If the applied horizontal force on a block on a rough surface is doubled, how does the acceleration change?

If the applied horizontal force on a block on a rough surface is doubled, how does the acceleration change?

• A. It doubles.
• B. It becomes zero.
• C. New acceleration a' = (2F − μ_k N)/m, which is not generally double.
• D. It is halved.

Answer: (C)

The main idea is that acceleration comes from the net horizontal force acting on the block. When the surface is rough and the block is sliding, the friction force opposite the motion is F_friction = μ_k N, with N ≈ mg on a horizontal surface. So the net force is F − μ_k N, and the acceleration is a = (F − μ_k N)/m. If the applied force is doubled, the new acceleration becomes a' = (2F − μ_k N)/m. This isn’t generally twice the original acceleration because the friction term μ_k N stays the same while the force term doubles. For example, with F = 5 N, μ_k mg = 3 N, and m = 2 kg, the original acceleration is 1 m/s^2, while after doubling the force it’s 3.5 m/s^2. Only in the special case where friction is negligible would the acceleration double.

The main idea is that acceleration comes from the net horizontal force acting on the block. When the surface is rough and the block is sliding, the friction force opposite the motion is F_friction = μ_k N, with N ≈ mg on a horizontal surface. So the net force is F − μ_k N, and the acceleration is a = (F − μ_k N)/m. If the applied force is doubled, the new acceleration becomes a' = (2F − μ_k N)/m. This isn’t generally twice the original acceleration because the friction term μ_k N stays the same while the force term doubles. For example, with F = 5 N, μ_k mg = 3 N, and m = 2 kg, the original acceleration is 1 m/s^2, while after doubling the force it’s 3.5 m/s^2. Only in the special case where friction is negligible would the acceleration double.