objectives:
1. Describe the concept of work.4.1. Introduction:
2. Describe the concepts of kinetic and potential energy.
3. Become familiar with various forms of energy.
4. Perform simple calculations involving energy conservation.
When the motion of an object acted on by a force is alreadyparallel to that force, the work done by the force is easily computed.For example, as a rock weighing 5 newtons falls 2 meters to the ground,the gravity force does an amount of work (5 N)´(2 m) = 10 J. But if the force and the motion are not parallel, the computationis harder and requires the use of trigonometry. Let us assume that a cartis constrained to move on a straight track. A 10-N force is applied continuouslyto the cart as it moves along the track a distance of 2 m. If the 10-Nforce is applied parallel to the track, the work done by the force is (10N)´ (2 m) = 20 J. On the other hand, ifthe force is perpendicular to the track, there is no net motion in thisperpendicular direction, and the work done by this force is zero. If theforce is at an angle with respect to the track, we first find the amountof movement parallel to the force and then multiply this component valueby the force. Figure 2-1 shows how to find this. First, a "line" is drawnalong the direction of the force at the starting point. Then a second "line"is drawn through the final position of the object so that this second lineis perpendicular to the first line. The effective parallel distance isthen the distance from the starting position to the second "line," as shown.Those readers who know about vectors or trigonometry may recognize thisas the component of the cart's displacement that is parallel to the directionof the force. If the force is opposite in direction to the motion, i.e.,if the cart above backs up as the force is applied, the work is considerednegative.
Case 2 A completely different effect of the workmight be to lift the book against the force of gravity, resulting not ina change in speed but rather in placing the book in an elevated positionso that the energy is "stored." Under this circ*mstance, the book is saidto have gravitational potential energy PE. The work done to lift the bookat constant velocity is the weight of the book (mg) times the height inmeters through which the book is lifted (h). This work is stored as gravitationalpotential energy PE = mgh, where g is the acceleration of gravity (g =9.8 m/s2). The energy is stored in thesense that if the book is dropped from its new elevated position, it willhave a KE of exactly the same amount when it reaches its original positionas if the work had gone directly into motion. This fact is a manifestationof the law of conservation of energy. The stored work is availableto be transferred without loss into an equivalent amount of KE.
Energy also can be stored in the form of chemical bondingof the atoms or molecules. The forces that act between atoms are complex.At certain separation distances the atoms repel each other, whereas atother separation distances the same atoms attract each other. This canresult in two or more atoms being bound together in a semi-stable configuration(i.e., a molecule) that when disturbed can be broken apart. When the bondsare broken, the atoms fly apart and release the same amount of energy asthe work originally done by Nature to build the molecule. The burning ofwood is an example. Finally, energy can be stored as temporary excitationsof the atoms or the molecules into their allowed energy states.
In the natural world, energy constantly changes from oneform to another. If this were not so, the world would be stagnant and lifeless.The truly wonderful principle of conservation of energy is a statementthat it is possible to design a conceptual accounting system in which theenergy flow can be tracked. The total energy contained within the systemstays constant unless external work is done on the system to add or subtractenergy. For a completely isolated system to which no work or other formsof energy are added, energy is said to be conserved. When energy is lostto one part of the system, we can be sure that the exact same amount ofenergy will reappear in another part of the system, although it may bein a different form. This is indeed a powerful reasoning tool. Mathematically,the law of energy conservation is
,
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iszero. This means that all of the KE gained by an object comes from a lossin gravitational PE.Some of the incident solar energy is reflected back outand has no effect on the earth's energy balance. Only the part of the solarenergy that is absorbed represents energy added. Over the long haul, inorder for the earth and its atmosphere to be in equilibrium, equal amountsof energy must be radiated back out to space. If more energy comes in thangoes out, Earth's average temperature would have to rise, unless the excesscould be stored (for example, by melting some of the ice cap). The energyis cycled in numerous ways before eventually being radiated back out tospace. Some of the energy, especially the X-ray and UV parts of the spectrum,is absorbed at great heights and heats the upper atmosphere. The visibleand much of the IR penetrate to the surface and heat the surface. The warmedsurface then re-radiates IR energy, which in turn is absorbed by watervapor in the air (and to some extent by carbon dioxide) to heat the lowerpart of the atmosphere. The warmed lower atmosphere re-emits energy, partof which gets re absorbed by the surface to be recycled and part of whichfinds its way back out to space. The cycling of energy is indeed a veryinvolved process that involves much detail. But the net effect is to poweratmospheric motions and cause weather systems and is a very important ideato keep in mind as we work our way through the course.
2-2. How much work is done when a 2.0-kg book is liftedfrom one end of a table carried to the other end 3.0 meters away and placedback on the table?
2-3. A 5.0-kg box slides at constant speed on a frictionless,horizontal tabletop. The gravity force, of course, pulls downward on thebox with a force of mg = (5 kg)´(9.8 m/s2) = 49 N. As the boxmoves 1/2 meter along the tabletop, what amount of work is done by thegravity force?
2-4. A box is pushed by a force F = 20 N up theincline shown.
b) What is the change in gravitational PE?
c) What must be the increase in KE of the box?
2-5. A woman who has a mass of 60 kg weighs 590 N.(Do you know why this is true?) If she climbs a flight of stairs 6 metershigh in 5 seconds, she is expending energy at a power level of
(1) 300 watts (2) 500 watts (3) 700 watts (4) 900 watts(5) 1,200 watts
2-6. A food calorie, also called a kilocalorie (kcal)is actually a form of chemical potential energy. (1.0 kcal = 4,184 J).A peanut butter sandwich contains about 300 kcal, assuming a normal amountof peanut butter. Estimate how many peanut butter sandwiches the womanof problem 5 must eat in order to supply her with enough energy to climba 1 km high mountain. Assume that half the food she eats goes into justmaintaining normal body function as she climbs.
2-7. One parcel of air having a relatively large densityis next to a second parcel of air having a lesser density. One would expectthe higher density air to tend to undermine the lesser density air so thatwhen all has settled down, the higher density air is on the bottom andthe lesser density air is on the top. Which one or more of the followingare true?
b) The lighter air gains PE as it rises above the denserair.
c) There will be a net gain in PE for the system.
(1) a (2) b (3) c (4) a and c (5) b and c
2-8. Figures 2.14 and 2.15 are very useful to study carefully.Based on these figures answer the following questions.
b) Which is more important at the earth's surface, energyabsorbed directly from the sun or energy absorbed by ground from IR emittedby the atmosphere?
c) What is the single most important mechanism by whichthe atmosphere is warmed?
d) Assume you are out in space looking at the earth andare equipped with instruments to measure the various types of radiant energy..From what source does virtually all of the visible light you see come?From what source does most of the IR you see with your IR sensors come?
