Asked by jake
Physics force of friction question
a 63kg water skier is pulled across a lake by a rope that exerts a force of 3.2x10^2 and makes an angle 17 degrees above the horizontal. if the coefficient between the skies and the lake is 0.27 at what rate will the skier accelerate?
a 63kg water skier is pulled across a lake by a rope that exerts a force of 3.2x10^2 and makes an angle 17 degrees above the horizontal. if the coefficient between the skies and the lake is 0.27 at what rate will the skier accelerate?
Answers
Answered by
drwls
acceleration = (Net horizontal force)/Mass
= (320 cos 17 - M g *0.27)/M
= 306/M -0.27 g = (4.86 - 2.65)
= 2.21 m/s^2
= (320 cos 17 - M g *0.27)/M
= 306/M -0.27 g = (4.86 - 2.65)
= 2.21 m/s^2
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