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A 651 N cross-country skier is moving over packed level snow at 6.6 ms The frictional force between the skies amd the snow is 6.0 N What distance will the skier travel before coming to a stop?
Kkcurls
The friction force of 6.0 N is the retarding force on the skier, so
you can use a straight application of [ F = m a ].
His mass is (weight/g) = 651/9.8 kg
F = m a
6 N = (651/9.8) a
a = (6 x 9.8)/651 m/s2
The time he continues to move while decelerating from 6.6 m/s to zero is
6.6 / a = (6.6 x 651) / (6 x 9.8) seconds.
His average speed during that time is (6.6/2) m/s.
Total distance = (average speed) x (time) = (6.6/2) x (6.6 x 651)/(6 x 9.8) =
241.14 meters (rounded)
during a pleasant glide lasting 73.1 seconds.
=======================================
Check:
His initial kinetic energy is [ 1/2 m v2 ] = 0.5 x (651/9.8) (6.6)2 = 1,446.81 joules.
The energy lost to friction = (force) x (distance) = (6.0 N) x (241.14 m) = 1,446.84 joules.
Close enough !
Gee. It would have been a lot easier to solve it this way in the first place.
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∙ 13y ago
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A 651 N cross-country skier is moving over packed level snow at 6.6 ms The frictional force between the skies amd the snow is 60.0 N What distance will the skier travel before coming to a stop?
The skier's mass is (weight/g) = 651/9.8 kg.His initial kinetic energy is [ 1/2 m v2 ] = 0.5 x (651/9.8) x (6.6)2 = 1,446.81 joules .That's the energy lost to friction during the glide to a stop.Energy burned by friction = (friction force) x (distance)Distance = (energy) / (force) = (1,446.81 joules) / (60 N) = 24.11 meters .
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