A 72-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 Mk = 0.25 (Fig. 4–74). (a) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal ten- sion force of magnitude FT = 240 N to the skier (0 = 0°)? (b) What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of FT = 240 N on the skier at an upward angle 0 = 12°? (c) Explain why the skier's acceleration in part (b) is greater than that in part (a). FT = 240 N Mk = 0.25 FIGURE 4-74 Problem 91.

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Chapter13: Gravitation
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A 72-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 Mk = 0.25
(Fig. 4–74). (a) What is the skier's acceleration if the rope
pulling the skier behind the boat applies a horizontal ten-
sion force of magnitude FT = 240 N to the skier (0 = 0°)?
(b) What is the skier's horizontal acceleration if the rope
pulling the skier exerts a force of FT = 240 N on the
skier at an upward angle 0 = 12°? (c) Explain why the
skier's acceleration in part (b) is greater than that in
part (a).
FT = 240 N
Mk = 0.25
FIGURE 4-74 Problem 91.
Transcribed Image Text:A 72-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 Mk = 0.25 (Fig. 4–74). (a) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal ten- sion force of magnitude FT = 240 N to the skier (0 = 0°)? (b) What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of FT = 240 N on the skier at an upward angle 0 = 12°? (c) Explain why the skier's acceleration in part (b) is greater than that in part (a). FT = 240 N Mk = 0.25 FIGURE 4-74 Problem 91.
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