The big idea: A force is a push or a pull (unit: newton, N).
The net force (also called the resultant force) is what you get when you add up all the forces on an object — direction included.
Newton's laws tell you what that net force does to the motion.
1st law — no net force, no change
With zero net force, an object stays still or keeps moving at constant velocity. Motion doesn't need a force — only a change in motion does.
2nd law — net force makes it accelerate
A net force gives an acceleration in the same direction: F = ma. Bigger force → bigger acceleration; bigger mass → smaller acceleration.
3rd law — forces come in pairs
If A pushes B, then B pushes A back equally hard, the opposite way. The two forces act on different objects, so they never cancel on one body.
Spot which law you need: Constant velocity or at rest? → 1st law: net force = 0.
Speeding up, slowing down or turning? → 2nd law: net force = ma.
Two objects pushing on each other? → 3rd law: an equal, opposite pair.
Newton's second law links the net force on an object to its acceleration. The same law can be written using momentum (momentum = mass × velocity): the net force equals how fast the momentum changes.
- net (resultant) force (N)
- mass (kg)
- acceleration (m s⁻²)
- change in momentum (kg m s⁻¹)
- time interval (s)
It's the NET force: The F in F = ma is the net force — every force on the object added together (direction matters).
Always find the net force first, then divide by the mass to get the acceleration.
[Diagram: phys-free-body] - Available in full study mode
Worked example — acceleration from a net force
A 4.0 kg trolley is pulled forward by a 30 N force while a 6.0 N friction force acts backward. Find its acceleration.
Solution
- Find the net force first (forward minus backward):
- Start with the given law:
- Put in the numbers (F = 24, m = 4.0) and solve — keep the unit:
Final answer
a = 6.0 m s⁻², in the direction of the pull.
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How this is tested: Most forces questions come down to drawing a free-body diagram and then applying F = ma.
- Paper 1A: quick net-force calculations — a block driven by an angled force, the acceleration of a single body. - Paper 2: connected systems — two masses joined by a string (find the tension), an elevator cable, or stacked blocks held together by friction.
Classic trap: plug in one force instead of the net force, or forget that connected objects share the same acceleration.
Connected bodies share an acceleration: When two objects are joined (by a string, or stacked so they move together), they have the same acceleration.
To find a connecting force (a string tension, or the friction between stacked blocks), apply F = ma to one object on its own.
| Set-up | Whole system | One body alone |
|---|---|---|
| Two masses on a string | a = (driving force) ÷ (total mass) | tension = (that body's mass) × a |
| Elevator going up/down | net force = T − mg | a = (T − mg) ÷ m |
| Stacked blocks | a = F ÷ (total mass) | friction on top = (top mass) × a |
IB-style question — (a) acceleration of the pair
Two blocks are joined by a light string on a smooth floor: a 2.0 kg block in front and a 3.0 kg block behind it. A force of 20 N pulls the front block forward. Find the acceleration of the pair.
Solution
- The blocks move together, so treat them as one mass (2.0 + 3.0 = 5.0 kg). Start with the given law:
- Put in the net force (20 N) and the total mass:
- Solve — keep the unit:
Final answer
a = 4.0 m s⁻² — the same for both blocks.
IB-style question — (b) tension in the string
Same set-up. Find the tension in the string joining the two blocks.
Solution
- The string only pulls the back block, so apply F = ma to that block alone:
- Its mass is 3.0 kg and it has the shared a = 4.0 m s⁻²:
- So the tension is:
Final answer
tension = 12 N — the net force needed to accelerate the 3.0 kg back block.