Energy Skate Park Post-Lab
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- Madeline Rosalind Robertson
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1 Energy Skate Park Post-Lab Directions: (a) (b) (c) Create a track for the skater that has at least one loop (you add track by clicking and dragging in pieces from the top left corner). DO NOT MKE YOUR TRCK TOO IG, OR THE PROGRM WILL CRSH!!! Once you track is set up answer the following questions. Once your track is created DO NOT PRESS THE RESET UTTON!!!! If your skater falls off the track, press the Return Skater button on the top right.
2 NOTES...LIVE ND IN COLOR!!!
3 I would suggest you take notes like this: C.O.E. (CONSERVTION OF ENERGY) Etot = 7500 J P.E. = 7500 J K.E. = 0 J v = 0 m /s Skater: mass = 75kg Etot = 7500 J P.E. = 3750 J K.E. = 3750 J v = 10 m /s C C C C 10 m 5 m 3 m Etot = 7500 J P.E. = 2250 J K.E. = 5250 J v = 11.8 m /s Point : h = 10 m P.E. = mgh P.E. = (75 kg)(10 m / s2)(10 m) P.E. = 7500 J Starts from rest v = 0 m / s K.E. = 1 / 2mv 2 K.E. = 1 / 2(75 kg)(0 m / s) 2 K.E. = 0 J E tot = P.E. + K.E. E tot = 7500 J + 0 J = 7500 J Point : h = 3 m P.E. = mgh P.E. = (75 kg)(10 m / s2)(3 m) P.E. = 2250 J E tot = 7500 J (C.O.E.) K.E. =??? E tot = P.E. + K.E J = 2250 J + K.E. K.E. = 7500 J 2250 J K.E. = 5250 J K.E. = 1 / 2mv J = 1 / 2(75 kg)v 2 (1) v = 11.8 m / s (2) v = C.O.E. 2(5250 J) 75 kg Caclulations Point C: h = 5 m P.E. = mgh P.E. = (75 kg)(10 m / s2)(5 m) P.E. = = 3750 J Etot = 7500 J (C.O.E.) K.E. =??? Etot = P.E. + K.E J = 3750 J + K.E. K.E. = 7500 J 3750 J = 3750 J K.E. = 3750 J K.E. = 1/2mv J = 1/2(75 kg)v 2 v = v = 10 m / s 2(3750 J) 75 kg
4 1) Start your skater at the top of your track and let your skater do his thing. Draw a picture of your skate ramp below. e sure to include where the ground is in relation to your track.
5 2) Click the Show Pie Chart box on the right. ) Place an on your track where the potential energy is the greatest. (There may be more than one place!!!) ) Place a on your track where the kinetic energy is the greatest. (There may be more than one place!!!!)
6 mg Our skater has mass, gravity, and a height above the ground to start, but he has no velocity! What does this mean? h
7 mg The skater has no kinetic or thermal energy. The total energy is all potential at the top. h
8 mg The skater has no kinetic or thermal energy. The total energy is all potential at the top. E tot = PE + KE + TE 0 0 h
9 75kg Let's use real values for emphasis 10m
10 75kg Let's use real values for emphasis 10m
11 75kg Point : PE = mgh PE = (75kg)(10m/s 2 )(10m) Let's use real values for emphasis PE = 7500 J 10m
12 75kg Let's use real values for emphasis Point : PE = mgh PE = (75kg)(10m/s 2 )(10m) PE = 7500 J This is E tot 10m NOTE: E tot is the same for all points on the track (C.O.E.)
13 PE = 7500 J The PE will be a maximum whenever the skater reaches the same height again
14 The KE will be at a maximum whenever the skater is at the lowest point on the track. The PE is at a minimum, and has been transformed to KE. #2 on Lab
15 How much PE does the rider have at these points on the track?
16 How much PE does the rider have at these points on the track? h=3m
17 How much PE does the rider have at these points on the track? h=3m
18 How much PE does the rider have at these points on the track? Point : PE = mgh h=3m
19 How much PE does the rider have at these points on the track? Point : PE = mgh PE = (75kg)(10m/s 2 )(3m) PE = 2250 J h=3m
20 How much KE does the rider have at these points on the track? h=3m
21 How much KE does the rider have at these points on the track? E tot = 7500 J PE = 2250 J h=3m
22 How much KE does the rider have at these points on the track? E tot = 7500 J PE = 2250 J h=3m
23 How much KE does the rider have at these points on the track? E tot = 7500 J PE = 2250 J Point : E tot = PE + KE h=3m
24 How much KE does the rider have at these points on the track? E tot = 7500 J PE = 2250 J Point : E tot = PE + KE h=3m 7500J = KE KE = 5250J
25 How fast is the skater going at these points?
26 How fast is the skater going at these points?
27 How fast is the skater going at these points? Point : KE = 5250J KE = 1 / 2 mv 2
28 How fast is the skater going at these points? Point : KE = 5250J KE = 1 / 2 mv 2 2KE/m = v 2 (2KE/m) = (v 2 ) [2(5250)/75] = v 11.8m/s = v
29 3) Click the Energy vs. Time button on the right... When are the kinetic and potential energy of the skater equal? Label these points as C your diagram above.
30 PE and KE are equal whenever the skater is located HLFWY to the ground. C C C C #3 on Lab
31 PE and KE are equal whenever the skater is located HLFWY to the ground. C C C C #3 on Lab
32 PE and KE are equal whenever the skater is located HLFWY to the ground. C C C C Point C: E tot = 7500 J
33 PE and KE are equal whenever the skater is located HLFWY to the ground. C C C C Point C: E tot = 7500 J PE = 3750 J KE = 3750 J
34 PE and KE are equal whenever the skater is located HLFWY to the ground. C C C C h=5m
35 PE and KE are equal whenever the skater is located HLFWY to the ground. C C C C Point C: PE = mgh PE = (75kg)(10m/s 2 )(5m) PE = 3750 J h=5m
36 Guess what? We can figure out how fast the skater is going at this point too! C C C C h=5m
37 Guess what? We can figure out how fast the skater is going at this point too! C C C h=5m
38 Guess what? We can figure out how fast the skater is going at this point too! Point C: KE = 3750 J KE = 1 / 2 mv 2 C C C C h=5m
39 Guess what? We can figure out how fast the skater is going at this point too! Point C: KE = 3750 J KE = 1 / 2 mv 2 C 2KE/m = v 2 (2KE/m) = (v 2 ) [2(3750)/75] = v 10 m/s = v C C C h=5m
40 I would suggest you take notes like this: C.O.E. (CONSERVTION OF ENERGY) Etot = 7500 J P.E. = 7500 J K.E. = 0 J v = 0 m /s Skater: mass = 75kg Etot = 7500 J P.E. = 3750 J K.E. = 3750 J v = 10 m /s C C C C 10 m 5 m 3 m Etot = 7500 J P.E. = 2250 J K.E. = 5250 J v = 11.8 m /s Point : h = 10 m P.E. = mgh P.E. = (75 kg)(10 m / s2)(10 m) P.E. = 7500 J Starts from rest v = 0 m / s K.E. = 1 / 2mv 2 K.E. = 1 / 2(75 kg)(0 m / s) 2 K.E. = 0 J E tot = P.E. + K.E. E tot = 7500 J + 0 J = 7500 J Point : h = 3 m P.E. = mgh P.E. = (75 kg)(10 m / s2)(3 m) P.E. = 2250 J E tot = 7500 J (C.O.E.) K.E. =??? E tot = P.E. + K.E J = 2250 J + K.E. K.E. = 7500 J 2250 J K.E. = 5250 J K.E. = 1 / 2mv J = 1 / 2(75 kg)v 2 (1) v = 11.8 m / s (2) v = C.O.E. 2(5250 J) 75 kg Caclulations Point C: h = 5 m P.E. = mgh P.E. = (75 kg)(10 m / s2)(5 m) P.E. = = 3750 J Etot = 7500 J (C.O.E.) K.E. =??? Etot = P.E. + K.E J = 3750 J + K.E. K.E. = 7500 J 3750 J = 3750 J K.E. = 3750 J K.E. = 1/2mv J = 1/2(75 kg)v 2 v = v = 10 m / s 2(3750 J) 75 kg
41 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Space: Remember Newton's 1st Law???
42 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Space: Remember Newton's 1st Law??? The skater is an object at rest with no unbalanced force acting on him until you pressed the arrow keys.
43 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Space: Remember Newton's 1st Law??? Once the skater is moving on the track, he has no gravity to affect his velocity, so he flies off forever.
44 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Moon:
45 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Moon: Since the skater's acceleration due to gravity is less on the Moon, he speeds up and slows down at a lesser rate. This makes him seem to move in slow motion.
46 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Jupiter: Jupiter:
47 4)... Now change the location of your skater to space, the moon, and Jupiter... Explain the changes in motion you observe when the skater goes to each new location. Jupiter: Jupiter's gravity is much greater than the Earth's so the skater speeds up and slows down at a greater rate (greater CCELERTION) #4 on Lab
48 5) Change your skater...using the pie graph, observe where the skater is on the track when potential and kinetic energies are equal. Label these points on your track as D. C C C C #5 and #6 on Lab
49 lthough the total energy is less with a smaller mass, the spots on the track that have equal PE and KE are the same. g=9.81 m/s 2 for all masses D D D D C C C C #5 and #6 on Lab
50 lthough the total energy is less with a smaller mass, the spots on the track that have equal PE and KE are the same. g=9.81 m/s 2 for all masses D C D D D C C C When: PE = KE mgh = 1 / 2 mv 2 #5 and #6 on Lab Mass has no affect
51 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? 20kg h=5m
52 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? 20kg h=5m
53 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? PE = mgh 20kg h=5m
54 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? PE = mgh PE = (20kg)(10m/s 2 )(5m) PE = 1000 J 20kg h=5m
55 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? E tot = 2000 J PE = 1000 J 20kg h=5m
56 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? E tot = 2000 J PE = 1000 J 20kg h=5m
57 If the bulldog has a total energy of 2000 J at the top of the track, how much PE and KE does he have at this point? E tot = 2000 J PE = 1000 J E tot = PE + KE 2000 J = 1000 J + KE KE = 1000 J 20kg h=5m
58 Guess what? We can figure out how fast the bulldog is going at this point too! 20kg h=5m
59 Guess what? We can figure out how fast the bulldog is going at this point too! 20kg h=5m
60 Guess what? We can figure out how fast the bulldog is going at this point too! KE = 1000 J KE = 1 / 2 mv 2 20kg h=5m
61 Guess what? We can figure out how fast the bulldog is going at this point too! KE = 1000 J KE = 1 / 2 mv 2 2KE/m = v 2 (2KE/m) = (v 2 ) [2(1000)/20] = v 10 m/s = v 20kg h=5m
62 7) Does changing the mass of the skater change the total energy in the skaterramp system? Explain the relationship. y lowering the mass of the skater, the total energy of the skater is less than before. Total Energy = PE top Total Energy = mgh
63 8) Finally, click the Track Friction button on the right. Move the slider to the right. What happens to the skater? s the skater moves through the track, what type(s) of energy does he possess now?
64 8) Finally, click the Track Friction button on the right. Move the slider to the right. What happens to the skater? s the skater moves through the track, what type(s) of energy does he possess now? The kinetic energy has been converted entirely to thermal energy Hello real life!
65 9) Move the skater around until you find where the skater has only thermal energy. What TWO things have to happen for the skater to have only thermal energy? What have we eliminated?
66 9) Move the skater around until you find where the skater has only thermal energy. What TWO things have to happen for the skater to have only thermal energy? PE = mgh 0 0 KE = 1/2mv 2 With no height (h) or velocity (v), all the energy has been converted to thermal energy h = 0 m v = 0 m/s
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