Asked by Shakedown78
A 30kg sled(with rider) slides without friction down a small, ice covered hill. On the first run the sled starts from rest and its speed at the bottom of the hill is 8.50m/s. On the second run the sled starts with a speed of 1.50m/s at the top. Find the speed of the sled at the bottom of the hill on the second run. Hint-find the total energy of the sled at the top of the hill for both cases.
Answers
Answered by
Henry
Mg*h = 0.5M*V^2.
h = 0.5V^2/g = 0.5*8.5^2/9.8 = 3.69 m.
V^2 = Vo^2 + 2g*h.
Vo = 1.5 m/s.
g = 9.8 m/s^2.
V = ?
h = 0.5V^2/g = 0.5*8.5^2/9.8 = 3.69 m.
V^2 = Vo^2 + 2g*h.
Vo = 1.5 m/s.
g = 9.8 m/s^2.
V = ?
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