It requires a 70.4 N force (parallel to the inclined plane) to pull a 5.86 kg box up a 58.1° inclined plane with a rope at a constant speed. (a) What is the coefficient of kinetic friction between the inclined plane and the box? (b) If the rope were to break, what acceleration would the box experience as it slid down the ramp?

0.667; 4.965

Explanation:

Look at the picture I attached for the force analysis.

a) The coefficient=Friction/Normal Force. Because it's at constant speed, the force of friction + mgsin58.1° (because it's on an inclined plane and has split forces) is equal to the applied force (70.4N). Normal force is not equal to weight force though, because the box is on an inclined plane; it's equal to mgcos58.1°.

b) If it were to break, then the box no longer has an applied force, and the direction of friction has changed to up the inclined plane. F=m/a, so acceleration = mgsin58.1°- Friction/mass

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