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In this presentation you will:

In this presentation you will:. Work-Energy Theorem. F. m. d. investigate quantities using the work-energy theorem. Next >. Introduction. Many words used in everyday conversation have very precise meanings in physics.

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  1. In this presentation you will: Work-Energy Theorem F m d • investigate quantities using the work-energy theorem Next >

  2. Introduction Many words used in everyday conversation have very precise meanings in physics. When you think of work, you may think of going to a place of work, or working hard, or even something like a computer working. Energy Work Force When you think of energy, you may think of how lively or tired you are. In physics, work and energy have very specific meanings. Next >

  3. What is work? Consider a force (F) acting on an object while it moves a distance (d). The object’s velocity will change; it will accelerate. F m a = F/m d Using the equations of motion, we can find the change in velocity. vf vi m m vf2 – vi2 = 2ad d Replacing a = F/m and multiplying both sides by m/2, we get: Fd = ½mvf2 – ½mvi2 Next >

  4. What is work? Fd = ½mvf2 – ½mvi2 The left side of the equation is the work done on the system. vf vi m m F d In physics, work has a precise definition: Work is the product of force × distance. Work = F × d W = ½mvf2 – ½mvi2 So: Next >

  5. Energy W = ½mvf2 – ½mvi2 The right side of the equation describes the change in quantity before and after the force acts. vf vi m m F d The quantity depends on the mass and the velocity of the object. This quantity is known as energy and as it is the energy of a moving body, kinetic energy. KE = ½mv2 So: W = KEf – KEi Next >

  6. Work-Energy Theorem Work is equal to the change in kinetic energy. W = ΔKE vf vi m m F In physics, the Δ symbol is used to represent a change in something. d The units of work and energy are the joule (J), named after the physicist James Joule, who discovered the relationship. 1 J = 1 Nm (Fd) = 1 kgm2/s2 (½mv2) Next >

  7. Calculating Work Work is calculated using the equation: F m W = Fd   d A force of 10 N is used to move an object over a distance of 10 m. How much work is done? W = Fd = 10 × 10 = 100 J Work = 100 J Next >

  8. Question 1 How much work is done when a force of 25 N is used to move an object a distance of 5 m? Give your answer as a number in J. Next >

  9. Question 1 How much work is done when a force of 25 N is used to move an object a distance of 5 m? W = F × d = 25 × 5 = 125 J Work = 125 J Give your answer as a number in J. 125 (J) Next >

  10. Friction Force In the real world, friction tends to act on a moving body to slow it down. Now we have a friction force (Ff) acting in the opposite direction. Ff F m Ff acts in the opposite direction, so it does negative work. d W = (F – Ff)d Ff m If the friction force Ff was the only force acting on the object when it is in motion, the object would slow down, and its kinetic energy would be reduced. d W = (0 - Ff)d = - Ff d Next >

  11. Question 2 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. How much work is done on the object? Give your answer as a number in J. Next >

  12. Question 2 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. How much work is done on the object? W = -Ff × d = -10 × 10 = -100 J Work = -100 J Give your answer as a number in J. -100 (J) Next >

  13. Question 3 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. What is its change in kinetic energy? Give your answer as a number in J. Next >

  14. Question 3 An object is in motion. A friction force of 10 N acts on it over a distance of 10 m. What is its change in kinetic energy? W = -Ff × d = -10 × 10 = -100 J Work = -100 J Work = ΔKE ΔKE = -100 J Give your answer as a number in J. -100 (J) Next >

  15. Question 4 A force of 100 N is used to move an object over a distance of 10 m. A friction force of 10 N is acting on the object to slow it down. What is the work done? Give your answer as a number in J. Next >

  16. Question 4 A force of 100 N is used to move an object over a distance of 10 m. A friction force of 10 N is acting on the object to slow it down. What is the work done? W = (F-Ff) × d = (100 – 10) × 10 = 900 J Work = 900 J Give your answer as a number in J. 900 (J) Next >

  17. Question 5 A moving object has a kinetic energy of 100 J. A force of 30 N is applied to the object in the same direction as its travel over a distance of 10 m. What is the total kinetic energy after 10 m? Give your answer as a number in J. Next >

  18. Question 5 A moving object has a kinetic energy of 100 J. A force of 30 N is applied to the object in the same direction as its travel over a distance of 10 m. What is the total kinetic energy after 10 m? W = F × d = 30 × 10 = 300 J Work = 300 J Work = ΔKE ΔKE = 300 J Total KE = 300 + 100 = 400 J Give your answer as a number in J. 400 (J) Next >

  19. Force at an Angle Enrichment To calculate force applied at an angle, the force is split into x and y components. F Only the component of the force that acts in the direction of the displacement is used. m d Fx θ Fy Fx = F cosθ F So, W = Fd cosθ The Fy component acts at right angles to the direction of displacement so no work is done. Next >

  20. Force at an Angle Enrichment For example, if the force of 10 N is acting at an angle of 30° over a distance of 10 m: F m W = Fd cosθ d Fx W = 10 × 10 × cos30° J θ Fy F W = 86.6 J Next >

  21. Summary In this presentation you have seen: • Work is an energy transfer calculated by the product of a force and distance. W = F × d • The kinetic energy of a moving body is related to its mass and velocity. KE = ½mv2 • The work-energy theorem relates work and energy. W = ΔKE End >

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