Answer:
about 0.567 m/s²
Step-by-step explanation:
You want the acceleration of a 72.5 kg skier up a 21.7° slope with µk = 0.120, towed by a rope exerting a force of 383 N.
Forces
The force up the slope is 383 N.
The forces down the slope will be the sum of the force due to gravity and the friction force.
Gravity
The force down the slope due to gravity is ...
F = m·g·sin(θ) = (72.5 kg)(9.8 m/s²)(sin(21.7°) ≈ 262.7 N
Friction
The force due to friction will be the product of µk and the force normal to the slope:
F = m·g·cos(θ)·µk = (72.5 kg)(9.8 m/s²)cos(21.7°)·0.120 ≈ 79.22 N
Net Force
The net force up the slope is ...
383 N -262.7 N -79.22 N ≈ 41.08 N
This will accelerate a mass of 72.5 kg in the amount of ...
A = F/m = 41.08 N/(72.5 kg) ≈ 0.567 m/s²
The acceleration of the skier is about 0.567 m/s² up the slope.
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Additional comment
We don't have to figure the forces. Rather we can figure the gross acceleration due to the tow rope, then subtract the accelerations due to gravity and friction. This saves a few math operations as we don't have to multiply, then divide, by 72.5 kg.
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