A 2 kg block on a horizontal surface has kinetic friction coefficient μk = 0.4. If an external horizontal force of 20 N is applied, what is the net horizontal force on the block?

A 2 kg block on a horizontal surface has kinetic friction coefficient μk = 0.4. If an external horizontal force of 20 N is applied, what is the net horizontal force on the block?

• A. 0 N
• B. 9.44 N to the right
• C. -15.68 N to the left
• D. 4.32 N to the right

Answer: (D)

On a horizontal surface with kinetic friction, the net horizontal force is the applied force minus the kinetic friction force. The friction force is μk times the normal force, and on a horizontal surface the normal force equals the weight, N = m g. Here the mass is 2 kg and g ≈ 9.8 m/s^2, so N = 2 × 9.8 = 19.6 N. The kinetic friction is F_friction = μk N = 0.4 × 19.6 = 7.84 N, opposing the motion. The applied force is 20 N to the right, so the net force is 20.0 − 7.84 = 12.16 N to the right. This means the block accelerates to the right with a = F_net / m = 12.16 / 2 ≈ 6.08 m/s^2. If you’ve seen 4.32 N as the net, that would require a larger friction force than given (or a different g or μk); with the provided values, the net is 12.16 N to the right.

On a horizontal surface with kinetic friction, the net horizontal force is the applied force minus the kinetic friction force. The friction force is μk times the normal force, and on a horizontal surface the normal force equals the weight, N = m g.

Here the mass is 2 kg and g ≈ 9.8 m/s^2, so N = 2 × 9.8 = 19.6 N. The kinetic friction is F_friction = μk N = 0.4 × 19.6 = 7.84 N, opposing the motion.

The applied force is 20 N to the right, so the net force is 20.0 − 7.84 = 12.16 N to the right. This means the block accelerates to the right with a = F_net / m = 12.16 / 2 ≈ 6.08 m/s^2.

If you’ve seen 4.32 N as the net, that would require a larger friction force than given (or a different g or μk); with the provided values, the net is 12.16 N to the right.