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The Law of Conservation of Energy states simply that the energy of a system is a constant, even if energy is transformed from one form to another within the system. In other words, the energy put into a system will equal the extra energy that will be available to do work by the system. The energy in a system is always accounted for; it never disappears. What the Law of Conservation of Energy does not state is that all that energy will be available to do work. In fact, in most cases, when an energy transformation occurs, some of the energy transforms into heat because of friction, air resistance, etc., a useless form of energy that cannot be harnessed effectively by humans (in most cases). We will deal first with the idealized situation in which all the energy in an energy transformation changes into a useful form such as kinetic energy or gravitational potential energy. Suppose a system is at rest (meaning no work has been done on it - in other words, a force through a distance has not been applied to it). If we do work on the system, we expend energy. This energy cannot disappear, and it doesn't. The energy is transferred to the system (it might even change forms). In formula form, any of the following apply: Win = Wout Ein = Eout Einitial = Efinal What does "work on the system" mean? EXAMPLE: Consider the life of a rock. The rock has no purpose in life if it is on the ground. If however, it is raised to a higher plane, it is given (gravitational) potential energy (GPE). How did it get up to that height? WORK (W) was done on it. Someone or something must have applied a force through a distance (W = F x d). Now the rock has the "potential" of doing something in life (like falling)! In formula form: h W in = GPEout F x d = mgh Suppose that you exert a force of 490 N upwards to lift the rock up to the ledge 3 meters above. By exerting a force of 490 Newtons through a distance of 3 meters, the following amount of work is done: F = 490 J Win =F x d = 490 N x 3 m = 1470 J How much GPE will the rock have after being lifted? Ein = Eout Win = GPEout 1470 J = GPEout Why, 1470 J, of course! The energy from Work has to go somewhere… Now suppose you wanted to find out the mass of the rock. Well: 1470 J = GPEout and…1470 J = mghout (because GPE = mgh) Solving for "m" and plugging in what we know - gravity is 9.8 m/s2 and the height raised is 3 m 1470 m gh 1470 m (9.8 m 2 )(3 m) s 50 kg = m PROBLEM SET 1: 1. 35 Joules of work is done to raise a 3 kg boulder to the top of a cliff. How high is the cliff? 2. If a crate of Naranjilla of mass 14 kg is raised 10 m, how much work must be done to accomplish this task? CONSERVATION OF ENERGY can be expressed in other forms besides Win = GPEout.. The energy can change from Potential Energy (energy due to a change in state) to Kinetic Energy (energy due to motion), for example. Following is an extremely effective strategy for solving Conservation of Energy problems: (a.) find from the problem statement what the initial state and final states of energy are; (b.) identify what quantity/variable the problem is asking you to find; (c.) identify and label your given information/values; (d.) set up the Conservation of Energy statement: Ein = Eout; decide what form(s) of energy there are initially and finally (e.) solve for the quantity in question (isolate it) without plugging in any values (numbers) given; (f.) plug in your given values once you have the quantity in question and solve; (g.) don't forget your units (kgs, Joules, m/s); Consider how energy is conserved in the following systems although it changes form: a. (a.) (b.) Ep Ep The system is at rest. No work has been done on the system. Kinetic energy and potential energy are zero. (a.) b. A hand has done work on the sytem. The work done equals the amount of elastic potential energy given to the spring (b.) Ek Ek c. (c.) (d.) Ep Ep Ek Ek (e.) Ep (f.) (g.) Ep Ep Ek Ek Ek Ep The spring is released. Some of the EPE of the spring transforms into TKE giving the spring a velocity. The total energy of the system: EPE + TKE remains constant. f-g. The spring oscillates back and forth in an endless cycle in which energy, although alternating between EPE and TKE, remains constant. b. A hand has done work on the system. The work done equals the amount of gravitational potential energy given to the pendulum bob. Ep Ek Ep Ek c. The pendulum bob is released. Some of the GPE of the pendulum bob transforms into TKE giving the pendulum a velocity (the pendulum has gained velocity but lost height). The total energy of the system: GPE + TKE remains constant. d. The pendulum, at the bottom of the swing, has only TKE (it is at maximum velocity) and no GPE (because it has no height). Consequently, the pendulum has reached its maximum velocity. e. The pendulum swings passed the middle and gains some GPE (because it has height). It begins to slow down (loses velocity) and so its TKE begins to lessen. TKE + GPE, however, is still a constant. (c.) d. The spring reaches equilibrium position. All of its energy is TKE. The spring experiences no forces (either to stretch it or compress it) for the moment. e. The spring is completely compressed and all of its energy becomes EPE again. The spring has no velocity therefore, the TKE is zero. Ek a. The system is at rest. No work has been done on the system. Kinetic energy and potential energy are zero. (d.) Ep Ek Ep Ek Ep Ek (e.) (f.) (g.) Ep Ek Ep Ek (h.) f.-h. The pendulum oscillates back and forth in an endless cycle in which energy, although alternating between GPE and TKE, remains constant. PROBLEM SET 2: 1. How does a system "at rest" gain energy? 2. If you do 25 Joules of WORK stretching a spring, (a.) how much elastic potential energy will it have available? (b.) what is the spring constant k of that spring if it was stretched 1.2 m? (c.) what is the fastest speed the spring will have when it is released? (d.) what would the elastic potential energy be when the spring is moving at its fastest speed? 3. If a pendulum bob has a mass of 2.5 kg, and its maximum velocity is 8 m/s, (a.) from what height was it released? (b.) what is the height at maximum velocity? (c.) do you need the mass to solve the problems above? Explain. (d.) how high is the pendulum bob when its velocity is 2 m/s? (e.) how fast is the pendulum bob moving when the height is 0.75 m? PROBLEM SET 3: 1. A 100 kg man jumps out of a burning building into a fire rescue net 15 m below. The net stretches 2.5 m before bringing the man momentarily to rest (and then tossing him back up). What is the potential energy of the stretched net if mechanical energy is conserved? icecube 2. An ice cube is released from the edge of a hemispherical frictionless bowl whose radius is 0.22 m. How fast is the ice cube moving at the bottom of the bowl? r 3. A frictionless roller-coaster car tops the first hill in the Figure below with speed vo m/s. What is the speed at (a.) (b.) (c.) (d.) Point A? Point B? Point C? How high will the car go on the last hill, which is too high for it to cross? vo first hill A B h h h/2 C 4. The figure at right shows an 8 kg stone resting on a spring. The spring is compressed 0.10 m by the stone. (a.) What is the spring constant k of the spring? (b.) If the stone is pushed down an additional 30 cm. and released, what is the potential energy of the compressed spring just before that release? (c.) How high above the release position will the stone rise? 5. A ball with a mass of m kg is attached to the end of a very light rod with B l A length "l" meters and a negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle (see Figure). The rod is held in the horizontal position and then given enough of a downward push so that the ball swings down and around and just reaches the vertically upright position, having zero speed there. (a.) What is the change in the potential energy of the ball from Point A to Point B? (b.) What initial speed must have been given to the ball? THE REAL WORLD Up to now we've been considering situations in IDEAL CONDITIONS where no friction or air resistance applies. We will now explore conservation of energy and its formulas in a more realistic scenario: when part of the energy of the system transforms into heat. When two objects are in contact or when the particles (or molecules) of a substance pass each other by or move, heat is generated. In the real world, heat is a byproduct of work and, when produced, is normally lost or diffused into the surrounding environment. Since it is lost, this heat generated is sometimes called "useless" energy. Nevertheless, it IS energy and must somehow figure into our formula for conservation of energy: Ein = Eout. Let us see how… EXAMPLE: If a block-spring system is set into motion on a surface on which friction exists, the back and forth oscillations will diminish until the blockspring system is at rest again. The mechanical energy has decreased and if we had a sensitive thermometer nearby, we would see that the temperature in the environment has risen, and consequently that heat energy has been transferred. The increase in the amount of heat energy will be precisely the decrease in the amount of mechanical energy. A byproduct of the work done by friction against the motion is heat! We will call this change in the thermal energy of the system Q…and we will stick it into our conservation of energy formula: EPEin input useful energy = EPEout + TKEout output useful energy + Q by-product useless energy EXAMPLE: A steel ball whose mass is 0.52 kg is fired vertically downward from a height h of 18 m with an initial speed vo of 14 m/s. If the ball hits the ground with a force of 2500 N, (a.) what is the total mechanical energy available to the ball right before it hits the ground? (b.) in the absence of friction/resistance, how far should the steel ball be able to penetrate into the surface? (c.) if the ball penetrates a distance of 0.025 m into the ground, how much of the ball's energy became heat? h = 18 m ? Ans. (a.) When shot, the ball has both TKE and GPE. Right before it hits the ground, all the ball's energy is kinetic. To find the total kinetic energy right before the ball hits the ground use the COE formula: Ein = Eout TKEin + GPEin = TKEout mghin + ½mvin2 = TKEout (0.52 kg)(9.8 m/s2)(18m) + ½(0.52 kg)(14 m/s)2 = TKEout 142.688 J = TKEout Ans. (b.) In the absence of friction/resistance, in other words, in the ideal situation, the steel ball does work penetrating the ground and that's it. So… By formula: Ein = Eout TKEin = Wout TKEin = F x d TKE in d F 142.688 J d 2500 N 0.0570752 m = d Ans. (c.) Since the ball penetrated somewhat less distance than it did originally, some of the ball's energy must have dissipated as heat. We need to find out with how much: Ein = Eout TKEin = Wout + Qout TKEin = F x d + Qout TKE in Q out F x d 142.688 J Q out 2500 x (0.025 m) 142.688 J Q out 62.5 m) 2.283008 J = Qout PROBLEM SET 4: 1. A dog of mass 6.0 kg runs onto a curved ramp as shown in the figure at right with a speed of 7.8 m/s. The initial perro height is 8.5 m. The dog slides down the ramp and comes to a momentary stop when he reaches a height of 11.1 m h1 from the floor at the other end of the ramp. What is the increase in heat energy of the dog/ramp due to the sliding? vo h2 2. A 0.075 kg FRISBEE is thrown from a height of 1.1 m above the ground with a speed of 12 m/s. When it has reached a height of 2.1 m, its speed is 10.5 m/s. (a.) How much gravitational potential energy did the FRISBEE gain? (b.) How much of the FRISBEE'S mechanical energy was dissipated by air drag? v = 12 m/s h = 2.1 m h = 1.1 m 3. A baseball player throws a baseball with an initial speed of 20 m/s. Just before the catcher catches the ball at the same height, the ball's speed has been reduced to 11.0 m/s. By how much did air drag reduce the mechanical energy of the ball? (the baseball has a mass of 0.09 kg). 4. A river descends 15 m through rapids. The speed of the river is 3.2 m/s upon entering the rapids and 13 m/s as it leaves the rapids. Consider 10 kg of water going down that river… (a.) What percentage of gravitational potential energy lost by the 10 kg of water becomes kinetic energy of water downstream? (b.) What happens to the rest of the energy? 5. Fozzy, a 25 kg bear slides, from rest, 12 m down a pine tree. Right before the bear hits the ground, it is moving with a speed of 5. 6 m/s. (a.) How much gravitational potential energy does Fozzy lose before hitting the ground? (b.) How much mechanical energy dissipates due to friction?