Asked by CHEESE🧀
a 60 kg skier with an initial velocity of 12 m/s coasts up a hill. at the top of the hill, the skier is traveling at 9.75 m/s. Assuming there is no friction force in the skis, how high was the hill
A) 7.35 m
B) 0.25 m
C) 10 m
D) 2.5 m
A) 7.35 m
B) 0.25 m
C) 10 m
D) 2.5 m
Answers
Answered by
R_scott
m V^2 - m v^2 = 2 m g h
12^2 - 9.75^2 = 2 * 9.8 * h
12^2 - 9.75^2 = 2 * 9.8 * h
Answered by
Ri
Okay so
GPE = KE (Gravitational Potential Energy = Kinetic Energy)
GPE = mgh (Gravitational Potential Energy = mass x gravity x height)
KE = 1/2mv^2
KE = 1/2(60) * (12)^2
KE = 4320
KE before hill: 4320 J
KE = 1/2(60) * (9.75)^2
KE = 2851.875
KE after hill: 2851.875 J
4320 - 2851.875 = 1468.125
1468.125 = mgh
1468.125 = (60)(9.81)(h)
1468.125/(60)(9.81) = h
1468.125/588.6 = h
2.494 = h
The answer is D. 2.5 m
c:
GPE = KE (Gravitational Potential Energy = Kinetic Energy)
GPE = mgh (Gravitational Potential Energy = mass x gravity x height)
KE = 1/2mv^2
KE = 1/2(60) * (12)^2
KE = 4320
KE before hill: 4320 J
KE = 1/2(60) * (9.75)^2
KE = 2851.875
KE after hill: 2851.875 J
4320 - 2851.875 = 1468.125
1468.125 = mgh
1468.125 = (60)(9.81)(h)
1468.125/(60)(9.81) = h
1468.125/588.6 = h
2.494 = h
The answer is D. 2.5 m
c:
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