A 65-kg water skier is being accelerated by a ski boat on a flat (\"glassy\") la

Question

A 65-kg water skier is being accelerated by a ski boat on a flat ("glassy") lake. The coefficient of kinetic friction between the skier's skis and the water surface is ?k = 0.25 . (Figure 1)

Part A

What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude FT = 270N to the skier (? = 0?)?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of FT = 270N on the skier at an upward angle ? = 12??

Express your answer to two significant figures and include the appropriate units.

A 65-kg water skier is being accelerated by a ski boat on a flat ("glassy") lake. The coefficient of kinetic friction between the skier's skis and the water surface is ?k = 0.25 . (Figure 1) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude FT = 270N to the skier (? = 0?)? Express your answer to two significant figures and include the appropriate units. What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of FT = 270N on the skier at an upward angle ? = 12?? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

1. Force equation:

F - uN = ma

N = mg

=> a = (270 - 0.25*650)/(65) = 1.654 m/s^2

2. Force equations:

FCos12 - uN = ma

FSin12 + N = mg

=> N = 593.86N

=> a = (270Cos12 - 0.25*593.86)/65 = 1.79 m/s^2

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