Forces: Friction, Hooke's Law and Resultant Forces

This lesson explains Forces: Friction, Hooke’s Law and Resultant Forces for Cambridge IGCSE Physics 0625. It separates the Core requirements from the additional Supplement work for Extended candidates. Focus on the cause-and-effect explanation and the exact quantities being compared. Work through the example before testing the same skill without notes.

What does a resultant force do?

A force is a push or a pull, measured in newtons. The resultant force is the single force that replaces all the forces acting on an object. If the resultant is zero, the object stays at rest or keeps a constant velocity. If the resultant is not zero, the object accelerates in the direction of that force.

In words: resultant force equals mass multiplied by acceleration. In symbols: $F = ma$.

Quantity	Symbol	Unit
Force	$F$	newton (N)
Mass	$m$	kilogram (kg)
Acceleration	$a$	m/s²
Spring constant	$k$	N/m or N/cm
Extension	$x$	m or cm

For forces along one line, add forces in the same direction and subtract opposing ones. A 500 N driving force against 300 N of friction gives a 200 N resultant forward. Friction acts between surfaces and opposes motion; air resistance is friction with air. Friction also heats the surfaces, and that energy transfer wording earns marks in 6-mark answers. Extended candidates must additionally find resultants of two perpendicular forces by scale drawing or calculation.

What is Hooke’s law and the limit of proportionality?

Stretch a spring and the extension is directly proportional to the load, up to the limit of proportionality. In words: force equals spring constant multiplied by extension. In symbols: $F = kx$. Extension means stretched length minus original length, never the stretched length itself. On a load-extension graph, Hooke’s law holds on the straight section through the origin; past the limit of proportionality the graph curves and $F = kx$ no longer applies.

Worked example

A spring is 4.0 cm long unstretched. With a load of 6.0 N, its length becomes 7.0 cm. (a) Calculate the spring constant $k$. [3] (b) The same load-extension behaviour continues. Calculate the spring’s length with a 10 N load. [2]

Solution (a). Extension: $x = 7.0 - 4.0 = 3.0\ \text{cm}$. Equation: $F = kx$, rearranged to $k = \dfrac{F}{x}$. Substitute: $k = 6.0 \div 3.0$. Answer: $k = 2.0\ \text{N/cm}$.

Solution (b). $x = \dfrac{F}{k} = 10 \div 2.0 = 5.0\ \text{cm}$. Length $= 4.0 + 5.0 = 9.0\ \text{cm}$.

Original marking points

• B1 (a): extension $= 3.0\ \text{cm}$.
• M1 (a): $k = \dfrac{F}{x}$ or $6.0/3.0$ seen.
• A1 (a): $2.0\ \text{N/cm}$ with unit.
• M1 (b): $x = 5.0\ \text{cm}$.
• A1 (b): length $= 9.0\ \text{cm}$.

These marking points belong to this original example. They are not an official Cambridge mark scheme.

Common errors and how to correct them

• Using length instead of extension in $F = kx$. Dividing 6.0 by 7.0 can cost the available marks. Fix: subtract the natural length first, every time.
• Forgetting friction when finding a resultant. Fix: list every force with its direction before adding.
• Saying zero resultant means “stopped”. It can also mean constant velocity. Fix: write both possibilities.
• Applying $F = kx$ beyond the limit of proportionality. Fix: check whether the graph is still straight at that load.
• Unit drift in $F = ma$. Grams in, wrong newtons out. Fix: convert mass to kg before substituting.

How to apply this in an exam

For any $F = ma$ question with several forces, draw the object as a box with labelled arrows first, then write “resultant = (forward forces) − (backward forces)” before touching the equation. Mark schemes can award the method mark for a correct resultant even when the final arithmetic slips, so this line protects two of the three marks.

Where this skill matters

This CS subtopic appears across both tiers. Papers 1 and 2 use MCQs on resultants, friction direction and spring graphs. Paper 3 sets resultant-force and load-extension graph questions; Paper 4 (Extended) sets the structured $F = ma$ and $F = kx$ calculations (both equations are Supplement only) and adds vector diagrams for perpendicular forces and questions past the limit of proportionality. Paper 6 can feature the spring practical: plotting load against extension, drawing the best-fit line and reading the unloaded length from the intercept. Core candidates need proportionality and one-line resultants; Extended candidates must also handle scale drawings, so practise them with a sharp pencil and stated scale.