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Force and translational dynamics practice problems AP Physics C: Mechanics, Unit 2

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 free-response question

Newton's laws, friction, inclines, connected objects, and resistive forces that depend on velocity. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for force and translational dynamics

ΣF = ma = m dv/dtNewton's second law
fk = μkNkinetic friction
Fdrag = −bvlinear drag
vT = mg/bterminal speed with linear drag

Use g = 9.8 m/s², as on the AP reference sheet. You can type answers like 4.9, 1/3 or 3.6e2. The full sheet is in our AP Physics C equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
A net force of 20 N acts on a 5 kg object. Find its acceleration.
m/s²

Hint

ΣF = ma.

Worked solution
  1. a = 20/5 = 4 m/s².
Q2Warm-up
A 10 kg box slides across a horizontal floor with coefficient of kinetic friction 0.3. Find the friction force.
N

Hint

On a horizontal floor with no other vertical forces, N = mg.

Worked solution
  1. N = 10 × 9.8 = 98 N.
  2. f = 0.3 × 98 = 29.4 N.
Q3Warm-up
The Earth pulls down on a book with its weight. What is the Newton's third law partner of this force?

Hint

Third-law pairs act on two different objects and are the same type of force.

Worked solution
  1. The pair of “Earth pulls book (gravity)” is “book pulls Earth (gravity)”.
  2. The answer is B. The table's push is a different force on the same object.

Exam-style questions

Questions at the level of the AP Physics C: Mechanics exam. The ones marked Challenge combine more than one idea.

Q4Exam-style
Masses of 4 kg and 6 kg hang on either side of a light frictionless pulley. Find the magnitude of their acceleration.
m/s²

Hint

Treat the two masses as one system pulled by the difference in weights.

Worked solution
  1. Net force on the system: (6 − 4)g = 19.6 N; total mass 10 kg.
  2. a = 19.6/10 = 1.96 m/s².
Q5Exam-style
A block slides down a frictionless incline at 30° to the horizontal. Find its acceleration.
m/s²

Hint

Only the component of gravity along the incline accelerates it.

Worked solution
  1. a = g sin 30°.
  2. a = 9.8 × 0.5 = 4.9 m/s².
Q6Exam-style
A 0.20 kg object falls through a fluid that exerts a drag force F = −bv with b = 0.50 kg/s. Find its terminal speed.
m/s

Hint

At terminal speed the acceleration is zero.

Worked solution
  1. At terminal speed: mg = bvT.
  2. vT = 0.20 × 9.8/0.50 = 3.92 m/s.
Q7Exam-style
A 2.0 kg object starts from rest and is pushed by a force F(t) = 6t N along a frictionless surface. Find its speed at t = 2 s.
m/s

Hint

a(t) = F/m. Integrate to find v(t).

Worked solution
  1. a = 6t/2 = 3t, so v = 1.5t2.
  2. v(2) = 1.5 × 4 = 6 m/s.
Q8Challenge
A 2.0 kg puck slides on a horizontal surface with a resistive force F = −bv, b = 0.40 kg/s, and no other horizontal forces. Its initial speed is 10 m/s. Find its speed after 5.0 s.
m/s

Hint

Write m dv/dt = −bv, separate the variables and integrate.

Worked solution
  1. m dv/dt = −bv gives dv/v = −(b/m) dt, so v = v0e−bt/m.
  2. bt/m = 0.40 × 5.0/2.0 = 1.0.
  3. v = 10e−1 = 3.68 m/s.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Mathematical routines style · 7 points · about 20 minutes

An object of mass m is released from rest and falls vertically. Air exerts a resistive force of magnitude bv, where b is a positive constant.

  1. (a) Derive an expression for the terminal speed vT.
  2. (b) Derive an expression for v(t).
  3. (c) For m = 0.50 kg and b = 2.0 kg/s, find the time to reach half the terminal speed.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

At terminal speed, mg − bvT = 0, so vT = mg/b.

Part (b)

m dv/dt = mg − bv. Separating: ∫ dv/(mg − bv) = ∫ dt/m, giving v = (mg/b)(1 − e−bt/m).

Part (c)

1 − e−bt/m = ½, so t = (m/b) ln 2 = 0.25 × 0.693 = 0.17 s.

Your mark: 0 of 7

Common mistakes in force and translational dynamics

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics C: Mechanics past papers, with official scoring guidelines. · All AP Physics C: Mechanics units