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Chapter 04 — Dynamics

Newton's Laws, friction, inclines, connected bodies

① Explanation

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Methods

Exam practice

Q1Easyfrom the 2010 Extreme PDF

A 5 kg box is pushed across a floor with a horizontal force of 20 N. The kinetic friction coefficient between the box and the floor is 0.2. Find the acceleration of the box. (g = 10 m/s²)

Hint: Normal force: N = mg = 5 × 10 = 50 N

SOLUTION

  1. 01Normal force: N = mg = 5 × 10 = 50 N
  2. 02Friction: f = μ_k N = 0.2 × 50 = 10 N
  3. 03Net force: F_net = 20 − 10 = 10 N
  4. 04Acceleration: a = F_net / m = 10 / 5 = 2 m/s²

a = 2 m/s²

Q2Mediumfrom the 2010 Extreme PDF

A 2 kg block slides down a 30° frictional incline with kinetic friction coefficient μ_k = 0.3. Find the acceleration of the block. (g = 10 m/s²)

Hint: Along the incline: mg sinθ pulls it down, friction μ_k mg cosθ resists motion.

SOLUTION

  1. 01Along the incline: mg sinθ pulls it down, friction μ_k mg cosθ resists motion.
  2. 02a = g(sinθ − μ_k cosθ)
  3. 03a = 10 × (sin30° − 0.3 × cos30°)
  4. 04a = 10 × (0.5 − 0.3 × 0.866) = 10 × (0.5 − 0.2598)
  5. 05a = 10 × 0.2402 ≈ 2.4 m/s²

a ≈ 2.4 m/s²

Q3Exam-levelfrom the 2010 Extreme PDF

Two blocks of mass 3 kg and 2 kg are connected by a light string over a frictionless, massless pulley (Atwood machine). Find the acceleration of the system and the tension in the string. (g = 10 m/s²)

Hint: Both masses share the same acceleration a; the heavier mass (3 kg) accelerates downward.

SOLUTION

  1. 01Both masses share the same acceleration a; the heavier mass (3 kg) accelerates downward.
  2. 02For 3 kg: 3g − T = 3a
  3. 03For 2 kg: T − 2g = 2a
  4. 04Add the equations: (3 − 2)g = 5a → a = g/5 = 10/5 = 2 m/s²
  5. 05Substitute back: T = 2(g + a) = 2 × (10 + 2) = 24 N

a = 2 m/s², T = 24 N

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