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Energy, Work, and Power

Unit 5. Energy, Work, and Power. Energy is the capacity for an object to do work For example, when a car moves, the engine performs work to get the car going. There are many different types of energy, including: electrical, kinetic, gravitational potential, and elastic potential to name a few.

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Energy, Work, and Power

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  1. Unit 5 Energy, Work, and Power

  2. Energy is the capacity for an object to do work For example, when a car moves, the engine performs work to get the car going. There are many different types of energy, including: electrical, kinetic, gravitational potential, and elastic potential to name a few. A more complete list can be found on p. 124 Types of Energy

  3. An energy transformation occurs whenever energy changes from one form into another. Examples of this would be a ball being held above the ground (gravitational potential) and then being released to fall to the ground (kinetic). Energy Transformation

  4. Work

  5. This is the energy transferred to an object The object must move a distance as a result of the force applied Does it matter what direction the object moves?? Work

  6. Work requires a force Work requires a distance This leads us to say: WαF and WαΔd This gives us: W = F Δd The units are Newton Meters (Nm) or, more commonly, Joules (J) How to calculate work

  7. For there to be work, Remember!!!!

  8. Any force applied in the same plane causes work to be done If the force makes the object increase in speed, then it is positive work If the force makes the object slow its speed, then it is negative work. All friction is negative work. Positive and Negative Work

  9. When we lift something up, we do work, why is this? • When we look at this type of work, we still must look at the force we are working with • Fg = mg • This lead to the following • W = Fgd • W = mgd Gravity

  10. If the force and displacement are perpendicular to each other, zero work is done. Zero work is also done if there is no force when there is a displacement, or then there is no displacement when there is an applied force. Ex. When carrying a bag forwards, no work is done.

  11. A bag of groceries of mass 8.1 kg is raised vertically without acceleration from the floor to a counter top, over a distance of 92 cm. Determine • The force needed to raise the bag without acceleration. • The work done on the bag against the force of gravity Example

  12. Mechanical Energy

  13. There are 2 types of mechanical energy • Gravitational Potential Energy • Kinetic Energy • Gravitational Potential Energy • This is energy that can be used to do work at a lower level • Kinetic Energy • This is the energy of motion Mechanical energy

  14. To hit a nail with a hammer, what must you do? • By lifting the hammer, Δh, you also need to apply a force. • The height is measured from a starting point or equilibrium position. • The force is found by lifting the mass against gravity • Ep = FΔh • Ep = mgh Determining Potential energy

  15. example • Assume that a 59 kg pole vaulter must raise their center of mass from 1.1 m off the ground to 4.6 m off the ground. What is the jumper’s gravitational potential energy at the top of the bar relative to where the jumper started to jump? • Ep = mgΔh • Ep = (59)(9.81)(4.6-1.1) • Ep = 2.0 x 103 J

  16. Grain Auger Pile Drivers Hydro Dams We use this in Red Lake everyday Applications of mechanical energy

  17. If you are interested in how the formula is generated, see p. 134 Kinetic energy is the energy of motion, so what do we need? Ek = ½ mv2 Determining kinetic energy

  18. example • Determine the amount of kinetic energy of a 48 g dart travelling at a speed of 3.4 m/s. • Ek = ½ mv2 • Ek = ½ (.048)(3.4)2 • Ek = 0.28 J

  19. Law of conservation of energy

  20. We know that there are many types of energy transformations When energy changes forms, energy is conserved What does this mean? Energy is never lost, it just changes form Energy conservation

  21. In this section, we will examine gravitational potential energy changing into kinetic energy and vice-versa. We will assume that energy does not change into any other form. In reality, friction converts mechanical energy into thermal energy all the time.

  22. The best way to approach problems involving the Law of Conservation of energy is to do the following: Set energy before equal to energy after Substitute Eg+Ek for E Solve for Eg and Ek

  23. example

  24. A comparison of the amount of energy put into a system compared to the energy output of a system Could also say a ratio of work you want done to the amount of work that you have to do. Efficiency

  25. Efficiency is never 100% because some mechanical energy is converted into thermal and sound energy (friction). Often, when lifting objects with pulleys or sliding objects up a ramp, the work out equals mgh and the work in equals the force that you apply (F) times the distance that you apply it Δd

  26. POWER

  27. Power Power is the rate of doing work or transforming energy P = W / t or P = ∆E / t The SI unit for Power is called a watt (W)

  28. Power Let’s say a hot plate produced 90J of energy in 30s. The power of the hot plate would be: P = 90 / 30 = 3W Lets say this same hot plate is used to heat water for 3 minutes. The amount of energy given to the water would be: E = P x t = 3W x 180s = 540 J

  29. Sample Problem A car on a level road acc. From 0 to 28m/s with an avg. acc. of 5.0 m/s2. The total friction on the car is 600N. Calculate the power of the engine if the car has a mass of 1400 kg Given: v1 = 0, v2 = 28m/s, a = 5.0 m/s2 m = 1400 kg, Ff = 600N

  30. Solution 2 Net force: Fnet = ma = 1400 kg x 5.0 m/s2 = 7000 N Appl. Force: FA = Fnet – Ff = 7000 N – (- 600N) = 7600 N Avg. speed = V1 + V2 / 2 = 0 + 28m/s / 2 = 14 m/s Power = P = W/t = Fd/t = Fvav = 7600 N x 14m/s = 1.1 x 102 kW

  31. Thermal Energy and Heat

  32. Thermal Energy and Heat Thermal Energy: is the total kinetic energy and potential energy of the atoms or molecules of a substance. Temperature: avg. kinetic energy of the atoms or molecules of a substance Heat: is the transfer of energy

  33. The Kinetic Molecular Theory of Heat Related to the motion of particles Scientists (Newton, Boyle and Bacon) suggested that heat was related to the motion of particles of matter. Boyle suggested that nails become heated when hit by a hammer, since the particles in the nail are set into violent motion

  34. The Kinetic Molecular Theory of Heat All matter is composed of many tiny particles called molecules The molecules are separated from one another by empty space, the distance is large compared to their size All molecules are constantly moving in some manner – therefore, possess kinetic energy When heat is added to matter, the molecules absorb the energy and move faster (their kinetic energy increases). When heat is removed, the molecules slow down (their kinetic energy is decreased)

  35. Heat Transfer by Conduction Conduction: transfer of heat through a substance, or by direct contact from one substance to another.

  36. Heat Transfer by Convection Convection: transfer of heat by the movement of fluid particles from a region of high to low temperature This occurs because of the change in density of the fluid. The fluid becomes less dense, in the case of hot air – it will rise.

  37. Heat Transfer by Radiation transfer between hot and cold objects in the absence of any kind of matter, by means of electromagnetic waves. ex. the sun transfers heat by radiation through the near vacuum of space to Earth

  38. Heat Conductors and Insulators Good heat conductors transfer heat rapidly. eg copper bottoms of pots or metal radiators in cars Heat insulators are poor heat conductors reduce heat loss. eg layers of fatty tissue beneath your skin

  39. Specific Heat Capacity (c) amount of energy needed to raise the temperature of 1.0 kg of a substance by 1.0° C. It is measured in J/kg°C. Q = mc∆t The principle of Heat Exchange: when heat is transferred from one body to another, the amount of heat lost by the hot body, equals the amount gained by the cold body Qlost + Qgained = 0

  40. Sample Problem Calculate the quantity of heat gained by a scrap of aluminum metal which has a mass of 2 kg. The aluminum metal has an initial temp. of 30° C and was heated to a temp. of 80° C. Given: m= 2 kg, ∆t = t2 – t1, c = 900 J/kg°C

  41. Solution Q = mc∆t = (2 kg) (900 J/kg°C) (50° C) = 90000 J

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