For a 4-kg block on a 30° incline with kinetic friction μk = 0.3, which of the following is true about the net force components? (use g = 9.8 m/s^2)

For a 4-kg block on a 30° incline with kinetic friction μk = 0.3, which of the following is true about the net force components? (use g = 9.8 m/s^2)

• A. N ≈ 40 N; F_par ≈ 20 N; F_f ≈ 12 N; F_net ≈ 8 N
• B. N ≈ 33.9 N; F_par ≈ 19.6 N; F_f ≈ 10.17 N; F_net ≈ 9.43 N
• C. N ≈ 20 N; F_par ≈ 5 N; F_f ≈ 1 N; F_net ≈ 4 N
• D. N ≈ 70 N; F_par ≈ 25 N; F_f ≈ 18 N; F_net ≈ 4 N

Answer: (B)

When a block sits on an incline, resolve the weight into components parallel and perpendicular to the surface. The parallel component pulls the block down the slope and equals mg sin(theta); the perpendicular component presses into the plane and equals mg cos(theta). Friction, being kinetic, has magnitude μk times the normal force and acts opposite to the motion along the surface. With mass 4 kg and g 9.8 m/s^2, the weight is 39.2 N. For a 30° incline, the down-slope component is 39.2 × sin(30°) = 19.6 N. The normal force is 39.2 × cos(30°) ≈ 33.9 N. The kinetic friction magnitude is μk N = 0.3 × 33.9 ≈ 10.17 N, acting up the slope. The net force along the incline is the down-slope component minus the friction: 19.6 − 10.17 ≈ 9.43 N down the incline. So the forces have: N ≈ 33.9 N, F_par ≈ 19.6 N, F_f ≈ 10.17 N, and F_net ≈ 9.43 N down the incline (which would give an acceleration of about 2.36 m/s^2 down the slope). The numbers above match the option that lists N ≈ 33.9 N, F_par ≈ 19.6 N, F_f ≈ 10.17 N, and F_net ≈ 9.43 N. The other choices don’t satisfy the relationships N = mg cos(30°), F_f = μk N, or F_par = mg sin(30°) with the given values.

When a block sits on an incline, resolve the weight into components parallel and perpendicular to the surface. The parallel component pulls the block down the slope and equals mg sin(theta); the perpendicular component presses into the plane and equals mg cos(theta). Friction, being kinetic, has magnitude μk times the normal force and acts opposite to the motion along the surface.

With mass 4 kg and g 9.8 m/s^2, the weight is 39.2 N. For a 30° incline, the down-slope component is 39.2 × sin(30°) = 19.6 N. The normal force is 39.2 × cos(30°) ≈ 33.9 N. The kinetic friction magnitude is μk N = 0.3 × 33.9 ≈ 10.17 N, acting up the slope. The net force along the incline is the down-slope component minus the friction: 19.6 − 10.17 ≈ 9.43 N down the incline.

So the forces have: N ≈ 33.9 N, F_par ≈ 19.6 N, F_f ≈ 10.17 N, and F_net ≈ 9.43 N down the incline (which would give an acceleration of about 2.36 m/s^2 down the slope).

The numbers above match the option that lists N ≈ 33.9 N, F_par ≈ 19.6 N, F_f ≈ 10.17 N, and F_net ≈ 9.43 N. The other choices don’t satisfy the relationships N = mg cos(30°), F_f = μk N, or F_par = mg sin(30°) with the given values.