2. Energy Energy = the potential to do work Units: Joules And, in reverse… work done = change in energy W = ∆ E

3. Energy Energy = the potential to do work Units: Joules And, in reverse… work done = change in energy Two major types of energy 1)Potential energy – energy that is stored due to an object’s position. Examples: gravitational potential energy elastic potential energy 2) Kinetic energy - energy of motion W = ∆ E

4. Gravitational Potential Energy Stored energy due to an object’s height. The higher the object, the more gravitational potential energy.** PE g = F g d = mgh  PE g = mgh

5. Elastic Potential Energy Energy stored due the compression or stretching of elastic material, such as a spring. F spring = kx (Hooke’s Law) where k is the spring constant and x is the distance the spring has been stretched or compressed. Pe spring = ½ kx 2

6. Potential Energy and Work We said earlier that W = Δ E and we know that potential energy depends only on the position of an object… THIS IMPLIES THAT WHEN DOING WORK, THE PATH DOESN’T MATTER --- ONLY THE END POSITION.

7. Work doesn’t depend on path. Imagine a mass m lifted a vertical distance h following two different paths. 1.straight up: –– the work done against gravitational force is: –– the work done against gravitational force is: W = F g h cos 0 0 W = F g h cos 0 0 W = mg h W = mg h 2.along a ramp a distance d (no friction) – minimum force needed is F 2 = mg sinθ. – minimum force needed is F 2 = mg sinθ. – the work done against gravitational force is: – the work done against gravitational force is: W = F 2 d cos 0 0 = mg sinθ d W = F 2 d cos 0 0 = mg sinθ d W = mg h W = mg h h d mg F2F2F2F2 θ W depends on VERTICAL distance moved, not on the path

8. Work doesn’t depend on path. Imagine a mass m lifted a vertical distance h following two different paths. 1.straight up: –– the work done against gravitational force is: –– the work done against gravitational force is: W = F g h cos 0 0 W = F g h cos 0 0 W = mg h W = mg h 2.along a ramp a distance d (no friction) – minimum force needed is F 2 = mg sinθ. – minimum force needed is F 2 = mg sinθ. – the work done against gravitational force is: – the work done against gravitational force is: W = F 2 d cos 0 0 = mg sinθ d W = F 2 d cos 0 0 = mg sinθ d W = mg h W = mg h h d mg F2F2F2F2 θ Also notice: W = the potential energy of the block! W = the potential energy of the block!

9. A ramp reduces the force needed – makes life easier 1. W = F  h = mg  h 1. W = F  h = mg  h 2. W = F  d = mg sinθ  h/sinθ = mg  h 2. W = F  d = mg sinθ  h/sinθ = mg  h θ h h d W = F  h or F  d (along the ramp)

10. Work and PE – You do a) A cart filled with bricks is pulled at constant speed up a ramp. The cart has a mass of 3.0kg and the height of the ramp is 0.45 m. What is the PE g of the cart at the top of the ramp? b) It required 14.7 N to pull the cart up the ramp, and the ramp is 0.9 m long. How much work was needed to pull the cart up the ramp?

11. Work and PE – You do a) A cart filled with bricks is pulled at constant speed up a ramp. The cart has a mass of 3.0kg and the height of the ramp is 0.45 m. What is the PE g of the cart at the top of the ramp? b) It required 14.7 N to pull the cart up the ramp, and the ramp is 0.9 m long. How much work was needed to pull the cart up the ramp? Both answers are 13.2 J!

12. Kinetic energy Kinetic energy is the energy of motion. KE = ½ m v 2 where v = velocity This means that a moving object possesses the capacity to do work. Example: A hammer by virtue of its motion can be used to do work in driving a nail into a piece of wood.

13. (1)A father pushes his child on a sled on level ice, a distance 5 m from rest, giving a final speed of 2 m/s. If the mass of the child and sled is 30 kg, how much work did he do? W = ∆ KE = ½ m v 2 – 0 = ½ (30 kg)(2) 2 = 60 J W = ∆ KE = ½ m v 2 – 0 = ½ (30 kg)(2) 2 = 60 J (2)What is the average force he exerted on the child? W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N

14. (3)A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the road does work on the car through a backward-directed friction force. How much work must this friction force do in order to stop the car? How much work must this friction force do in order to stop the car? d v i = 45 km/h F fr (work done by friction force) v f = 0

15. (3)A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the road does work on the car through a backward-directed friction force. How much work must this friction force do in order to stop the car? How much work must this friction force do in order to stop the car? W = ∆ KE = 0 – ½ m u 2 = – ½ (1000 kg) (45 x1000 m/3600 s) 2 W = ∆ KE = 0 – ½ m u 2 = – ½ (1000 kg) (45 x1000 m/3600 s) 2 W = – 78125 J = – 78 kJ W = – 78125 J = – 78 kJ (the – sign just means the work leads to a decrease in KE) (the – sign just means the work leads to a decrease in KE) d v i = 45 km/hr F fr (work done by friction force) v f = 0

16. (4)Instead of slamming on the brakes the work required to stop the car is provided by a tree!!! provided by a tree!!! What average force is required to stop a 1000-kg car going at 45 km/h if the car collapses one foot (0.3 m) upon impact? What average force is required to stop a 1000-kg car going at 45 km/h if the car collapses one foot (0.3 m) upon impact? corresponds to 29 tons hitting you OUCH Do you see why the cars should not be rigid. Smaller collapse distance, gretaer force, greater acceleration. For half the distance force would double!!!! OUCH, OUCH The net force acting on the car is F W = – 78125 J W = – F d = – ½ m u 2 F = 260 x 10 3 N F = 260 x 10 3 N a = (v 2 - u 2 )/2d = - 520 m/s 2 HUGE!!!!

17. KE and Work – You do  Determine the KE of a 625 kg roller coaster that is moving with a speed of 18.3 m/s.  What would the KE be if the roller coaster doubled its speed?

18. KE and Work – You do  Determine the KE of a 625 kg roller coaster that is moving with a speed of 18.3 m/s. 1.05 X 10 5 J  What would the KE be if the roller coaster doubled its speed? 4.19 X 10 5 J