A 4.0 kg block is pulled across a rough horizontal floor by a constant horizontal force of 28 N. If the coefficient of kinetic friction between the block and the floor is 0.20 (take g = 9.8 m/s^2), what is the acceleration of the block?

Answer and explanation

Correct answer: B. 5.04 m/s^2

  1. Step 1: Calculate the normal force on the block: F_N = m g = 4.0 kg 9.8 m/s^2 = 39.2 N.
  2. Step 2: Compute the opposing force of kinetic friction: f_k = mu_k F_N = 0.20 39.2 N = 7.84 N.
  3. Step 3: Find net horizontal force: F_net = F_pull - f_k = 28 N - 7.84 N = 20.16 N.
  4. Step 4: Apply Newton's Second Law: a = F_net / m = 20.16 N / 4.0 kg = 5.04 m/s^2.
Answer
5.04 m/s^2.
Why not the others: '7.00 m/s^2' ignores friction entirely (28 / 4.0); '3.08 m/s^2' doubles the friction value; '1.96 m/s^2' computes f_k / m.

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