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Force and translational dynamics practice problems AP Physics 1, Unit 2

Written by STEM-Tutor's tutors · Updated October 2026 · 10 questions and 1 free-response question

Free-body diagrams, Newton's three laws, friction, springs, gravitation and circular motion. 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

a = ΣF / mNewton's second law, for a system
Ff ≤ μFNfriction: equal to μkFN when sliding
Fs = −kΔxspring force (Hooke's law)
Fg = Gm1m2 / r2gravitational force between two masses
ac = v2 / rcentripetal acceleration

All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 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 12 N acts on a 3.0 kg trolley. What is its acceleration?
m/s²

Hint

Newton's second law links net force, mass and acceleration.

Worked solution
  1. a = ΣF / m.
  2. a = 12 / 3.0 = 4.0 m/s², in the direction of the net force.
Q2Warm-up
A book rests on a table. By Newton's third law, the force paired with “the Earth's gravitational force on the book” is:

Hint

A third-law pair involves the same two objects, swapped, and the same type of force.

Worked solution
  1. The original force is gravitational, exerted by the Earth on the book.
  2. Its partner must be gravitational too, exerted by the book on the Earth.
  3. The normal force on the book balances the weight, but that is Newton's first law, not a third-law pair.
  4. The answer is B.
Q3Warm-up
A 5.0 kg box sits on a level floor. The coefficient of static friction is 0.40. What is the smallest horizontal force that will start it moving?
N

Hint

On a level floor with only a horizontal push, the normal force equals the weight.

Worked solution
  1. Normal force: FN = mg = 5.0 × 9.8 = 49 N.
  2. Maximum static friction: μsFN = 0.40 × 49.
  3. The push must just exceed 19.6 N to start the box moving.
Q4Warm-up
A lift moves upward at a constant speed. How does the tension in its cable compare with the lift's weight?

Hint

Constant speed in a straight line: what does that tell you about the net force?

Worked solution
  1. Constant velocity means zero acceleration.
  2. By Newton's first law the net force is zero, so the upward tension balances the weight.
  3. The answer is B, whatever the constant speed. The tension is larger only while the lift speeds up going upward.

Exam-style questions

Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.

Q5Exam-style
A 2.0 kg block slides down a frictionless ramp inclined at 30° to the horizontal. What is its acceleration?
30°mgFNfriction
Free-body diagram for a block on a 30° slope. Weight mg acts straight down, the normal force FN acts at right angles to the slope, and friction acts up the slope if the block tends to slide down. Resolve weight into mg sinθ along the slope and mg cosθ into it.
m/s²

Hint

Resolve the weight into components along and perpendicular to the ramp.

Worked solution
  1. Along the ramp the only force is the component of weight, mg sinθ.
  2. a = mg sinθ / m = g sinθ, so the mass cancels.
  3. a = 9.8 × sin 30° = 4.9 m/s², down the ramp.
Q6Exam-style
A 3.0 kg mass and a 5.0 kg mass hang on either side of a light string over a frictionless pulley. What is the size of their acceleration?
m/s²

Hint

Treat both masses and the string as one system. What is the net force that drives it?

Worked solution
  1. The net force on the system is the difference in weights: (5.0 − 3.0) × 9.8 = 19.6 N.
  2. The total mass being accelerated is 8.0 kg.
  3. a = 19.6 / 8.0 = 2.45 m/s².
Q7Exam-style
For the two masses in the previous question (3.0 kg and 5.0 kg over a pulley), what is the tension in the string?
N

Hint

Apply Newton's second law to the 3.0 kg mass alone. Which way does it accelerate?

Worked solution
  1. The 3.0 kg mass accelerates upward at 2.45 m/s².
  2. Newton's second law for it: T − mg = ma, so T = m(g + a).
  3. T = 3.0 × (9.8 + 2.45) = 36.8 N. Check: it lies between the two weights, 29.4 N and 49 N, as it must.
Q8Exam-style
A 1200 kg car drives round a flat bend of radius 50 m at a steady 15 m/s. What friction force must the road supply?
N

Hint

On a flat bend, friction is the only horizontal force, so it provides all of the centripetal force.

Worked solution
  1. The net force must point to the centre and equal mv2/r.
  2. F = 1200 × 152 / 50 = 1200 × 225 / 50.
  3. F = 5400 N, towards the centre of the bend.
Q9Challenge
A 0.50 kg ball on a 0.80 m string is swung in a vertical circle. At the lowest point its speed is 4.0 m/s. What is the tension in the string there?
N

Hint

At the lowest point, tension acts up and weight acts down. The net force must point up, towards the centre.

Worked solution
  1. Net force towards the centre (upward): T − mg = mv2/r.
  2. mv2/r = 0.50 × 16 / 0.80 = 10 N, and mg = 4.9 N.
  3. T = 4.9 + 10 = 14.9 N.
Q10Exam-style
The distance between two planets doubles. What happens to the gravitational force between them?

Hint

The force depends on 1/r2.

Worked solution
  1. Fg = Gm1m2/r2, an inverse-square law.
  2. Doubling r multiplies r2 by 4, so the force is divided by 4.
  3. The answer is B, one quarter.

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.

Translation between representations style · 8 points · about 22 minutes

A block of mass m is pulled along a rough horizontal floor by a rope that makes an angle θ above the horizontal. The tension in the rope is F, and the coefficient of kinetic friction is μk.

mFθrough surface, coefficient μk
A block pulled by a force F at an angle θ above the horizontal.
  1. (a) Draw a free-body diagram showing every force on the block, as a labelled arrow starting on the block.
  2. (b) Derive an expression for the acceleration of the block in terms of F, θ, m, μk and g.
  3. (c) Calculate the acceleration when F = 20 N, θ = 30°, m = 4.0 kg and μk = 0.25.
  4. (d) A student says: “Pulling at a larger angle always gives a larger acceleration, because the friction gets smaller.” Do you agree? Explain.

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)

Four forces: weight mg straight down; normal force FN straight up; tension F along the rope at θ above the horizontal; kinetic friction backward, opposite the motion.

Part (b)

Vertically: FN + F sinθ = mg, so FN = mg − F sinθ. Friction = μk(mg − F sinθ). Horizontally: ma = F cosθ − μk(mg − F sinθ), so a = [F cosθ − μk(mg − F sinθ)] / m.

Part (c)

FN = 39.2 − 10 = 29.2 N, friction = 7.3 N, F cos 30° = 17.3 N. a = (17.3 − 7.3) / 4.0 = 2.5 m/s².

Part (d)

No. A larger angle does reduce the normal force and so the friction, but it also reduces the forward component F cosθ. At small angles the acceleration rises, but beyond a certain angle the loss of forward force wins and the acceleration falls.

Your mark: 0 of 8

Common mistakes in force and translational dynamics

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units