A 10 N horizontal force is applied to a 5 kg block on a horizontal surface with μk = 0.3. What is the resulting acceleration?

A 10 N horizontal force is applied to a 5 kg block on a horizontal surface with μk = 0.3. What is the resulting acceleration?

• A. a ≈ 0.60 m/s^2 to the right
• B. a ≈ 0.30 m/s^2 to the right
• C. a ≈ 0.06 m/s^2 to the left
• D. a ≈ 0.06 m/s^2 to the right

Answer: (D)

In this situation, the motion of the block depends on the balance between the applied push and the friction that resists motion. The normal force on a horizontal surface is N = m g = 5 kg × 9.8 m/s^2 ≈ 49 N. The maximum static friction is f_s,max = μs N, and the problem provides the kinetic coefficient μk = 0.3. Since μs is at least as large as μk, the maximum static friction is at least μk N ≈ 0.3 × 49 ≈ 14.7 N. The applied horizontal force is only 10 N, which is less than the available static friction. That means the frictional force can adjust to exactly balance the push, keeping the block at rest. When the block doesn’t move, the net force is zero and the acceleration is 0 m/s^2. If the block somehow were sliding, the kinetic friction would be about 14.7 N to the left, giving a net force of 10 − 14.7 ≈ −4.7 N and an acceleration of about −0.94 m/s^2 to the left, but that scenario requires F to exceed the kinetic friction, which it does not here. So the proper result is zero acceleration.

In this situation, the motion of the block depends on the balance between the applied push and the friction that resists motion. The normal force on a horizontal surface is N = m g = 5 kg × 9.8 m/s^2 ≈ 49 N. The maximum static friction is f_s,max = μs N, and the problem provides the kinetic coefficient μk = 0.3. Since μs is at least as large as μk, the maximum static friction is at least μk N ≈ 0.3 × 49 ≈ 14.7 N.

The applied horizontal force is only 10 N, which is less than the available static friction. That means the frictional force can adjust to exactly balance the push, keeping the block at rest. When the block doesn’t move, the net force is zero and the acceleration is 0 m/s^2.

If the block somehow were sliding, the kinetic friction would be about 14.7 N to the left, giving a net force of 10 − 14.7 ≈ −4.7 N and an acceleration of about −0.94 m/s^2 to the left, but that scenario requires F to exceed the kinetic friction, which it does not here.

So the proper result is zero acceleration.