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Chapter 4

Chapter 4. Work & Energy Dr. Ali. CHAPTER OUTLINE. WORK. Who is doing more work?. WORK. Work is defined as the product of the net force acting on a body and its displacement (in the direction of the force). Work =. force x parallel distance. W =. F x d.

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Chapter 4

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  1. Chapter 4 Work & Energy Dr. Ali

  2. CHAPTER OUTLINE

  3. WORK Who is doing more work?

  4. WORK • Work is defined as the product of the net force acting on a body and its displacement (in the direction of the force). Work = force x parallel distance W = F x d Unit of Energy: 1 Joule = 1 J = 1 Nm = 1 kgm2/s2

  5. WORK • Only the component of the applied force, F, in the direction of the motion of the lawn mower, FHorizontal, is used to do work on the lawn mower. For work to be done, force and motion must be in the same direction

  6. Example 1: An object is moved with a force of 15 N across a horizontal surface. How much work is done if the object is moved 50 m? W = F x d F = 15 N d = 50 m W = ??? W = (15 N)(50 m) W = 750 J

  7. Example 2: 650 J of work is done in moving a desk a horizontal distance of 5 m. How much force is used to move the desk? F = ??? d = 5 m W = 650 J F = 130 N

  8. Example 3: How much work is done in lifting a 10 kg box 1.5 m off the floor? F = w = m x g m = 10 kg d = 1.5 m F = ??? W = ??? F = 10 kg x 10 m/s2 = 100 N W = F x d = 100 N x 1.5 m W = 150 J

  9. Example 4: How much work is done while walking 5.0 m holding an object with mass of 3.0 kg? No work is being done, since force and motion are not in the same direction.

  10. POWER • Power is the rate at which work is done. SI Unit for Power: 1 watt = 1 J/s

  11. Example 1: A force of 150 N is used to push a motorcycle 10 m along a road in 20 s. Calculate the power in watts. F = 150 N d = 10 m t = 20 s P = ??? P = 75 watts

  12. Example 2: An 80 kg man runs up a flight of stairs 5.0 m high in 10 seconds. What is the man’s power output in watts? F = w = m x g m = 80 kg d = 5.0 m t = 10 s P = ??? F = 80 kg x 10 m/s2 = 800 N P = 400 watts

  13. Example 3: A pump lifts 30 kg of water a vertical distance of 20 m each second. What is the power output? F = w = m x g m = 30 kg d = 20 m t = 1 s P = ??? F = 30 kg x 10 m/s2 = 300 N P = 6000 watts

  14. Example 4: A crane uses 750 kwatts of power to lift a car 0.5 m in 12 seconds. How much work is done? P = 750 kwatts d = 0.5 m t = 12 s W =??? W = P x t = 750 kwatts x 12 s W = 9000 kJ

  15. Example 4: A crane uses 750 kwatts of power to lift a car 0.5 m in 12 seconds. How much work is done? What force is used by the crane? W = 9000 kJ d = 0.5 m F =??? F = 18000 kN

  16. Example 5: Which person does more work, 1 or 2? Which person has greater power, 1 or 2? F1 = F2 d1 = d2 W1 = W2 t1 > t2 1 2 P1 < P2

  17. ENERGY • Energy is the ability to do work The brick and the hammer possess energy and thus can do work on the nail • The SI units for energy is joules (J). 1J = 1 joule = 1 kgm2/s2

  18. ENERGY • There are two types of energy here: Energy of motion Stored energy Potential Kinetic

  19. ENERGY • Energy can be converted from one type to another. Potential E Kinetic & potential Kinetic E

  20. KINETICENERGY • Kinetic energy is energy of motion. Velocity has greater effect on KE than mass • KE is a scalar quantity • (Note: Energy is never a vector)

  21. Example 1: What is the kinetic energy of a 60 kg girl on skis traveling at 20 m/s? m = 60 kg v = 20 m/s KE = ??? KE = 12000 J

  22. Example 2: A sports car is moving at 4.0 m/s. If the mass of the car is 800 kg, how much KE does it have? m = 800 kg v = 4.0 m/s KE = ??? KE = 6400 J

  23. Example 3: Two identical cars are moving, one with twice the velocity of the other. How much more kinetic energy does the faster car possess? v2 = 2 v1 Faster car has 4 times greater KE

  24. POTENTIALENERGY • Potential energy is stored energy. The compressed spring has potential energy because when released it can do work on the mass, m.

  25. POTENTIALENERGY • Gravitational potential energy is energy of position. Potential energy = Weight x height High PE PE = m g h Low PE

  26. Example 1: A mass of 100 kg is lifted a distance of 50 m. How much PE does it possess? PE = m g h m = 100 kg h = 50 m PE = ??? = (100 kg) (10 m/s2) (50 m) PE = 50000 J

  27. Example 2: A 70-kg diver standing on a diving platform possesses 35000 J of PE. How high is the platform? m = 70 kg h = ??? PE = 35000 J h = 50 m

  28. CONSERVATIONOF ENERGY • The sum of KE and PE in a system is constant, in the absence of friction. KE + PE = constant KE + PE = ETotal

  29. CONSERVATIONOF ENERGY Energy cannot be created or destroyed PE It may be transformed from one form to another  ETotal in a system remains constant. KE

  30. Example 1: A 10-kg boulder rests at the edge of a 100-m cliff. How much PE does the rock possess? m = 10 kg h = 100 m PE = ??? PE = m g h = (10 kg) (10 m/s2) (100 m) PE = 10000 J

  31. Example 1: The rock rolls off the cliff and falls to the bottom. How much KE does the rock possess at the bottom of the cliff? KEbottom = PEtop = 10000 J

  32. Example 1: What speed does the rock have just before hitting the ground? m = 10 kg v = ??? KE = 10000 J v2 = 2000 m2/s2 Thus, v = 45 m/s

  33. Example 2: A 60-kg boy and his sled (neglect the mass of the sled) are at a 10 m high slope. How much PE do the boy and sled possess? PE = mgh m = 60 kg h = 10 m PE = ??? = (60 kg)(10 m/s2)(10 m) PE = 6000 J

  34. Example 2: What kinds of energy, and how much of each, do they possess halfway down the slope? @ halfway PE = KE PE = KE = 3000 J

  35. THE END

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