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Study Guide: How to Solve: Newton’s Laws Problems
Source: https://www.fatskills.com/k-12-assessment-tests/chapter/how-to-solve-newtons-laws-problems

How to Solve: Newton’s Laws Problems

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~7 min read

How to Solve: Newton’s Laws Problems

For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script

Introduction

"If you can solve Newton’s Laws problems, you can predict why a car skids on ice, why rockets launch, and how to lift a heavy box without breaking your back—plus, you’ll crush 20% of your physics exam. Let’s get started."

What You Need To Know First

Before tackling Newton’s Laws, you must already understand: 1. Forces as vectors – Forces have both size (magnitude) and direction. You must add them using vector rules (tip-to-tail or components). 2. Free-body diagrams (FBDs) – A sketch showing only the object and all forces acting on it (not forces it exerts on others). 3. Basic algebra – Solving for unknowns in equations like F = ma.

If any of these are shaky, pause here and review them first.

Key Vocabulary

Term Plain-English Definition Quick Example
Force (F) A push or pull on an object. Measured in Newtons (N). Pushing a book across a table.
Net Force (Fnet) The total force acting on an object after adding all forces (vector sum). If you push a box right with 10 N and friction pushes left with 4 N, Fnet = 6 N right.
Mass (m) How much "stuff" is in an object. Measured in kilograms (kg). Does not change with location. A 2 kg brick has the same mass on Earth and the Moon.
Weight (W) The force of gravity pulling on an object. W = mg (g = 9.8 m/s² on Earth). A 5 kg object weighs 49 N on Earth.
Inertia An object’s resistance to changes in motion. More mass = more inertia. A bowling ball is harder to stop than a tennis ball.
Normal Force (N) The support force a surface exerts on an object. Always perpendicular to the surface. A book resting on a table: the table pushes up with a normal force equal to the book’s weight.

Formulas To Know

Formula Variables Notes
Newton’s 1st Law If Fnet = 0, then a = 0 (object stays at rest or moves at constant speed). MEMORISE THIS. No net force = no acceleration.
Newton’s 2nd Law Fnet = ma MEMORISE THIS. Fnet is the vector sum of all forces.
Newton’s 3rd Law FA on B = -FB on A MEMORISE THIS. Forces come in pairs: equal size, opposite direction.
Weight W = mg g = 9.8 m/s² on Earth (given on exam sheet).
Friction (static/kinetic) f = μN μ = coefficient of friction (given), N = normal force.

Step-by-Step Method

Follow these steps exactly for every Newton’s Laws problem.

  1. Read the problem carefully. Underline what’s given and what’s asked.
  2. Draw a free-body diagram (FBD).
  3. Sketch the object as a dot or box.
  4. Draw all forces acting on the object as arrows.
  5. Label each force (e.g., Fgravity, Ftension, Ffriction).
  6. Choose a coordinate system.
  7. Align axes with the direction of acceleration (if any).
  8. Example: If a box slides down a ramp, tilt the x-axis along the ramp.
  9. Write Newton’s 2nd Law for each axis.
  10. Fnet,x = max
  11. Fnet,y = may (usually ay = 0 if no vertical acceleration).
  12. Plug in known values and solve for the unknown.
  13. Use algebra to isolate the variable you need.
  14. Check units and reasonableness.
  15. Forces should be in Newtons (N), mass in kg, acceleration in m/s².
  16. Does your answer make sense? (e.g., a 10 kg object shouldn’t accelerate at 1000 m/s²).

Worked Example Using the Steps

Problem: A 5 kg box is pulled across a frictionless table by a 20 N horizontal force. What is its acceleration?

  1. Given: m = 5 kg, Fpull = 20 N, friction = 0 (frictionless).
    Asked: a = ?
  2. FBD:
  3. Box with arrow right: Fpull = 20 N
  4. Arrow down: Fgravity = mg = 49 N
  5. Arrow up: Fnormal = 49 N (balances weight)
  6. Coordinate system: x-axis horizontal (right), y-axis vertical (up).
  7. Newton’s 2nd Law:
  8. x-axis: Fnet,x = max
    20 N = (5 kg)(ax)
  9. y-axis: Fnet,y = may = 0 (no vertical acceleration)
    Fnormal - Fgravity = 0 → 49 N - 49 N = 0
  10. Solve:
    20 N = 5 kg × ax
    ax = 20 N / 5 kg = 4 m/s²
  11. Check:
  12. Units: N/kg = m/s² ✔️
  13. Reasonable: 4 m/s² is a moderate acceleration for a 5 kg box.

Answer: a = 4 m/s² to the right.

Worked Examples

Example 1 – Basic: Elevator Problem

Problem: A 60 kg person stands on a scale in an elevator. The elevator accelerates upward at 2 m/s². What does the scale read?

  1. Given: m = 60 kg, a = 2 m/s² upward, g = 9.8 m/s².
    Asked: Scale reading (apparent weight) = ?
  2. FBD:
  3. Person with arrow up: Fscale (normal force)
  4. Arrow down: Fgravity = mg = 588 N
  5. Coordinate system: y-axis upward.
  6. Newton’s 2nd Law (y-axis):
    Fnet,y = may
    Fscale - Fgravity = ma
    Fscale - 588 N = (60 kg)(2 m/s²)
  7. Solve:
    Fscale = 588 N + 120 N = 708 N
  8. Check:
  9. Units: N ✔️
  10. Reasonable: Scale reads more than weight (708 N > 588 N) because elevator accelerates upward.

What we did and why: We used Fnet = ma to find the normal force (scale reading). The scale shows the normal force, not the true weight, because the elevator is accelerating.

Example 2 – Medium: Inclined Plane with Friction

Problem: A 10 kg box slides down a 30° ramp. The coefficient of kinetic friction is 0.2. What is the box’s acceleration?

  1. Given: m = 10 kg, θ = 30°, μk = 0.2, g = 9.8 m/s².
    Asked: a = ?
  2. FBD:
  3. Box on ramp.
  4. Arrow down ramp: Fparallel = mg sinθ
  5. Arrow up ramp: Ffriction = μkN
  6. Arrow perpendicular to ramp: Fnormal = mg cosθ
  7. Coordinate system: x-axis down the ramp, y-axis perpendicular to ramp.
  8. Newton’s 2nd Law:
  9. x-axis: Fnet,x = max
    mg sinθ - Ffriction = ma
    mg sinθ - μkN = ma
  10. y-axis: Fnet,y = 0 (no acceleration perpendicular to ramp)
    N - mg cosθ = 0 → N = mg cosθ
  11. Substitute N into x-axis equation:
    mg sinθ - μk(mg cosθ) = ma
    a = g(sinθ - μk cosθ)
    a = 9.8 m/s² (sin30° - 0.2 cos30°)
    a = 9.8 (0.5 - 0.2 × 0.866) = 9.8 (0.5 - 0.173) = 9.8 × 0.327 ≈ 3.2 m/s²
  12. Check:
  13. Units: m/s² ✔️
  14. Reasonable: Acceleration is less than g sinθ (4.9 m/s²) because friction opposes motion.

What we did and why: We broke forces into components along the ramp and perpendicular to it. Friction depends on the normal force, which is mg cosθ on an incline.

Example 3 – Exam Style: Connected Objects

Problem: Two blocks are connected by a light string over a frictionless pulley. Block A (3 kg) is on a frictionless table. Block B (2 kg) hangs vertically. What is the acceleration of the system?

  1. Given: mA = 3 kg, mB = 2 kg, g = 9.8 m/s², friction = 0.
    Asked: a = ?
  2. FBDs:
  3. Block A (table):
    • Arrow right: T (tension)
    • Arrow down: Fgravity,A = 29.4 N
    • Arrow up: Fnormal,A = 29.4 N
  4. Block B (hanging):
    • Arrow up: T (tension)
    • Arrow down: Fgravity,B = 19.6 N
  5. Coordinate system:
  6. Block A: x-axis right.
  7. Block B: y-axis down (to match acceleration direction).
  8. Newton’s 2nd Law:
  9. Block A (x-axis): T = mAa
  10. Block B (y-axis): Fgravity,B - T = mBa
  11. Substitute T from Block A into Block B:
    mBg - mAa = mBa
    19.6 N - 3a = 2a
    19.6 = 5a
    a = 3.92 m/s²
  12. Check:
  13. Units: m/s² ✔️
  14. Reasonable: Acceleration is less than g (9.8 m/s²) because Block A’s mass slows the system.

What we did and why: We treated the two blocks as a single system with the same acceleration. Tension is the same throughout the string (Newton’s 3rd Law).

Common Mistakes

Mistake Why it Happens Correct Approach
Forgetting to draw an FBD Students try to solve problems in their head and miss forces. Always draw an FBD first. Label every force.
Mixing up mass and weight Using kg for weight or N for mass. Weight = mg (in N). Mass is in kg.
Ignoring vector directions Adding forces as if they’re all in the same direction. Assign + and – signs based on your coordinate system.
Assuming normal force = weight Forgetting that normal force changes on inclines or with vertical acceleration. Normal force = mg cosθ on inclines. In elevators, N = m(g ± a).
Using the wrong friction formula Using f = μmg instead of f = μN. Friction depends on normal force, not weight. f = μN always.

Exam Traps

Trap How to Spot it How to Avoid it
Hidden forces Problem mentions "smooth surface" (no friction) or "light string" (no mass). Read carefully! "Smooth" = no friction. "Light" = ignore mass.
Non-inertial frames Problem involves accelerating reference frames (e.g., elevators, cars turning). Stick to Fnet = ma in the ground frame. Don’t add "fake forces."
Newton’s 3rd Law confusion Question asks for the force on one object but shows forces by it. Remember: Newton’s 3rd Law pairs act on different objects.

1-Minute Recap

"Alright, let’s lock this in. Newton’s Laws problems follow the same 6 steps every time: 1. Read the problem. 2. Draw an FBD—every force, only on the object. 3. Pick axes (usually along acceleration). 4. Write Fnet = ma for each axis. 5. Plug in numbers and solve. 6. Check units and reasonableness.

Memorise these formulas: - Fnet = ma (2nd Law) - W = mg (weight) - f = μN (friction)

Common traps? Forgetting the FBD, mixing up mass and weight, and ignoring vector directions. On exam day, slow down, draw the diagram, and trust the steps. You’ve got this!


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