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Work

Work. The change in energy. Defining Work. How much the energy changes If a car has 100J of Ek when the brakes are applied, and has 20J of Ek after, the brakes have done 80J of work on the car

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Work

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  1. Work The change in energy

  2. Defining Work • How much the energy changes • If a car has 100J of Ek when the brakes are applied, and has 20J of Ek after, the brakes have done 80J of work on the car • If an object has 100J of GPE at the top of a hill and 70J of KE when it gets to the bottom friction did 30J of work on the car

  3. To Change an Object’s Kinetic Energy… • A force must be applied over distance! • If we push a stationary object and it doesn’t move, there is no work done • Work = Force * distance • The units of work are Joules • 1J of work is defined as 1N of force applied over a distance of 1m

  4. In our LOL graphs… • The circle is where we keep track of how much work is done • The object can gain energy or lose energy • A force causing an object to speed up adds energy to the object • A force causing an object to slow down removes energy from the object

  5. An object slowing down

  6. An object speeding up

  7. What happened?

  8. Practice Problem #1 • A 25kg car at rest speeds up to 60m/s. • Draw an LOL graph • How much work was done? • How much force did the engine apply if this was accomplished over a distance of 50m?

  9. Problem #1 Solution • Work = Change in energy • = ½ * 25kg (60m/s)2 -OJ • =12.5kg 3600m2s2 • = 45,000J • Work = Force *distance • 45,000J = F * 50m • F = 900N

  10. Practice Problem #2 • A 5kg ball moving 10m/s comes to a hill. It comes to a stop at a height of 4m on the hill. • Draw an LOL graph • How much work did friction do on the ball? • If the force of friction was 5N, how much distance did the ball travel while rolling uphill?

  11. Problem #2 Solution • Only KE to start • ½ * 5kg * (10m/s)2 = 250J • When the ball stops, only GPE • 5kg*10N/kg * 4m = 200J • The object lost 50J of energy = 50J of work • 50J = F * d • 50J = 5N * d • d= 10m

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