A 10 kg block on a horizontal surface with μ_s = 0.4 and μ_k = 0.3 is at rest. If a horizontal force of 25 N is applied, will it move? If it does move, what is the acceleration using μ_k?

A 10 kg block on a horizontal surface with μ_s = 0.4 and μ_k = 0.3 is at rest. If a horizontal force of 25 N is applied, will it move? If it does move, what is the acceleration using μ_k?

• A. Yes; the block will move with a = (F − μ_k m g)/m.
• B. Yes; the block will move with some acceleration.
• C. No movement; a = 0.
• D. No movement; F is less than μ_s m g, so friction prevents motion.

Answer: (D)

The main idea here is static friction: on a horizontal surface, static friction can adjust up to a maximum value of μs N to oppose the push. The normal force N equals m g, so the maximum static friction is μs m g. Compute it: m = 10 kg, g ≈ 9.8 m/s², μs = 0.4, so μs m g ≈ 0.4 × 10 × 9.8 ≈ 39 N. The applied horizontal force is 25 N, which is less than that maximum. Therefore the friction force can counter all of the push without needing motion, and the block stays at rest. The acceleration is zero. If the force had exceeded μs m g, the block would start moving and you’d use the kinetic friction value μk m g to find the acceleration via a = (F − μk m g)/m. In this case, since F is below the threshold, motion does not occur, which is why the statement about no movement due to F being less than μs m g is the best answer.

The main idea here is static friction: on a horizontal surface, static friction can adjust up to a maximum value of μs N to oppose the push. The normal force N equals m g, so the maximum static friction is μs m g.

Compute it: m = 10 kg, g ≈ 9.8 m/s², μs = 0.4, so μs m g ≈ 0.4 × 10 × 9.8 ≈ 39 N. The applied horizontal force is 25 N, which is less than that maximum. Therefore the friction force can counter all of the push without needing motion, and the block stays at rest. The acceleration is zero.

If the force had exceeded μs m g, the block would start moving and you’d use the kinetic friction value μk m g to find the acceleration via a = (F − μk m g)/m. In this case, since F is below the threshold, motion does not occur, which is why the statement about no movement due to F being less than μs m g is the best answer.