Posted by James on Sunday, February 27, 2011 at 12:29pm.
They want you to neglect friction and use conservation of energy. The masses will not affect the result.
g*L*sin30 = V^2/2
L = 143 m
g = 9.81 m/s^2
Solve for V, and round to nearest m/s
The m/s number that you get will correspond to well over 100 miles per hour - a very unsafe value.
There will actually be appreciable friction unless they are on an ice-covered luge run, and using no brakes or foot dragging.
This is a rather unrealistic problem.
THANKYOU
Thank you drwls, there is another question that I am not getting really well right now. It is this: Two skaters are in the exact center of a circular frozen pond. Skater 1 pushes skater 2 off with a force of 100 N for 1.4 seconds. If skater 1 has a mass of 30 kg and skater 2 has a mass of 74 kg, what is the relative velocity (v1 - v2) after the push to the nearest hundredth of a m/s? After reaching the other shore, how fast, to the nearest tenth of a m/s, must skater 1 run around the lake to meet skater 2 at the opposite shore?
Thanks
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