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Chapter 10: Energy, Work and Simple Machines

Chapter 10: Energy, Work and Simple Machines. Barry Latham Glencoe Science 2005. Impulse = Force x time Work = Force x distance W= Fd W= Fd cos q Vector quantity- quantity and direction Work is always in the same direction as the force Measures in Joules (derived unit) (N)(m) = J

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Chapter 10: Energy, Work and Simple Machines

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  1. Chapter 10: Energy, Work and Simple Machines Barry Latham Glencoe Science 2005

  2. Impulse = Force x time • Work = Force x distance • W=Fd • W=Fdcosq • Vector quantity- quantity and direction • Work is always in the same direction as the force • Measures in Joules (derived unit) • (N)(m) = J • Larger values can be measured in kJ (kilo=1,000) • q= angle between force and direction of displacement Section 10.1

  3. If force is constant, then the work can be easily calculated. W=Fd • If the force varies, then we need to solve the problem graphically. The area under the curve of a graph will tell us work. • So we can divide it into smaller and smaller rectangles and find the area of each. (That’s calculus!) Work & Calculus

  4. Kinetic energy is the energy of motion. • Radiant Energy- light, x-rays, gamma rays, radio • Thermal Energy- heat • Sound- longitudinal waves (earthquake waves as well) • Electrical Energy- flowing electrons, lightening, electricity • Motion- movement of objects • The faster an object is moving or the larger it is (or both), the more kinetic energy it has. • The kinetic energy of an object at constant velocity • KE = ½mv2 Kinetic Energy

  5. Work is done when energy is transferred from one object into another. • Pushing a shopping cart • Throwing a bowling ball • Firing an arrow from a bow • W=DKE • Fdcosq=1/2 mvf2-1/2 mvi2 Work-Energy Theorem

  6. 1. Sketch the system • 2. Draw the force and displacement vectors. • 3. Find the angle, q, between each force and displacement. • 4. Calculate the work done by each force using W=Fdcosq • 5. Calculate the net work done. • Practice Problems 1-3 • HW p. 265 q. 15-22 Problem Solving

  7. Power is the rate by which work is done. Rates always include a time element. • P = average power = work/time • = (energy transformed)/(time) = J/s = Watt (W) • P=W/t • Although we are used to this being an electrical term, it is derived from the conversion of mechanical energy into electrical energy. • The power of a horse refers to how much work it can do per unit of time. One horsepower (hp) is defined as 550ft lbs/s (British system), which equals 746Watts in SI. Power

  8. Transparency 10-1 • 1. 150.0 N • 2. 8.0 m • 3. Increasing force from 0.0 m-3.0 m, the constant force from 3.0 m-6.0 m, then increasing force from 6.0 m-10.0 m • 4. W=Fd • 5. W=350.0 J • 6a. W=800.0 J • 6b. Read from graph • 6c. Increases in a linear fashion • 7. W=1475 J Quiz Force, Distance, and Work

  9. 9. A box that weighs 575N is lifted a distance of 20.0m straight up by a cable attached to a motor. The job is done in 10.0s. What power is developed by the motor in Watts and kW? • P=W/t=Fd/t • P=(575N)(20.0m)/(10.0s) • =1150 Watts • =1.15 kWatts Practice Problems 9

  10. A winch is designed to be mounted on a truck is advertised as being able to exert a 6,800 N force and to develop a power of 0.3kW. How long would it take the truck and winch to pull an object 15 m? • P=W/t=Fd/t • t=Fd/P • t=(6800N)(15m)/(300W) • t=340 s = 5.7 minutes Practice Problem 13

  11. Machine- eases work by changing the magnitude or direction of a force • Wi=Input work, what you do • Wo=Output work, what you get, ≤Wi • Mechanical Advantage (MA) • Effort force (by you), Fe=Wi/de • Resistance force (by machine), Fr=Wo/dr • MA=Fr/Fe Section 10.2 Machines

  12. If Wi=Wo, then Fede=Frdr • Recall that MA=Fr/Fe • So, IMA=de/dr • For an ideal machine, where there is no loss due to friction or other forms of resistance Ideal Mechanical Advantage (IMA)

  13. Efficiency=ratio of output to input of work • e=(Wo/Wi)(100) • Ideal machine: e=1 • Wo/Wi=(Frdr)/(Fede) • MA=Fr/Fe • IMA=de/dr • Therefore, e=(MA/IMA)(100) Efficiency

  14. Compound Machine- more than one simple machine combined • Six Simple Machines • Lever- opposite direction & DF • Pulley- Ddirection & DF • Wheel & Axle- Ddirection & DF • Inclined Plane- Ddirection & DF • Wedge- direction & splits force in half • Screw- direction & DF • Transparency 10-2 Compound Machines

  15. A sledgehammer is used to drive a wedge into a log to split it. When the wedge is driven 0.20m into the log, the log is separated a distance of 5.0cm. A force of 1.7x104N is needed to split the log, and the sledgehammer exerts a force of 1.1x104N. • a. What is the IMA of the wedge? • b. What is the MA of the wedge? • c. Calculate the efficiency of the wedge. Practice Problem 25

  16. MA=Fr/Fe • IMA=de/dr • e=(MA/IMA)(100) • IMA=de/dr=(0.20m)/(0.05m)=4 • MA=Fr/Fe=(1.7x104N)/(1.1x104N)=1.5 • e=(MA/IMA)(100)= (1.5/4)(100)=38% • Transparency 10-3 & 10-4 Practice Problem 25, continued

  17. Work in groups of four • Re-draw and complete Data Table • Complete “Conclude & Apply” • Three graphs for each group • Work vs. Time • Power vs. Work • Power vs. Time Stair Climbing and Power, p. 274

  18. Energy w.s. • W-E w.s. • Conserv. w.s. • Power w.s. • Test Prep 10 • Study Guide 10 Final Assessments

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