Hi folks,
Today we did a bit more work with the relationships between kinetic energy, potential energy and work. We also did a fairly complex lab where we had to use our formulas and ingenuity to calculate the energy efficiency of a ramp and ball system.
Read the Potential vs. Kinetic energy chapter in the eBook.
Don't forget, those photos are due on Monday.
Use these formulas to solve the following Olympic energy problems. Answers below.
gPE = MgH
KE = ½ MV^2 (^2 means squared)
W = FD
1. A ski jumper who has a mass of 70 kg stands at the top of a ski jump that is 20m tall.
a. What is his potential energy?
b. If there's no friction, what would his speed be as he launches from the jump?
c. If his speed is 14 m/s, what is the energy efficiency of the ramp?
d. What work did friction do on our ski hero?
e. What was the average force of friction between the ramp and his skis if he travels
75 m down the ramp
2. a. If the Jamaican bob sled with a mass of 300 kg reaches a top speed of 150 km/h (42 m/s), what was the least amount of energy the bob sled started with?
b. What was the lowest height the bob sled started at?
c. If the track is 2200m long, what work did friction do on our Jamaican bob sledders by the time the team comes to a complete rest at the bottom?
d. If the height of the track was really 200 m, what was the energy efficiency of the bob sled?
1. a. gPE = Mgh
70kg x 20 m x 10 m/s^2 = 14,000J
b. KE = ½ MV^2
14,000 J = ½ 70 kg x v^2
v = 20 m/s
c. KE = ½ 70kg x (14)^2
KE = 6860 J
Work out/ Work in x 100
6860/14000 x 100 = 49% Efficient
d. 14,000J - 6860J = 7140 J
e. W = FD
7140J = F x 75m
F = 95.2 N
2. a. KE = ½ MV^2
KE = ½ 300kg (42 m/s)^2
KE = 264,600 J
b. gPE = MgH
264,600 = 300kg x 10m/s^2 x H
H = 88.2 m
c. W = FD
264,600 J = F x 2200 m
F = 120.27 N
d. gPE = MgH
gPE = 300 kg x 10 m/s^2 x 200 m
gPE = 600,000 J That's the maximum amount of energy. Since their energy was 264,600 J their energy efficiency was…
Work out / Work in x 100
264,600/ 600,000 x 100 = 44%
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