1. Class WORK 4/28/16 Throughout the PowerPoint ( review of Energy there were 7 questions Follow the PowerPoint lecture to answer the questions. The 7 questions were collected at the end of the classroom for points.

2. Warm UP 1) Calculate the change in potential energy of 8million Kg of water dropping 50 m over Niagara Falls.

3. Calculate the change in potential energy of 8 million Kg of water dropping 50 m over Niagara Falls PE = m g Δh = (8,000,000 kg ) (9.8 m/s 2 ) ( 50 m) = 3,920,000,000 = 3.92 x 10 9 J

4. Definition of Work Work done by a Force = Force × displacement W = Fd Unit for work is Joules ( J ) Unit for work is Joules ( J ) Note that the Force and the displacement have to be in the same direction

5. Work Two things occur whenever work is done: application of force movement of something by that force

6. Work against a force Work done to change speed.

7. Work W = Fd Examples: Twice as much work is done in lifting 2 loads 1 story high versus lifting 1 load the same vertical distance. Reason : force needed to lift twice the load is twice as much. Twice as much work is done in lifting a load 2 stories instead of 1 story. Reason : distance is twice as great.

8. UNITS: Force (F) = Newton Distance (d) = Meters Work = N*m or Joule (J) Time (t) = Seconds (s)

9. 2) Bud, a very large man of mass 130 kg, stands on a pogo stick. How much work is done as Bud compresses the spring of the pogo stick 0.05 m?

10. Bud, a very large man of mass 130 kg, stands on a pogo stick. How much work is done as Bud compresses the spring of the pogo stick 0.050 m? First, find Bud’s weight, which is the force with which he compresses the pogo stick spring Given: m=130 kg Unknown: w? g =9.8 m/s 2 equation: w =mg (130 kg)(9.8 m/s 2 )=1,274 N

11. Bud, a very large man of mass 130 kg, stands on a pogo stick. How much work is done as Bud compresses the spring of the pogo stick 0.05 m? Now use this weight to solve for the work done to compress the spring Given: F =1,274 N Unknown: W=? d = 0.050 m equation: W =Fd W = (1,274 N)(0.050 m)=64 J

12. Warm UP Bud, a very large man of mass 130 kg, stands on a pogo stick. How much work is done as Bud compresses the spring of the pogo stick 0.50 m? w =mg means weight measured in ( N) W=Fd means work measured in (J) Also look at the context of the exercise

13. Energy: Energy: The ability to do work. SI unit is Joules ( J)

15. Potential Energy Potential energy is energy which results from position or configuration ( stored energy) Gravitational potential energy -the capacity for doing work as a result the height.

16. Gravitational Potential Energy

17. Potential Energy Gain of gravitational potential energy when it is lifted from one level to a higher level change in potential energy or ΔPE, which is proportional to the change in height, Δh.

18. gravitational potential energy relies only upon the vertical change in height, Δh, and not upon the path taken Δh = change in height

19. Potential Energy gravitational potential energy =(mass)(acceleration due to gravity)(Δheight) PE = m g Δh

20. Kinetic Energy Kinetic energy is energy of motion

21. KE = Kinetic Energy an expression of the fact that a moving object can do work on anything it hits; the amount of work the object could do as a result of its motion SI UNIT is ( JOULES)

25. Kinetic Energy Kinetic energy and work KE equal to the work required to bring an object from rest to that speed, or the work the object can do while being brought to rest Equation: net force  distance  kinetic energy Fd  1/2 mv 2

26. Conservation of Energy Law of conservation of energy Energy cannot be created or destroyed; it may be transformed from one form into another, but the total amount of energy never changes. Energy is not lost, its converted to heat due to friction.

27. Conservation of Energy Example: Energy transforms without net loss or net gain in the operation of a pile driver. The Total Mechanical Energy = KE + PE

31. Review Work refers to an activity involving a force and movement in the direction of the force. A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work. Energy is the capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy. Power is the rate of doing work or the rate of using energy, which are numerically the same. If you do 100 joules of work in one second (using 100 joules of energy), the power is 100 watts.

32. 3) If a mouse and an elephant both run with the same kinetic energy, can you say which is running faster? Explain in terms of equation for KE.

33. If a mouse and an elephant both run with the same kinetic energy, can you say which is running faster? Explain in terms of h equation for KE KE = ½ mv 2, a small m must mean a large v if the KE is to be the same. The mouse must be running much faster than the elephant to have the same KE.

34. 4) An astronaut in full space gear climbs a ladder on the earth. Later, the astronaut makes the same climb on the moon. In which location does the gravitational potential energy of the astronaut change more ?

35. Earth; greater g means greater PE ( PE = mgh) moon is 1/4 the size of Earth g is 5/6 less on the moon 100 pounds = 16 pounds on the moon

37. Total Energy = Kinetic (KE) + Potential Energy (PE) = constant

38. As the skater free-falls, his gravitational energy is converted into kinetic energy. Total Energy = Kinetic (KE) + Potential Energy (PE) = constant 5) Suppose the mass of the skater is 70kg. What is the maximum height?

39. Suppose the mass of the skater is 70kg. What is the maximum height? PE=mgh PE/mg = h ( 3,000 J)/ ( 70 kg x 10m/s 2 ) = 4.3 m

40. 6) What is the final velocity when the mass of the skater is 70kg.?

41. Two ways to find Velocity final Using KEUsing PE

43. 7) The mass of the coaster is about 500kg. Approximately, what is the total mechanical energy of the car at any point along the track?

44. Using PE = mgh Given: m = 500 kg g = 9.8 m/s 2 h = 72 m ( max height) PE = (500kg) (9.8 m/s 2 )( 72 m) = 353,000 J Using KE = ½ mv 2 Given : m = 500 kg v = 37 m/s ( max speed) KE = ½ ( 500kg) (37) 2 = 353,000 J

45. 8) Brittany is changing the tire of her car on a steep hill 20.0 m high. She trips and drops the 10.0-kg spare tire, which rolls down the hill with an initial speed of 2.00 m/s. What is the speed of the tire at the top of the next hill, which is 5.00 m high? (Ignore the effects of rotation KE and friction)

46. Brittany is changing the tire of her car on a steep hill 20.0 m high. She trips and drops the 10.0-kg spare tire, which rolls down the hill with an initial speed of 2.00 m/s. What is the speed of the tire at the top of the next hill, which is 5.00 m high? h initial = 20 m h final = 5 m m = 10 kg v initial = 2 m/s a= g = 10 m/s 2 ( or 9.8 m/s 2 ) v final = ? v final = ? = 17.4 m/s