various forms.

*Every*physical process that occurs in the Universe involves energy and energy transfers or transformations. Unfortunately, despite its extreme importance, energy cannot be easily defined. The variables in previous chapters were relatively concrete; we have everyday experience with velocities and forces, for example. The notion of energy is more abstract, although we do have*experiences*with energy, such as running out of gasoline, or losing our electrical service if we forget to pay the utility bill.The concept of energy can be applied to the dynamics of a mechanical system without resorting to Newton’s laws. This “energy approach” to describing motion is especially useful when the force acting on a particle is not constant; in such a case, the acceleration is not constant, and we cannot apply the constant acceleration equations that were developed in Chapter 2. Particles in nature are often subject to forces that vary with the particles’ positions. These forces include gravitational forces and the force exerted on an object attached to a spring. We shall describe techniques for treating such situations with the help of an important concept called

*conservation of energy*. This approach extends well beyond physics, and can be applied to biological organisms, technological systems, and engineering situations. Our problem-solving techniques presented in earlier chapters were based on the motion of a particle or an object that could be modeled as a particle. This was called the

*particle model*. We begin our new approach by focusing our attention on a*system*and developing techniques to be used in a*system model*.**7.1 Systems and Environments**

In the system model mentioned above, we focus our attention on a small portion of the Universe—the system—and ignore details of the rest of the Universe outside of the system. A critical skill in applying the system model to problems is

*identifying the system.*A valid system may

• be a single object or particle

• be a collection of objects or particles

• be a region of space (such as the interior of an automobile engine combustion

cylinder)

• vary in size and shape (such as a rubber ball, which deforms upon striking a

wall)

Identifying the

*need*for a system approach to solving a problem (as opposed to a particle approach) is part of the “categorize” step in the General Problem-Solving Strategy outlined in Chapter 2. Identifying the particular system and its nature is part of the “analyze” step.No matter what the particular system is in a given problem, there is a system boundary, an imaginary surface (not necessarily coinciding with a physical surface) that divides the Universe into the system and the environment surrounding the system.

As an example, imagine a force applied to an object in empty space. We can define the object as the system. The force applied to it is an influence on the system from the environment that acts across the system boundary.

We shall find that there are a number of mechanisms by which a system can be influenced by its environment. The first of these that we shall investigate is

*work*.**7.2 Work Done by a Constant Force**

Almost all the terms we have used thus far—velocity, acceleration, force, and so on—convey a similar meaning in physics as they do in everyday life. Now, however, we encounter a term whose meaning in physics is distinctly different from its everyday meaning—

*work*.Let us examine the situation in Figure 7.2, where an object undergoes a displacement along a straight line while acted on by a constant force F that makes an angle Ɵ with the direction of the displacement.

The work

*W*done on a system by an agent exerting a constant force on the system is the product of the magnitude*F*of the force, the magnitude ∆*r*of the displacement of the point of application of the force, and cos Ɵ, where Ɵ is the angle between the force and displacement vectors:W = F ∆r cos θ

The sign of the work also depends on the direction of F relative to ∆r. The work done by the applied force is positive when the projection of F to ∆r is in the same direction as the displacement. For example, when an object is lifted, the work done by the applied force is positive because the direction of that force is upward, in the same direction as the displacement of its point of application. When the projection of F onto ∆r is in the direction opposite the displacement,

*W*is negative. For example, as an object is lifted, the work done by the gravitational force on the object is negative. The factor cos ! in the definition of*W*(Eq. 7.1) automatically takes care of the sign.If an applied force F is in the same direction as the displacement ∆r, then Ɵ=0 and cos 0 = 1. In this case, Equation 7.1 gives :

*W*=

*F*∆

*r*

Work is a scalar quantity, and its units are force multiplied by length. Therefore, the SI unit of work is the newton! meter (N· m). This combination of units is used so frequently that it has been given a name of its own: the joule ( J).

An important consideration for a system approach to problems is to note that work is an energy transfer. If

*W*is the work done on a system and*W*is positive, energy is transferred*to*the system; if*W*is negative, energy is transferred*from*the system. Thus, if a system interacts with its environment, this interaction can be described as a transfer of energy across the system boundary. This will result in a change in the energy stored in the system.**7.3 The Scalar Product of Two Vectors**

Because of the way the force and displacement vectors are combined in Equation 7.1, it is helpful to use a convenient mathematical tool called the scalar product of two vectors. We write this scalar product of vectors A and B as A.B. (Because of the dot symbol, the scalar product is often called the dot product.)

In general, the scalar product of any two vectors A and B is a scalar quantity equal to the product of the magnitudes of the two vectors and the cosine of the angle Ɵ between them:

A.B≡AB cos θ

Comparing this definition to Equation 7.1, we see that we can express Equation 7.1 as a scalar product:

W = F ∆r cos θ = F ∆r

From the right-hand side of Equation 7.2 we also see that the scalar product is commutative.3 That is :

A.B = B.A

Finally, the scalar product obeys the distributive law of multiplication, so that:

A (B+C) = A.B + A.C

The dot product is simple to evaluate from Equation 7.2 when A is either perpendicular or parallel to B. if A is perpendicular to B (θ = 90˚), then A.B = 0. (The equality A.B = 0 also hols in the more trivial case in which either A or B is zero). If vector A is parallel to vector B and two point in the same direction (θ=0), then A.B = AB. If vector A is parallel to vector B but the two point in opposite direction (θ = 180˚), then A.B = - AB. The scalar product is negative when 90˚<θ≤180˚.

The unit vector î, ĵ, and ǩ, which were definded in Chapter 3, lie in the positive x, y and z directions, respectively, of a right-handed coordinate system. Therefore, it follows from the definition of A.B that scalar product of these unit vectors are

Î . î = ĵ . ĵ = ǩ . ǩ = 1 (7.4)

Î . ĵ = î . ǩ = ĵ . ǩ = 0 (7.5)

Equation 3.18 and 3.19 state that two vectors A and B can be expressed in component vector from as

A = Axǐ + Ayǰ + Azǩ

B = Bxǐ + Byǰ + Bzǩ

Using the information given in equation 7.4 and 7.5 shows that the scalar product of A and B reduces to :

A.B = AxBx + AyBy + AzBz (7.6)

(Details of the derivation are left for you in problem 6). In the special case in which A=B, we see that ;

A.A = Ax

^{2}+ Ay^{2}+ Az^{2}= A^{2}**7.4 Work Done by a Varying Force**

Consider a particle being displaced along the

*x*axis under the action of a force that varies with position. The particle is displaced in the direction of increasing*x*from x=x_{i}to x=x_{f}. In such a situation, we cannot use W = F ∆r cos θ to calculate the work done by the force because this relationship applies only when F is constant in magnitude and direction. However, if we imagine that the particle undergoes a very small displacement ∆x , shown in Figure 7.7a, the*x*component*Fx*of the force is approximately constant over this small interval, for this small displacement, we can approximate the work done by the force as :W= F ∆x

If we imagine that the

*Fx*versus*x*curve is divided into a large number of such intervals, the total work done for*the displacement from**xi*to*xf*is approximately equal to the sum of a large number of*such terms:**W =*∑

^{xf}_{xi }F

_{x}∆x

If the size of the displacements is allowed to approach zero, the number of terms in the sum increases without limit but the value of the sum approaches a definite value equal to the area bounded by the

*Fx*curve and the*x*axis:Therefore, we can express the work done by

*Fx*as the particle moves from*xi*to*xf*as :Work Done by a Spring

A model of a common physical system for which the force varies with position is shown in Figure 7.10. A block on a horizontal, frictionless surface is connected to a spring. If the spring is either stretched or compressed a small distance from its unstretched (equilibrium) configuration, it exerts on the block a force that can be expressed as :

Where

*x*is the position of the block relative to its equilibrium (*x*= 0) position and*k*is a positive constant called the force constant or the spring constant of the spring. In other words, the force required to stretch or compress a spring is proportional to the amount of stretch or compression*x*. This force law for springs is known as Hooke’s law.Because the spring force always acts toward the equilibrium position (

*x=*0), it is sometimes called a*restoring force.*If the spring is compressed until the block is at the point –*x*max and is then released, the block moves from -*x*max through zero to +*x*max. If the spring is instead stretched until the block is at the point +*x*max and is then released, the block moves from +*x*max through zero to -*x*max. It then reverses direction, returns to +*x*max, and continues oscillating back and forth. Suppose the block has been pushed to the left to a position -*x*max and is then released. Let us identify the block as our system and calculate the work*Ws*done by the spring force on the block as the block moves from*xi*= -*x*max to*xf*= 0. Applying equation 7.7 and assuming the block may be treated as a particle, we obtain:If the block undergoes an arbitrary displacement from

*x*=*xi*to*x*=*xf*, the work done by the spring force on the block is :The work done by an applied force on a block–spring system between arbitrary positions of the block is :

Notice that this is the negative of the work done by the spring as expressed by Equation 7.11. This is consistent with the fact that the spring force and the applied force are of equal magnitude but in opposite directions.

**7.5 Kinetic Energy and the Work–Kinetic Energy Theorem**

Using Newton’s second law, we can substitute for the magnitude of the net force ∑F = ma and then perform the following chain-rule manipulations on the integrand:

∑ W = ½mv

_{f}^{2}- ½mv_{i}^{2 }(7.14)**In general, the kinetic energy**

*K*of a particle of mass

*m*moving with a speed

*v*is defined as :

K ≡ ½mv

^{2} Kinetic energy is a scalar quantity and has the same units as work. It is often convenient to write Equation 7.14 in the form :

∑W = K

_{f}– K_{i}=∆K The work–kinetic energy theorem indicates that the speed of a particle will

*increase*if the net work done on it is*positive,*because the final kinetic energy will be greater than the initial kinetic energy. The speed will*decrease*if the net work is*negative,*because the final kinetic energy will be less than the initial kinetic energy.
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