Answer to Question #85211 in Mechanics | Relativity for Maria

Question

A 71-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 μk = 0.26 .
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 = 270 N to the skier (θ = 0∘)?
Part B
What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of FT = 270 N on the skier at an upward angle θ = 12∘?

Answer 1

m=71 kg

μ_k=0.26

F_T=270 N

θ=0°

θ=12°

a-?

Solution.

According to the Newton’s second law of motions:

F ⃗_T+F ⃗_f=ma ⃗

Case A:

F_T-F_f=ma

Force of friction:

F_f=μ_k N=μ_k mg

Than the skier's acceleration:

a=(F_T-μ_k mg)/m=(270-0.26∙71∙9.8)/71=1.25 (m/s)

Case B:

F_T cosθ-F_f=ma

a=(F_T cosθ-μ_k mg)/m=(270∙0.9781-0.26∙71∙9.8)/71=1.17 (m/s)