Choose the Best Answer. A Car Traveling at Constant Speed Has a Net Work of Zero Done on It.

Learning Objectives

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

• Explain work as a transfer of energy and net work every bit the work washed by the net forcefulness.
• Explain and apply the piece of work-energy theorem.

Piece of work Transfers Free energy

What happens to the work washed on a system? Energy is transferred into the organisation, simply in what grade? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Figure 1a is pushed just hard plenty to keep information technology going at a abiding speed, and so free energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, piece of work done on the briefcase by the person carrying information technology up stairs in Figure 1d is stored in the briefcase-Earth system and tin can be recovered at whatsoever time, every bit shown in Figure 1e. In fact, the edifice of the pyramids in aboriginal Egypt is an instance of storing energy in a organisation past doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the rock-Earth organisation and has the potential to do work.

Effigy one. examples of work. (a) The piece of work done past the force F on this backyard mower isFd cosθ. Notation thatF cosθ is the component of the strength in the management of motility. (b) A person holding a briefcase does no work on information technology, because there is no movement. No free energy is transferred to or from the briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by conveying information technology upward stairs at constant speed, because there is necessarily a component of force F in the direction of the move. Energy is transferred to the briefcase and could in turn be used to practice work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into an electrical generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because F and d are in reverse directions.

In this section we begin the report of various types of work and forms of free energy. Nosotros will find that some types of work exit the free energy of a system constant, for example, whereas others modify the arrangement in some way, such as making it move. We will besides develop definitions of important forms of energy, such as the energy of motion.

Internet Work and the Work-Free energy Theorem

We know from the study of Newton's laws in Dynamics: Force and Newton'due south Laws of Motion that internet force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process nosotros will as well detect an expression for the free energy of move.

Permit us start by considering the total, or net, work washed on a system. Internet work is divers to be the sum of work done by all external forces—that is, net work is the piece of work done by the net external force F _net. In equation form, this is W _internet =F _internet d cosθ where θ is the angle between the force vector and the displacement vector.

Figure 2a shows a graph of forcefulness versus displacement for the component of the forcefulness in the direction of the displacement—that is, anF cosθ vs. d graph. In this instance,F cosθ is constant. You lot can see that the area under the graph isFd cosθ, or the work done. Figure 2b shows a more general procedure where the force varies. The area under the curve is divided into strips, each having an average force (F cosθ) _{i (ave)}. The work done is (F cosθ) _{i (ave)} d_i  for each strip, and the full work washed is the sum of the Westward _i . Thus the total piece of work washed is the total expanse under the curve, a useful holding to which we shall refer later.

Figure 2. (a) A graph of F cosθ vs. d, when F cosθ is abiding. The area nether the bend represents the piece of work done past the force. (b) A graph of F cosθ vs. d in which the force varies. The work washed for each interval is the area of each strip; thus, the total expanse under the curve equals the total work washed.

Net piece of work will be simpler to examine if nosotros consider a one-dimensional state of affairs where a strength is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller chugalug conveyor system shown in Effigy 3.

Figure iii. A packet on a roller chugalug is pushed horizontally through a distance d.

The force of gravity and the normal force acting on the package are perpendicular to the deportation and do no piece of work. Moreover, they are also equal in magnitude and opposite in direction and so they cancel in calculating the cyberspace force. The net strength arises solely from the horizontal applied strength F _app and the horizontal friction force f. Thus, every bit expected, the net force is parallel to the displacement, so that θ= 0º and cosθ = 1, and the cyberspace work is given pastW _net =F _net d.

The effect of the net force F _net is to accelerate the package from v ₀ to v. The kinetic free energy of the parcel increases, indicating that the net work done on the system is positive. (Come across Example 1.) By using Newton's second law, and doing some algebra, we can accomplish an interesting conclusion. Substituting F _net =ma  from Newton'due south second constabulary givesW _net =mad.

To go a relationship between net piece of work and the speed given to a arrangement by the net force acting on it, nosotros take d=10−x ₀ and use the equation studied in Motility Equations for Constant Acceleration in One Dimension for the change in speed over a distance d if the acceleration has the abiding value a; namely, 5 ² =v ₀ ⁱⁱ + 2ad (note that a appears in the expression for the net work). Solving for acceleration gives [latex]\displaystyle{a}=\frac{five^ii-v_0^two}{2d}\\[/latex]. When a is substituted into the preceding expression for W _net, nosotros obtain

[latex]\displaystyle{W}_{\text{net}}=g\left(\frac{v^two-v_0^ii}{2d}\right)d\\[/latex]

The d cancels, and we rearrange this to obtain

[latex]W=\frac{one}{2}mv^2-\frac{i}{2}mv_0^2\\[/latex].

This expression is called the work-energy theorem, and it actually applies in full general (even for forces that vary in direction and magnitude), although we have derived information technology for the special instance of a abiding force parallel to the displacement. The theorem implies that the internet work on a system equals the change in the quantity [latex]\frac{1}{2}mv^2\\[/latex]. This quantity is our outset case of a form of energy.

The Work-Free energy Theorem

The net work on a system equals the change in the quantity [latex]\frac{1}{2}mv^ii\\[/latex].

[latex]W_{\text{net}}=\frac{ane}{two}mv^two-\frac{1}{2}mv_0^two\\[/latex]

The quantity [latex]\frac{ane}{ii}mv^2\\[/latex] in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass thousand moving at a speed five. (Translational kinetic free energy is distinct from rotational kinetic energy, which is considered subsequently.) In equation class, the translational kinetic energy, [latex]\text{KE}=\frac{1}{2}mv^two\\[/latex], is the energy associated with translational motion. Kinetic energy is a form of energy associated with the move of a particle, single body, or system of objects moving together.

We are aware that information technology takes free energy to become an object, like a motorcar or the packet in Figure three, up to speed, but information technology may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a auto traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.

Example 1. Calculating the Kinetic Free energy of a Bundle

Suppose a 30.0-kg package on the roller belt conveyor system in Figure 3 is moving at 0.500 k/southward. What is its kinetic energy?

Strategy

Because the mass m and speed v are given, the kinetic energy can be calculated from its definition as given in the equation [latex]\text{KE}=\frac{1}{ii}mv^two\\[/latex].

Solution

The kinetic energy is given past [latex]\text{KE}=\frac{1}{2}mv^2\\[/latex].

Entering known values gives KE = 0.five (30.0 kg)(0.500 k/s)ⁱⁱ, which yields

KE = 3.75 kg ⋅ thousand²/south² = 3.75 J.

Give-and-take

Annotation that the unit of kinetic energy is the joule, the same as the unit of work, every bit mentioned when work was commencement defined. It is also interesting that, although this is a fairly massive packet, its kinetic energy is not large at this relatively low speed. This fact is consistent with the ascertainment that people can move packages like this without exhausting themselves.

Instance 2. Determining the Work to Advance a Package

Suppose that you button on the thirty.0-kg bundle in Figure 3 with a constant force of 120 North through a distance of 0.800 thou, and that the opposing friction force averages 5.00 N.

1. Calculate the net work done on the parcel.
2. Solve the same problem equally in part ane, this time past finding the work done by each force that contributes to the net force.

Strategy and Concept for Part 1

This is a motion in one dimension problem, because the downward force (from the weight of the packet) and the normal force have equal magnitude and contrary direction, and so that they abolish in calculating the net strength, while the practical force, friction, and the displacement are all horizontal. (See Figure 3.) As expected, the net piece of work is the internet force times altitude.

Solution for Function 1

The cyberspace force is the push force minus friction, or F _net = 120 N – 5.00 Northward = 115 N. Thus the internet work is

[latex]\begin{array}{lll}W_{\text{net}}&=&F_{\text{net}}d=(115\text{ N})(0.800\text{ thou})\\\text{ }&=&9.20\text{ N}\cdot{\text{m}}=92.0\text{ J}\end{array}\\[/latex]

Word for Function 1

This value is the net work washed on the parcel. The person really does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal free energy. The cyberspace work equals the sum of the work done by each private force.

Strategy and Concept for Function 2

The forces acting on the packet are gravity, the normal force, the forcefulness of friction, and the applied strength. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

Solution for Part 2

The applied force does piece of work.

[latex]\begin{assortment}{lll}W_{\text{app}}&=&F_{\text{app}}d(\cos0^{\circ})=F_{\text{app}}d\\\text{ }&=&(120\text{ N})(0.800\text{ one thousand})\\\text{ }&=&96.0\text{ J}\end{array}\\[/latex]

The friction force and displacement are in contrary directions, and then that θ=180º, and the work washed past friction is

[latex]\begin{array}{lll}W_{\text{fr}}&=&F_{\text{fr}}d(\cos180^{\circ})=F_{\text{fr}}d\\\text{ }&=&-(5.00\text{ North})(0.800\text{ m})\\\text{ }&=&-4.00\text{ J}\end{array}\\[/latex]

So the amounts of work done past gravity, by the normal force, past the practical force, and by friction are, respectively,

[latex]\begin{array}{lll}W_{\text{gr}}&=&0,\\W_{\text{Northward}}&=&0,\\W_{\text{app}}&=&96.0\text{ J},\\W_{\text{fr}}&=&-4.00.\text{ J}\finish{array}\\[/latex]

The full work done equally the sum of the piece of work done by each force is and then seen to beW _total =W _gr +W _N +Due west _app +West _fr = 92.0 J.

Discussion for Role 2

The calculated total work West _total as the sum of the work by each force agrees, as expected, with the work W _net done by the net strength. The work washed by a collection of forces acting on an object can be calculated past either arroyo.

Example 3. Determining Speed from Work and Energy

Find the speed of the packet in Figure iii at the end of the push button, using work and energy concepts.

Strategy

Here the piece of work-energy theorem can be used, because we accept simply calculated the internet piece of work, W _net, and the initial kinetic energy, [latex]\frac{1}{2}mv_0^2\\[/latex]. These calculations allow us to detect the terminal kinetic energy, [latex]\frac{1}{2}mv^2\\[/latex], and thus the final speed 5.

Solution

The work-energy theorem in equation form is

[latex]W_{\text{net}}=\frac{1}{2}mv^2-\frac{ane}{2}mv_0^two\\[/latex].

Solving for

[latex]\frac{1}{two}mv^ii\\[/latex] gives [latex]\frac{1}{2}mv^2=W_{\text{internet}}+\frac{1}{ii}mv_0^2\\[/latex].

Thus,

[latex]\frac{1}{2}mv^2=92.0\text{ J}+iii.75\text{ J}=95.75\text{ J}\\[/latex].

Solving for the terminal speed equally requested and entering known values gives

[latex]\begin{array}{lll}v&=&\sqrt{\frac{ii(95.75\text{ J})}{\text{m}}}=\sqrt{\frac{191.v\text{ kg}\cdot\text{thou}^two\text{south}^ii}{30.0 \text{ kg}}}\\\text{ }&=&2.53\text{ m/due south}\end{assortment}\\[/latex]

Word

Using work and energy, nosotros non only arrive at an answer, nosotros run across that the last kinetic energy is the sum of the initial kinetic energy and the net work washed on the package. This means that the piece of work indeed adds to the energy of the package.

Example 4. Piece of work and Free energy Can Reveal Distance, Too

How far does the package in Figure iii coast subsequently the push, assuming friction remains abiding? Employ work and energy considerations.

Strategy

We know that one time the person stops pushing, friction will bring the package to residuum. In terms of energy, friction does negative piece of work until it has removed all of the parcel's kinetic energy. The piece of work done past friction is the forcefulness of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.

Solution

The normal force and force of gravity cancel in computing the net strength. The horizontal friction strength is then the net forcefulness, and information technology acts reverse to the displacement, and then θ= 180º. To reduce the kinetic energy of the package to zip, the work Westward _fr by friction must be minus the kinetic energy that the package started with plus what the packet accumulated due to the pushing. Thus W _fr=−95.75 J. Furthermore, W _fr = f d′ cosθ = −fd′, where d′ is the distance it takes to finish. Thus,

[latex]\displaystyle{d}\prime number=-\frac{W_{\text{fr}}}{f}=-\frac{-95.75\text{ J}}{5.00\text{ N}}\\[/latex]

thend′=19.two m.

Discussion

This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the piece of work done by friction is negative (the forcefulness is in the opposite direction of motion), so it removes the kinetic energy.

Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this state of affairs. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics lonely.

Section Summary

• The cyberspace work W _internet is the work done past the net force interim on an object.
• Piece of work done on an object transfers energy to the object.
• The translational kinetic free energy of an object of massm moving at speed v is [latex]\text{KE}=\frac{1}{2}mv^{2}\\[/latex].
• The work-energy theorem states that the internet workW _net on a system changes its kinetic free energy, [latex]{West}_{\text{internet}}=\frac{1}{two}mv^{two}-\frac{ane}{2}{mv}_0^ii\\[/latex] .

Conceptual Questions

1. The person in Figure 4 does work on the lawn mower. Nether what conditions would the mower gain free energy? Under what weather condition would it lose energy?

   

   Figure iv.

2. A person pushing a lawn mower with a strength F. Force is represented by a vector making an angle theta below the horizontal and altitude moved by the mover is represented past vector d. The component of vector F along vector d is F cosine theta. Work done by the person, W, is equal to F d cosine theta.
   Work washed on a system puts free energy into information technology. Work done by a system removes energy from information technology. Give an example for each argument.
3. When solving for speed in Example three, nosotros kept only the positive root. Why?

Issues & Exercises

1. Compare the kinetic energy of a xx,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.
2. (a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the motility of larger animals would relate to metabolic rates.
3. What is the value for the kinetic energy of a 90,000-ton shipping carrier at 30 knots? You will need to look up the definition of a nautical mile (1 knot = 1 nautical mile/h).
4. (a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the automobile hits a concrete abutment at total speed and is brought to a end in ii.00 m. Calculate the force exerted on the car and compare information technology with the force found in office (a).
5. A car'due south bumper is designed to withstand a 4.0-km/h (1.one-m/due south) collision with an immovable object without damage to the body of the car. The bumper cushions the stupor by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 1000 while bringing a 900-kg car to rest from an initial speed of 1.1 one thousand/s.
6. Battle gloves are padded to lessen the force of a blow. (a) Summate the force exerted by a boxing glove on an opponent'due south face, if the glove and confront compress 7.50 cm during a accident in which the 7.00-kg arm and glove are brought to balance from an initial speed of 10.0 m/s. (b) Calculate the forcefulness exerted by an identical blow in the gory quondam days when no gloves were used and the knuckles and face would shrink only two.00 cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to crusade harm even though information technology is lower than the force with no glove?
7. Using free energy considerations, calculate the average force a sixty.0-kg sprinter exerts astern on the track to accelerate from 2.00 to 8.00 m/south in a distance of 25.0 thousand, if he encounters a headwind that exerts an average force of 30.0 N against him.

Glossary

net piece of work: work done by the cyberspace force, or vector sum of all the forces, acting on an object

piece of work-free energy theorem: the result, based on Newton's laws, that the cyberspace work done on an object is equal to its change in kinetic free energy

kinetic energy: the energy an object has by reason of its motion, equal to [latex]\frac{1}{2}{\text{mv}}^{ii}\\[/latex] for the translational (i.due east., non-rotational) motion of an object of mass m moving at speed v

Selected Solutions for Bug & Exercises

one. [latex]\frac{1}{250}\\[/latex]

iii. 1.1 × 10¹⁰

5. two.8 × 10³ Due north

7. 102 N

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