CHAPTER 6: IN AN ISOLATED SYSTEM, ENERGY IS TRANSFERRED FROM ONE OBJECT TO ANOTHER WHENEVER WORK IS DONE 6.1 Work and Energy In science, work is done when a force acts over a displacement; energy is transferred. Both the force and displacement are necessary for work to be done, and the force must act in the direction of the displacement. The simplest formula for work is W = F d, with force in newtons (N), displacement in meters (m) and work in joules (J.) Because energy is transferred when work is done, it should seem reasonable that the units for work are the same as the units for energy. If the force doing the work acts at an angle to the resulting displacement, only the force component in the displacement direction does work. This gives the more general formula for work, with θ being the angle between the force and displacement: W = (F cos θ) d EXERCISE 6.1 1. A student lifts a box with a weight of 100 N from the floor (a distance of 0.850 m) and carries it horizontally a distance of 4.50 m across a room. Where is work done, in the lifting of the box, the carrying of the box, or both? How much work is done in total? (85.0 J) 2. Describe a situation where a force acts on an object but no work is done. Describe a second situation where an object undergoes a displacement but no work is done. 3. The sketch shows a wagon being that is pulled across a horizontal driveway by a child. The child pulls the handle of the wagon with a force of 55.0 N to overcome the force of kinetic friction acting against the motion. If the wagon moves a distance of 15.0 m, find the work done. 42Þ 46.0 (573 J) PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 1
4. When an object is lifted vertically against the force of gravity, work is done against gravity. If the object is released and allowed to fall, gravity does work on the object to increase its velocity and its kinetic energy (energy of motion.) Clearly, lifting an object against gravity stores energy in some form. Name this type of stored energy; state the equation including units. 5. Find the E p when a box with a mass of 25.0 kg is lifted from ground level to a height of 1.50 m. 6. Consider the sketch to the right. Calculate E p when the box shown is lifted from the ground level (II) to level III. Calculate the E p when the box is lifted from level III to level IV. Finally, calculate the E p when the box is lifted directly from the ground level to level IV. Does the total change in E p from ground level to level IV change if you go from ground level to level III and then to level IV? (368 J) 15.0 kg box IV 5.00 m 7.00 m III (736 J; 1.03 kj; 1.77 kj; 1.77 kj) GROUND II 7. Again consider the sketch to the right. What E p will the box have at level IV, measured with respect to level I? ( Measured with respect to identifies the starting or reference point.) What I E p will the box have at level III, measured with respect to level I? How does the change in E p between levels III and IV compare to the value you found in the previous question? Does it matter what you use as a reference point when calculating changes in E p? 3.00 m (2.21 kj; 1.18 kj; 1.03 kj) PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 2
Hooke s Law To stretch or compress a spring, force is required; because this force acts through a distance, work is done. For a linear spring (also called a Hooke s Law spring), proportionally more force is required to stretch a spring a greater distance. For example, as sketched at right, if a particular spring stretches 2.0 cm (from its normal length) when a force of 5.0 N is applied, it should stretch to 4.0 cm if the force is increased to 10 N, and to 6.0 cm for a force of 15 N. 0 2.0 cm 4.0 cm 6.0 cm 5.0 N Generally, the force needed to stretch a spring depends on how strong the spring is, and how far it 10 N is stretched. The strength of a spring is given by the spring constant (or elastic constant) k the value 15 N of k ranges from around 2 N/m for the spring in a retractable pen, to around 150 N/m for the spring cable at a bungie jump, to 2500 N/m for the suspension springs in a car. The relationship between force F (in newtons), stretch x (in meters) and constant k (in N/m) is given by Hooke s Law: F = kx When work is done to stretch a spring, energy is stored. This stored energy is called elastic potential energy. To calculate the stored energy in the spring above when it is stretched 6.0 cm, note that the average 0 +15 force needed between zero stretch and 6.0 cm is or 7.5 N; this average force moved the end of the 2 spring a distance of 6.0 cm (0.060 m.) So the work done in stretching the spring (and therefore the energy stored) is equal to (7.5)(0.060) or 0.45 J. Generally, for any spring, the elastic potential energy can be found in a similar manner. Work done = energy stored Energy stored = average force times stretch For a spring with constant k and stretch x from zero, the force at this stretch is given by F = kx. At zero stretch, the force is zero. So the work done (elastic potential energy stored) is: This is the average force; zero force at zero stretch added to force kx at stretch x, divided by two E p = 0 + kx x 2 E p = 1 2 kx2 This is x, or how far the force acts (the displacement in the work equation) EXAMPLE A spring with an elastic constant of 12.0 N/m is stretched an amount of 10.0 cm. How much force is needed to do this, and how much elastic potential energy is stored in the stretched spring? Use F = kx to find the force: F = (12.0)(0.100) F = 1.20 N Use E p = 1 2 kx2 to find the stored energy: Ep = 1 2 (12.0)(0.100)2 Ep = 0.0600 J PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 3
8. A spring is 25.0 cm long. When a force of 40.0 N is applied to it, the spring stretches by 20.0 cm. What is the elastic constant k of the spring? (200 N/m) 9. A spring-operated ping-pong ball launcher requires a force of 24.0 N to fully compress the spring, which has an elastic constant of 12.0 N/m. How far does the spring compress, and how much elastic potential energy is stored in the compressed spring? (66.7 cm; 2.67 J) 10. State the formula for translational kinetic energy (E k ), with correct units. What other form of kinetic energy is there? 11. Find the kinetic energy of a cyclist riding at 25.0 km/h, if the rider and his bicycle have a total mass of 86.0 kg. (2.07 kj) 12. An arrow in flight has a kinetic energy of 170 J. If the arrow has a mass of 65.0 g, what is its speed? (72.3 m/s) 13. A 1920 kg car brakes from a speed of 100 km/h to 50.0 km/h. How much does its kinetic energy change? Where do you suppose this energy goes? (556 kj) PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 4
6.2 Mechanical Energy EXERCISE 6.2 1. Name the three forms of energy referred to by mechanical energy. 2. A runner with a mass of 78.0 kg follows a horizontal trail along a cliff overlooking a lake. The trail is 13.0 m above the lake, and the runner is moving at 4.50 m/s. What is the runner s total mechanical energy? (The lake is the reference point.) (10.7 kj) To change the total mechanical energy of an object, it must lose energy by doing work on something, or something must do work on it to increase its energy. The amount the object s energy changes must equal the net work that is done. This is a law of energy conservation called the work-energy theorem, and can be stated as: The net work done on a system is equal to the sum of the changes in the potential energy (including gravitational and elastic) and in the kinetic energy of the system:s W = E k + E p. EXAMPLE A 2400 kg truck traveling at 18.0 m/s (about 65 km/h) reaches the bottom of a steep hill. As it moves up the hill, the truck s speed drops, finally to 11.1 m/s (about 40 km/h) at it reaches the top of the 35.0 m high hill. How much useful work does the truck s engine do from the bottom to the top of the hill? In driving up the hill, the truck gains gravitational potential energy, but loses kinetic energy. The total of these changes will be the useful work done by the engine. E k = 1 2 mv2 E k1 = 1 2 (2400)(18.0)2 E k2 = 1 2 (2400)(11.1)2 E p = (2400)(9.81)(35.0) E p = mgh E k1 = 388 800 J E k2 = 147 852 J E p = +824 040 J E k = E k2 - E k1 = -240 948 J W = E k + E p W = 240 948 + 824 040 W = +583 092 J W = +583 kj 3. A cyclist traveling on a horizontal roadway at 35.0 km/h brakes to a stop. If the cyclist and her bike have a total mass of 73.0 kg, how much work is done by forces of friction? (3.45 kj) PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 5
6.3 Mechanical Energy in Isolated and Non-isolated Systems In physics, one or more objects that have and exchange amounts of energy form what is called a system. If a system is defined or designed so that no mechanical energy can be exchanged with any other object outside the system, it is said to be an isolated system. (Recall from Science 10 that in a closed system nothing, no matter nor any form of energy, can be exchanged with the outside.) Isolated systems are frictionless (because friction would generate heat, which could leave the system) and have no net external forces acting on any object in the system. In such a carefully-defined isolated system, total mechanical energy is constant. That is, within the system, mechanical energy can be transferred from one form to another, or traded among the objects in the system, but the total number of joules of kinetic, gravitational potential and elastic potential remains the same. This is the law of conservation of mechanical energy. It s somewhat the same as the work-energy theorem above, but describes energy changes occurring within a system, rather than connecting work done by outside forces to changes in energy. Many common forces (like kinetic friction) cause mechanical energy to be transferred to another form, typically heat. Such forces are called non-conservative forces. Other forces (like the force of gravity) can act without changing the overall mechanical energy of a system; these are conservative forces and they are associated with isolated systems. (Note that a system that includes gravity must include the earth as one of the objects in the system.) In the real word, just about any system has some non-conservative forces that act, and is not completely isolated from its surroundings. To allow us to solve problems using the law of mechanical energy conservation, we often assume a system is isolated and has only conservative forces and we recognize that our answers might be slightly different if we accounted for energy losses due to friction, etc. Such assumptions are often indicated by the phrase friction can be neglected. Real systems involving non-conservative forces can also be analyzed. For example, if a known amount of energy is lost through frictional forces, the total amount of mechanical energy in the system at the start can be reduced by this amount, before equating total energy before to total energy after and solving for an unknown. In some problems, the amount of energy lost to friction may be the actual unknown. In summary, to solve energy conservation problems: Calculate the total amount of mechanical energy (kinetic plus gravitational potential and spring potential) at the start, before anything happens to objects in the system If a known amount of energy is lost through friction, subtract this value from the total at the start; if an unknown amount is lost and this is what you are trying to find, include a variable for this value in the energy equation (i.e., write E f on the before or starting side) Calculate the total amount of mechanical energy after something occurs Equate the two total energy values: energy before equals energy after (remember, mechanical energy is conserved) and solve for the unknown EXAMPLES 1. A car with a mass of 1250 kg is parked at the top of a 25.0 m long driveway. The end of the driveway where the car is parked is 0.850 m above the level of the street. Due to a mechanical malfunction, the car slips out of park, rolls down the driveway and across the street. Friction can be neglected. How fast is the car rolling as it crosses the street? At the start, the car has only gravitational potential energy. This will be the total before something (in this case, rolling down the driveway) occurs. When the car reaches the level street, only kinetic energy remains. This will be the total energy after. Start: E T = E p E p = mgh E p = (1250)(9.81)(0.850) E p = 10 423 J Across driveway: E T = E k E k = 1 2 mv2 E k = 1 2 (1250)v2 Make the two totals equal and solve: 10 423 = 1 2 (1250)v2 v = 4.0837 m/s v = 4.08 m/s PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 6
2. An arrow of mass 85.0 g is fired straight upwards with an initial speed of 48.0 m/s. If friction is neglected, find how high it goes. Equate the kinetic energy at the start (before) to the gravitational potential energy at the highest point (after remember that at the highest point, the arrow will have no kinetic energy why?) and solve for h. E k = 1 2 mv2 1 = E p = mgh 2 (0.0850)(48.0)2 = (0.085)(9.81)h h = 117.43 m h = 117 m 3. The arrow in the previous question actually rises to a measured height of 76.0 m. How much energy is lost through air resistance (a kind of frictional force) as the arrow rises? How fast should the arrow be traveling after it falls back down to ground level? E k E f = E p 1 2 (0.0850)(48.0)2 E f = (0.085)(9.81)(75.0) E f = 35.381 J E f = 35.4 J To find the speed on the way back down, note that the arrow will lose an additional 35.381 J as it falls back to the ground. Use the highest point at 76.0 m as the start (before energy will all be gravitational potential energy) and calculate after at ground level (when energy less that lost to friction will all be kinetic energy.) E p E f = E k (0.085)(9.81)(75.0) (35.381) = 1 2 (0.0850)v2 v = 25.278 m/s v = 25.3 m/s EXERCISE 6.3 1. A flowerpot of mass 2.20 kg falls from a second floor window to the ground 7.50 m below. Friction can be neglected. How fast was the flowerpot traveling just before it hit the ground? (12.1 m/s) 2. A pendulum with a mass of 250 g is released from rest from a height of 10.0 cm above its lowest point. How fast will the pendulum be moving as it passes through the lowest point? (1.40 m/s) 3. A 5.00 g pellet is placed in the barrel of a toy gun and is propelled by a spring with a force constant of 50.0 N/m that has been compressed 20.0 cm and then released. Friction can be neglected. a) Calculate the maximum velocity of the pellet when it is fired horizontally. b) If the pellet is fired vertically, how high will it go? (20.0 m/s) PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 7
(20.4 m) c) If 0.350 J of mechanical energy are lost from the system due to air resistance, how high will the pellet go? (13.3 m) 4. A frictionless roller coaster car arrives at point X at right with a speed of 1.60 m/s. The car plus occupants has a mass of 545 kg. Find the speed of the car as it passes point Y and point Z. X 33.5 m 26.0 m Z Y (25.7 m/s; 12.2 m/s) 5. The speed of the roller coaster car in the previous question is actually 9.40 m/s at point Z. How much energy is lost due to frictional forces between points X and Z? 6.4 Work and Power (16.7 kj) The same amount of energy can be transferred over a short or long amount of time. The rate of energy transfer is power; in SI, power is measured in watts (W.) One watt is a rate of energy transfer of one joule every second. Historically in the British system of measurement (still in common use in the United States, and also still found in Canada) the unit for power is the horsepower (HP.) These units can be converted using the equivalence 1.00 HP = 746 W. P(watts) = W(joules) time(seconds) or P(watts) = E(joules) time(seconds) ; P = E t EXERCISE 6.4 PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 8
1. The crane shown lifts a load of logs with a mass of 450 kg a vertical distance of 3.50 m in a time of 10.5 s. Find the power developed by the crane s engine in lifting the logs, in watts and horsepower. (1.47 kw; 1.97 HP) 2. A off-road truck with a total mass of 3250 kg travels at a constant speed from the bottom to the top of a hill in a time of 1.40 minutes. If the top of the hill is 22.5 m higher than the bottom, find the power of the truck s engine in watts and horsepower. (Note that non-conservative forces are acting in this problem; the actual work done by the engine and the power produced are definitely larger than the value you found.) (8.54 kw; 11.5 HP) 3. In question #2, suppose the truck started at the bottom of the hill with a speed of 7.00 m/s and reached the top with a speed of 13.6 m/s. Find the power produced by the truck in watts. (11.2 kw) 4. Two students helping a neighbor drag a small refrigerator on a pallet a horizontal distance of 21.5 m from the door of a house to the back of a truck. The total horizontal force the students exert is 855 N. a) If moving the refrigerator takes 3.50 minutes, how much power do the students generate? (87.5 W) b) The refrigerator has a mass of 44.0 kg. If the students lift the refrigerator from the pallet into the truck through a vertical distance of 75.0 cm, how much work do they do? If lifting the refrigerator takes 2.50 s, what power is generated? (324 J; 129 W) PHYSICS 20N NOTES AND OUTLINE QUESTIONS CHAPTER 6 REVISED JANUARY 08 PAGE 9