A 3 kg mass sits on a 60° incline with kinetic friction coefficient μk = 0.15. Which statement best describes the forces along the plane and the motion?

A 3 kg mass sits on a 60° incline with kinetic friction coefficient μk = 0.15. Which statement best describes the forces along the plane and the motion?

• A. The block will not move because friction balances gravity along the plane.
• B. The component of gravity along the plane is about 25.4 N, the kinetic friction is about 2.20 N, and the block will move down the plane with an acceleration of roughly 7.7 m/s^2.
• C. The normal force is 44.1 N and the friction is 2.20 N.
• D. The friction equals the gravitational component along the plane.

Answer: (B)

When an object sits on a slope with kinetic friction, motion along the plane is determined by the net force down the incline: the downslope component of gravity minus the friction opposing the motion. The gravity component along the plane is m g sinθ, and the normal force is m g cosθ, with the kinetic friction equal to μk times the normal force. For this case, the downslope gravitational component is about 3 × 9.8 × sin60 ≈ 25.5 N. The normal force is 3 × 9.8 × cos60 ≈ 14.7 N, so the kinetic friction is μk N ≈ 0.15 × 14.7 ≈ 2.2 N, acting up the plane. Since 25.5 N downslope exceeds 2.2 N of friction, the net force along the plane is roughly 23.3 N down the plane. The acceleration follows from a = Fnet / m ≈ 23.3 / 3 ≈ 7.8 m/s^2 down the plane. Thus the block will move down the plane with an acceleration about 7.7–7.8 m/s^2. Note that the normal force is not 44.1 N (it’s 14.7 N), friction does not balance gravity along the plane, and friction does not equal the gravitational component along the plane.

When an object sits on a slope with kinetic friction, motion along the plane is determined by the net force down the incline: the downslope component of gravity minus the friction opposing the motion. The gravity component along the plane is m g sinθ, and the normal force is m g cosθ, with the kinetic friction equal to μk times the normal force.

For this case, the downslope gravitational component is about 3 × 9.8 × sin60 ≈ 25.5 N. The normal force is 3 × 9.8 × cos60 ≈ 14.7 N, so the kinetic friction is μk N ≈ 0.15 × 14.7 ≈ 2.2 N, acting up the plane. Since 25.5 N downslope exceeds 2.2 N of friction, the net force along the plane is roughly 23.3 N down the plane. The acceleration follows from a = Fnet / m ≈ 23.3 / 3 ≈ 7.8 m/s^2 down the plane.

Thus the block will move down the plane with an acceleration about 7.7–7.8 m/s^2. Note that the normal force is not 44.1 N (it’s 14.7 N), friction does not balance gravity along the plane, and friction does not equal the gravitational component along the plane.