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Physics: Motion to AP Physics 1 · Module 3 of 8

Forces & Newton's Laws

Newton's three laws, free-body diagrams, friction, inclines, and apparent weight.

In this module
  1. Newton's Three Laws
  2. Free-Body Diagrams & Friction
  3. Inclines & Apparent Weight
  4. Check yourself

Newton's Three Laws

  • First law (inertia): an object keeps its velocity (including staying at rest) unless a net external force acts on it.
  • Second law: F_net = ma. The net force and the acceleration point in the same direction.
  • Third law: forces come in equal and opposite pairs that act on DIFFERENT objects (you push the wall, the wall pushes you), so they never cancel each other.
Worked example

A 1,200 kg car accelerates at 2.5 m/s². What net force acts on it?

Solution. F_net = ma = (1,200)(2.5) = 3,000 N in the direction of the acceleration.

Common mistake: Moving at constant velocity does not require a net force. If the velocity is constant, the forces are balanced (F_net = 0).

Free-Body Diagrams & Friction

  • Draw every force on ONE object as an arrow from its center: weight (mg, down), normal force (perpendicular to the surface), tension, applied forces, and friction.
  • Kinetic friction f = μₖN opposes sliding. Static friction adjusts up to a maximum of μₛN to prevent sliding.
  • Choose axes along the acceleration, write ΣF = ma for each axis, and solve.
Free-body diagram
10 kgN = 98 Nmg = 98 NF = 50 Nf = 19.6 N
The 10 kg box from the example: weight (98 N) down, normal force (98 N) up, the 50 N pull to the right, and 19.6 N of kinetic friction to the left. Net force = 30.4 N right, so a ≈ 3.0 m/s².
Worked example

A 10 kg box is pulled across a level floor by a 50 N horizontal force. μₖ = 0.20. Find its acceleration.

Solution. N = mg = 98 N, so friction = 0.20 × 98 = 19.6 N. F_net = 50 − 19.6 = 30.4 N, so a = 30.4/10 ≈ 3.0 m/s².

Inclines & Apparent Weight

  • On an incline at angle θ, split the weight: mg sin θ acts down the slope and mg cos θ presses into it (so N = mg cos θ if nothing else pushes perpendicular).
  • Apparent weight is the normal force. In an elevator accelerating upward, N = m(g + a); accelerating downward, N = m(g − a).
Worked example

A 60 kg student stands on a scale in an elevator accelerating upward at 2.0 m/s². What does the scale read?

Solution. N = m(g + a) = 60(9.8 + 2.0) = 708 N, more than the student's true weight of 588 N.

Check yourself

2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.

1. A hockey puck slides at constant velocity across frictionless ice. The net force on it is:

2. A 5 kg object has a net force of 20 N on it. Its acceleration is:

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