A 65kg skier is moving at 6.50 m/s on a frictionless,horizontal snow-covered pla
ID: 1680893 • Letter: A
Question
A 65kg skier is moving at 6.50 m/s on a frictionless,horizontal snow-covered plateau when she encounters a rough patch3.50m long. The coefficient at kinetic friction between thsi patchand her skis is 0.300. After crossing the rough patch and returningto friction-free snow, she skis down an icy frictionless hill 2.50mhigh,a) How far the skier moving when she gets the bottom of thehill?
b) How much internal enrgy was generated in crossing the roughpatch? A 65kg skier is moving at 6.50 m/s on a frictionless,horizontal snow-covered plateau when she encounters a rough patch3.50m long. The coefficient at kinetic friction between thsi patchand her skis is 0.300. After crossing the rough patch and returningto friction-free snow, she skis down an icy frictionless hill 2.50mhigh, a) How far the skier moving when she gets the bottom of thehill? b) How much internal enrgy was generated in crossing the roughpatch?
Explanation / Answer
mass m = 65 kg Initial velocity u = 6.5 m / s length of rough patch L = 3.5 m The coefficient at kinetic friction between thsi patch and herskis = 0.300 height h = 2.50m velocity after passing the rough patch v = ? from the relation v ^ 2 -u ^ 2 = 2aL v = [ u ^ 2 + 2aL ] where a = accleration = - g = -2.94 m / s^ 2 plug the values we get v = 4.655 m / s Let the height reached be H then mgH = ( 1/2) mv ^ 2 from this required height H = v ^ 2/ ( 2g) = 1.105 m a) the skier moving when she gets the bottom of the hill= 1.05 m b) Internal enrgy was generated in crossing the roughpatch = work done due to frictional force = frictional force * lengt of the patch = mg * L = 668.85 JRelated Questions
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