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An adventurer pulls a loaded sled with a mass of 56 kg along a horizontal meadow at a constant speed. The adventurer applies 260 N of force the rope and the angle of the rope between the sled and the adventurer is 55°. What is the coefficient of friction between the sled and the ground?

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Answer:his tension is 262 N

Explanation:1) Call T the tension of the rope2) Split T into its horizontal and vertical components.3) Horizontal component of the tension, Tx = T cos(55°)4) Vertical component of the tension, Ty = T sin (55°)5) Force equilibrium in the vertical direcction:∑Fy = 0Ty + normal force - weight of the sled = 0Call N the normal forceTy + N - 56 kg * 9.8 m/s^2Ty + N = 56 kg * 9.8 m/s^2Ty + N = 548.8NT sin(55) + N = 548.8N6) Force equilibrium in the horizontal directionconstant velocity => ∑Fx = 0Tx - Fx = 0Tcos(55) - Fx = 07) Fx is the friction force.The friction force and the normal force are related by the kinetic friction coefficient.Call μk the friction coefficientFx = μk N=> Tcos(55) - μk N = 0Tcos(55) - 0.45N = 08)  Solve the system of two equations:Eq (1) T sin(55) + N = 548.8Eq (2) T cos(55) - 0.45N = 0Eq(1) 0.819T + N = 548.8Eq(2) 0.574T - 0.45N = 0The solution of the system is T = 262.01 N and N = 334.21 NThen T ≈ 262N

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