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Chapter 10 Glencoe Physics Energy, Work, and Simple Machines 2006 A. What is work? • 1. Webster’s Definitions – a. “energy expended by a natural phenomenum” – b. “activity in which one exerts strength to do something” 2. Physics Definition of Work • a. Product of a force acting on an object and the displacement of the object in the direction of the force • b. W = F s – (1) work is a scalar quantity – (2) displacement is critical - w/o movement no work is done c. SI unit of work is the Newton-meter or JOULE – (1) A joule is the work done by a force of one newton as it acts through a distance of one meter along the line of the force – (2) Also use the joule to measure energy • d. Other units of work include the foot- pound, the erg, and the electron-volt (eV) 3. Examples of work calculations • a. A person exerts a vertical force of 30.0N on a pail as the pail is carried a horizontal distance of 8.0 m at a constant speed. How much work does the 30N force do on the pail? • You do no work, as the force is not along the lines of the distance moved. It would have to have been moved vertically for work to be done. b. How much work do you do on an object of weight mg as: • (1) you lift it a distance of h meters straight up at constant speed; and, (2) you lower it through this same distance at a constant speed? • To lift you pull up on the object with a force F = mg to counteract the weight – W=F s=mgh • When you lower the object F and s are in opposite directions, so W = m g (-h) = -mgh 4. Work and the Direction of Force • a. Only the component of the force in the direction of movement does work. The component of force perpendicular to the direction of motion does NO work. • Key Point to Remember - What is critical is to determine the component of force in direction of motion b. Note as seen in a previous example you can have “negative” work • (1) although work is a scalar quantity, negative work means the force is applied in the opposite direction from the object’s motion. • (2) Work is also cumulative, so negative work will offset positive work. • c. Example. Motion is in the x -direction Fy F Fx Motion s w = Fx s = (F cos ) s = F s cos is the angle between the applied force’s direction and the direction of motion d. Special Work cases • (1) = 0o, cos = 1 and W = F s • (2) =180o, cos = -1 and W = -F s • (3) = 90o, cos = 0 and W = 0 • (4) Note that if you push an object from point A to point B and then back to A, you would have no displacement, so net work done is zero. e. Example: A man cleaning his apartment pulls a vacuum cleaner with a force of 50N and angle of 30o to the horizontal. A frictional force of 40 N retards his motion, and the vacuum is pulled a distance of 3 meters. • (1) Calculate the work done by the 50N force • (2) Calculate the work done by the frictional force • (3) Calculate the net work done on the vacuum by all forces acting on it. • First draw a picture or free body diagram FN FA = 50N Ff=40N motion W=mg • W app force = (F cos)s = (50N cos30o)(3 m)= = 130 Joules • Wf = (Ff cos) s = (40 cos 180o)(3m) = = -120 Joules (work is being done on the vacuum by rug) • m g and FN do no work, so Wnet = Wf + WAF FN FA = 50N Ff=40N =30 motion W=mg 5. Kinetic Energy • a. energy itself is the ability of an object to produce a change in itself or in the world around it • b. KE is the energy an object has due to its motion • c. KE = ½ mv2 – Mass, m, is in kilograms – Velocity , v, is in m/s – KE is measured in Joules 6. The Work-Energy Theorem • a. work is equal to the change in energy that an object undergoes • b. work is the transfer of energy by mechanical means. • c. first established by Joule in nineteenth century • d. energy transfer can go both ways – from work on an object that increase the objects energy or work done by the object so that it transfers energy to its surroundings. 7. Force versus displacement Curve • a. A graph which shows how force is applied to an object with respect to the object’s displacment from the starting point. • b. Area under the curve is the work done by the force over that distance. Note that a negative force means it is being applied in the opposite direction from motion. • c. Area is cumulative. Negative offsets positive B. Power • 1. Power is the work done in a unit of time Power = Work Done time to do work P=W/t • 2. Basic units - work is in joules, time is in seconds, and power is in watts – a. 1 kilowatt = 1000 watts – b. 1 horsepower = 746 watts Please note that power is the timed rate at which work is being done. • 3. Examples – a. A motor lifts a 200 kg object straight up at a constant speed of 3.00 cm/sec. What power is being developed by the motor? Express your answer in watts and horsepower. • Givens: – Constant velocity so Fmotor = m g – m = 200 kg – v = 3.00 cm/sec = 0.0300 m/s so in one second the object moves 0.0300 m • s = 0.0300 m • t = 1 sec – w = m g = 1960 N • Find P. P = w / t and W = F s • W = 58.8 J • P = W / t = 58.8 J / 1 sec = 58.8 watts = 58.8 watts ( 1 HP/ 746 watts) =0.0788 HP • NOTE Pavg = W / time = F (s)/ t = F vavg so power can be determined if you know a force causes an object to move a given distance at a constant speed b. What average power is developed by an 800N physics teacher while running at a constant speed up a flight of stairs rising 6 meters if he takes 8 seconds to complete the climb? • Givens: – F = w = 800 N – v = constant – s = 6 meters – t = 8 sec • P=Fs/t • P = 600 watts c. A 50 kg student climbs a rope 5 m in length at a uniform speed and stops at the top. (1) What must her average speed have been in order to match the power output of a 200 watt bulb; (2) how much work did she do; and (3) how long did it take her to climb the rope? • Givens: – m = 50 kg –s=5m – constant velocity so F = m g – P = 200 watts • P=Fv so v = P / F • v = P / F = 0.41 m/s • W = F s = 490 N (5 m) = 2450 Joules • t = W / P = 2450 Joules / 200 watts = 12.5 sec C. Machines 1. What is a machine? • a. A device by which the magnitude, direction, or method of application of a force is changed so as to achieve some advantage. • b. NOTE: A machine does not change the amount of work done and does not make the amount of work done less, just makes it easier for someone to do a job. c. Simple machines • lever - three classes determined by the relationship between the applied force, load, and fulcrum point • pulley • inclined plane (wedge) • wheel and axle • screw d. complex machines are made up of simple machines 2. Mechanical Advantage (MA) • a. Use a machine NOT to lessen work but usually to lessen the force required to do the work, to make the output work easier to achieve • b. Concerned with the work into a machine and the work out of the machine Ideal Machine Wout Win Win = Wout c. In a real machine, however, this does not occur and Wout < Win • (1) Reason - input work energy is converted to heat due to friction • (2) Wout Win Real Machine Thermal Energy Win > Wout d. Mechanical Advantage = MA = Fout / Fin • (1) Force ratio, so unit-less quantity • (2) If MA > 1, then machine has increased the force you applied to the object moved • (3) NOTE: the machine did not lessen the work output or input in any way. e. Ideal Mechanical Advantage (IMA) • (1) IMA = Din / D out • (2) This is the Distance Ratio, and again it is unitless • (3) Design a machine to have a given IMA, then work to maximize efficiency and MA 3. Efficiency • (1) A ratio between MA and IMA expressed as a percent • (2) Efficiency = MA / IMA x 100% = W out / W in x 100% 4. Examples: • a. A bottle opener required a force of 35N to lift the cap of a bottle 0.90cm. The opener has an IMA of 8.0 and an efficiency of 75%. • (1) What type of simple machine is the basis for the opener? • (2) What is the MA of the opener? • (3) What force is applied to the bottle cap? • (4) How far does the handle of the opener move? • hardest part in any problem is to identify what items are what • “what force is applied to the bottle cap?” is asking for the Force out of the machine • And, “How far does the handle of the opener move?” is asking for the Distance in • If you get confused, remember a machine is used to lessen the force in, and it does this by trading distance for force. So the distance in will always be greater than distance out and Force out > force in. • Givens: – Fin = 35N – D out = 0.90 cm = 0.0090m – IMA = 8.0 – Efficiency = 75% • eff = MA / IMA x 100% – MA = eff x IMA = .75 (8.0) = 6.0 note efficiency was converted to a decimal – MA = F out / F in so F out = MA (F in) = (6.0)(35N) = 210N • IMA = D in / D out so D in = IMA (D out) = 8.0 (0.90cm) = 7.2 cm b. A worker uses a pulley to raise a 225 N carton 10 meters. A force of 110 N is required and the rope is pulled 30 m. • (1) What is the MA of the system? (2) What is the IMA of the system? (3) What is the efficiency of the system? • Givens: – F out = 225N F in = 110N – D out = 10 m D in = 30 m • MA = 2.05 IMA = 3.0 eff = 68% QUIZ - on separate sheet of paper • What is the efficiency of a pulley if a force of 650 N acting through 15 meters is required to lift a 11,000 N crate up a distance of 0.750 meters?

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