Work, Power, Energy

Documents you may need:

Physics Work Review of Labs (pdf or google doc) (This is Activity #12 and needs to be completed in your Notebook)

- Copy this drawing of an example showing Work being done at an Angle, this is part of Activity #12 (image)

Energy Skate Park Lab (simulation is available here and free to download or run off their site)

- Lab write-up (pdf or google doc) (This is Activity #13 and needs to be completed in your Notebook)

Roller-coaster Worksheet (pdf) (This is Activity #14 and needs to be completed in your Notebook)

- Solutions to Roller-coaster Worksheet (pdf)

Ch. 10 & 11 Review sheet (pdf or google doc) (Solutions)

Energy Review Problems (did together in class on 2/14 and 2/15) (google doc) (Solutions)

Conservation of Energy Lab (pdf or google doc)

Energy Concepts worksheet (pdf or google doc)

Work Practice Worksheet (pdf or google doc)

Power & Machines Worksheet (pdf or google doc)

Lecture Notes you may want to refer to (you will need to be signed in to your PUSD Google account in order to download them):

Energy Demos & Calculations (This is Activity #15 and needs to be completed in your Notebook)

Conservation of Energy Notes (This is Activity #16 and needs to be completed in your Notebook)

Work and Power Notes (These are just for reference.)

Energy Notes (These are just for reference.)

Homework Help:

February 6th, 2017: Ch. 10 & 11 Review Worksheet

1. Given that distance and the total amount of work done, solve for the size of the force using W=Fd. (>100)

2. a. Use the Impulse-Momentum Theorem to solve for the size of the Force. (<8E5)

b. Given that force and the distance that the car crushes you can calculate the amount of Work done. (<2E5)

3. Use the weight of the student and the distance they move their Center of Mass to calculate the amount of work done for one push-up. Given that 2007 J of work was done, divide that by the amount of work done for one push-up and you will find how many push-ups they did.

5. Calculate the KE of the car and then set that equal to the KE of the truck to solve for its velocity. Hint: the larger truck should be going slower to have the same amount of KE. (<20)

6. Find the PE of the car at that height using PE=mgh. (<70,000) If the car travels to the lowest point then all the PE becomes KE and then you can solve for the velocity of the car. (<50)

7. a. The force applied to pull the bowstring back is applied through 0.7 m. Use that information to solev for the work done. (<300)

b. Refer to the Work-Energy Theorem.

8. Remember that how ever far you move is the effort distance and how ever hard you have to push/pull is the effort force. The resistance (or "result") distance and resistance force is what you get out of the machine. If we have 100% efficiency then MA=IMA and Wout=Win. Plug in what you know and solve for the unknown effort force. (>20)

9. Calculate the PE of the person on top of the slide using PE=mgh. (<1700) Use the velocity to calculate the KE at the bottom of the slide. (<1400) Compare those two to find what was lost.

10. You can solve this problem two ways: solve the energy at each part or solve for it all together. In the beginning, at hi, there is only PE. At hf, there is both PE and KE. The decrease in PE is equal to the increase in KE because the total amount of energy must remain the same. PEi=PEf+KEf (<5)

11. a. Win = Fapplieddramp(1500); c . Compare Work out and work in.

January 11th, 2016: pg. 287 #2, pg. 291 #6, 10, pg. 297 #18, pg. 307 #57

2. Remember W=(delta)E so if you find the change in energy, you will have found the work.(<6.4E5)

18. First find the amount of PE that the first diver has that steps off the platform. All of his PE becomes KE when he hits the water. The second diver would have to have more KE at the bottom than the first diver. Using the PEi=KEf of the first diver, find out what height the second diver would have to fall from in order to make that much PE. Notice that the second diver is smaller, what does that mean his height must be to have the same amount of PE?

6. The desk is h=0 so the difference between them is the height. (<30)

10. TWO calculations: First, the PE at 425 m (>1E5). And then in order to know the change from 425 m to 225 m you need to find the PE at 225 m. The difference between them is the change in PE. (<5E4)

57. a. (>2000) b. His new KE will not be half as much! c. Ratio = (a):(b)

January 6th, 2016: pg. 272 #25, 26; pg. 273 #30; pg. 280 #81, 82

25. a. the effort distance is the distance the wedge is driven down (0.2) and the resistance distance is how far the log is split (0.05). (4)

b. The force needed to split the log is the resistance distance, the force exerted by the sledge hammer is the effort distance. (<4)

c. Use the efficiency equation with MA and IMA since you already have it. (<40%)

26. a. You are given the effort force but have to calculate the resistance force. Since you know the mass of the crate you can find its force of gravity; the resistance force is equal but opposite to the force of gravity. (<2)

b. Solve for the IMA first. The effort distance is how far the rope has to be pulled and the resistance distance is how far the crate actually moves. (2) Then use the efficiency equation with MA and IMA since you already have it. (>90%)

30. The worker can only exert an effort force of 875 N. We know that the effort distance is how far the worker moves the rope and the resistance distance is how far the rope moves. YOU HAVE TO ASSUME THAT IT IS 100% EFFICIENT IN ORDER TO SOLVE THIS PROBLEM. Set the Work in (Fede) equal to the work out (Frdr) and solve for the resistance force. (>5000)

81. Try this one on your own: a. (4), b. (<4), c. (<90)

82. a. You are given the effort force but have to calculate the resistance force. Since you know the mass you can find its force of gravity; the resistance force is equal but opposite to the force of gravity. (<4) b. (4), c. (<90)

January 4th, 2016: pg. 261 #3, pg. 264 #10, 13, pg. 265 #17-20, pg. 278 #63

3. Sometimes the Force that is applied is against the Force of Gravity. You may need to find an object's weight (Force of Gravity) and assuming its being raised at a constant velocity you know the applied force is equal but opposite to that weight. So for some of your problems: W=Fd=-Fgd=-mgd a. (>600), b. (<6000), c. (>3)

10. a. Since W=Fd and P=W/t, you can substitute and get P=Fd/t or calculate them individually. b. Think of your practice problem today.

13. Since W=Fd and P=W/t, you can substitute and get P=Fd/t or calculate them individually. (>300 s)

18. Similar to #3. (<35,000)

19. Similar to #19 (<4)

20. Use W=Fd to find the Force applied. If lifted at a constant velocity the Force applied would be equal in magnitude but opposite in direction to the Force of Gravity. Use that information to find the mass of the box. (<600)

63. Apply what you did before in #3 and #10. (>2000)