The same 60.0 kg skier on the 25 degree slope now meets 45.0 N of friction. What is her acceleration down the slope?

1. Friction is parallel to the slope and opposes the motion, so it subtracts from the parallel weight component. 2. The net parallel force is the mass times the acceleration due to gravity times the sine of 25 degrees, minus the friction. 3. Substitute the values: (60.0 kg)(9.80 m/s^2)(0.4226) = 248.5 N along the slope. 4. Subtract the friction: 248.5 N - 45.0 N = 203.5 N. 5. Newton's second law gives the acceleration as that net force divided by the mass. 6. Substitute again: (203.5 N) / (60.0 kg) = 3.39 m/s^2. 7. The result is smaller than the frictionless 4.14 m/s^2, exactly as friction demands. 8. Subtracting 45.0 from the acceleration rather than from the force is the trap here. _Source: OpenStax College Physics (CC BY 4.0), Ch 4 "Dynamics: Force and Newton's Laws of Motion", section 4.5 Normal, Tension, and Other Examples of Forces_