Chapter 6: Conservation of Energy CQ: 1, 3. Problems: 1, 3, 5, 11, 17, 25, 29, 32, 49, 51, 53, 69, 71. Law of Conservation of Energy Work done by a constant.

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Presentation transcript:

Chapter 6: Conservation of Energy CQ: 1, 3. Problems: 1, 3, 5, 11, 17, 25, 29, 32, 49, 51, 53, 69, 71. Law of Conservation of Energy Work done by a constant force Kinetic energy Gravitational potential energy Work by variable force, Hooke’s law Elastic potential energy 1

2 Energy & Work Energy is the capacity to do work, Energy is position & speed dependent Unit: joule = newton·meter (J = N·m) Work = force x distance (Fd) when force is in direction of motion (or opposite to motion) Ex. 50N pushes distance of 4 meters. W = (50N)(4m) = 200 J / 2

Work & Force Work is energy transferred by part of force in line of motion Ex. Force 60° above path of motion 3

4 Machines change an applied force by increasing it, decreasing it, or changing its direction. Types: inclined plane, screw, wedge pulley, wheel lever 4

levers Work input F d = Work output F d Ex. Your hand moves 100m, causes car to rise 0.10m. The force amplification factor is, F F d d = __ 5

6 inclined plane Weight x height change = Force x distance along plane Force along ramp less than Weight Ramp distance greater than height change ADA Standards: Ramp must be at least 12x longer than vertical rise Ex. A 1ft vertical rise requires 12ft of ramp. 6

ADA Ramp 77

Energy of Motion Called Kinetic Energy (KE) KE = ½(mass)(velocity) 2 = ½mv 2. Ex. 2000kg car moving at 2m/s. KE = ½ (2000)(2) 2 = 4000J. Position Dependent Energies are called Potential Energies “PE” or “U” / 88

Gravitational Potential Energy U = weight x height (mgh) 1kg at 1m height: U = (1kg)(9.8N/kg)(1m) = 9.8J Energy released in falling / 9

Hooke’s Law The restoring force an object exerts is proportional to the amount it has been deformed (F = -kx) 10

Elastic Potential Energy PE-elastic = average force x distance 10N compresses a spring 1m. U = (avg. force, 5 N)(1m) = 5 J k = spring constant = force/distance U = (½kx)x = ½kx 2 11

Work Energy Theorem Let direction of motion be +x 12

13 Power Power is the rate work is performed Power = work/time = Force x velocity Unit: watt = joule/second = J/s Other Unit: horsepower 1 horsepower = 746 watts / 13

Energy & Power Energy = power x time Ex. A toy car has 1000 J of energy at full charge. How long can it run at 100 watts? At 10 watts? Time = Energy/power = 1000J/100watts = 10 seconds = 1000J/10watts = 100 seconds/ 14

15 Conservation of Energy E = K + U = constant Ex. Falling: Kinetic ↑ as Potential ↓ /

16

Energy Summary PE-gravitational = mgy PE-elastic = ½kx 2 KE = ½mv 2 Mechanical Energy ME = KEs + PEs E = ME + Thermal Energy 17

18 EnergyE1E2E3 Kinetic0½mv U-grav00mgh U-elastic½kx 2 00 U-therm 000 Totals  ½kx 2 ½mv 2 2 mgh

19 Energy h y Kinetic0½mv 2 U-gravmghmgy U-elast00 U-therm00 Totals  mgh½mv 2 + mgy Energy and speed are same at height y Accelerations are not same

20 EnergyEiEf Kinetic½mv i 2 0 U-grav00 U-elastic00 U-therm0fkdfkd Totals  ½mv i 2 fksfks Ex. Sled slides to a stop d

21 A 2.00kg ball is dropped from rest from a height of 1.0m above the floor. The ball rebounds to a height of 0.500m. A movie- frame type diagram of the motion is shown below. TypeE1E2E3E4E5 gravita- tional mg(1)000mg(1/2) kinetic0½ m(v2) 2 0½ m(v4) 2 0 elastic00U-el00 thermal00U-therm

22 Terminology E: total energy of a system E-mech = total energy minus the thermal energy E-mech = E – U therm. Mech. Energy conserved in a frictionless system

23 Power: The time rate of doing work. SI Unit: watt, W = J/s Example: How much average power is needed to accelerate a 2000kg car from rest to 20m/s in 5.0s? work =  KE

24 Another equation for Power: Ex: A car drives at 20m/s and experiences air- drag of 400N. The engine must use (400N)(20m/s) = 8,000 watts of engine power to overcome this force. 8,000 watts = 10.7 hp.

Work Example mg mgsin  d mg h  (mgsin  )d = mg(dsin  ) = mgh Moving down an inclined plane 25

26 Summary Energy: Kinetic + Potential + Thermal, is conserved. Mech. Energy: Kinetic + Potential, conserved in frictionless systems Work is energy transfer (+ or -) Power is rate of energy transfer

Vehicle Efficiency 1 gallon gasoline has 138,000,000 J Engines only get a fraction of this: Ex. A 25% efficient car gets (0.25)(138,000,000 J) = 34,500,000J out of 1 gallon. A 20% efficient car gets 27,600,000J. 27

Vehicle Frictional Work = Total Frictional Force x distance Ex. 400N friction for 1600 meters (1 mile) Work = (400N)(1600m) = 640,000J for one mile traveled / 28

Mpg 29

Ex. Mpg 20% Efficiency, f = 400N Engine gets 27,600,000 J/gal Frictional Work/Mile = 640,000J/mile = 43 mpg (at constant speed) 30

31 Horsepower: 1 hp = 746 watts For the previous example:

32 What size electric motor is needed to raise 2000lbs = 9000N of bricks at 10cm/s? Minimum Power: P avg = Fv avg = (9000N)(0.1m/s) P = 900 W = 1.2 hp

33

34

35

36 Similar to gram bullet moves at 200m/s and goes 10cm into a tree. What is the average force on the bullet? Tree? Wnet on bullet = -Fd = change in K Change in K = 0 – ½ (0.003kg)(200)(200) -F(0.1m) = - 60Nm F = 600N

37 By energy conservation, the sum of all energies in each column is the same, = E1 = mg(1) = 19.6J Calculate v2: (use 1st and 2nd columns) mg(1) = ½ m(v2)2. g = ½ (v2)2. v2 = 4.43m/s Calculate PE-thermal: (use 1st and 5th columns) mg(1) = mg(1/2) + PE-thermal mg(1/2) = PE-thermal PE-thermal = 9.8J 37

38 Calculate PE-elastic: (use 1st and 3rd columns) PE-elastic + PE-thermal = mg(1) PE-elastic = 19.6 PE-elastic = 9.8J Calculate v4: (use 1st and 4th columns) ½ m(v4)2 + PE-thermal = mg(1) ½ m(v4) = 19.6 ½ m(v4)2 = 9.8 (v4)2 = 2(9.8)/2 v4 = 3.13m/s 38

EnergyABC Kinetic½m2 2 ½mv B 2 ½mv C 2 U-gravmg7mg3mg0 U-elastic000 U-therm Totals  39