What this covers
This standard sits within Force and Motion I. In a one-to-one lesson we make sure the idea is clear first, then move straight to applying it in the exact way SPM asks, with correct units and full working.
Formulas you may need
How it is examined
It can appear in Paper 1 (objective) and Paper 2 (structured), and where an experiment applies, in Paper 3. We do not predict which questions appear; we prepare the technique for all of them.
A common mistake
How to study it
Learn the definition precisely, practise one or two SPM-style questions with full working, and link it to the rest of Force and Motion I. If it keeps costing marks, a one-to-one lesson fixes exactly that.
Force and Motion I · Formulas · Exam Papers
What you need to know
Force is a push or pull that can change the shape, speed, or direction of motion of an object. Newton's second law of motion states that the net force acting on an object equals the rate of change of momentum, which for constant mass simplifies to F = ma, where F is the net force in newtons (N), m is mass in kilograms (kg), and a is acceleration in m s⁻².
The newton is itself a derived unit, defined as the force needed to give a 1 kg mass an acceleration of 1 m s⁻², so 1 N = 1 kg m s⁻².
Only a net, or resultant, force, the combined effect of all forces acting on an object, produces acceleration; if the forces on an object are balanced, the net force is zero and the object either remains at rest or continues at constant velocity, consistent with Newton's first law. For example, a resultant force of 10 N acting on a 2 kg object produces an acceleration of a = F / m = 10 N ÷ 2 kg = 5 m s⁻².
Force is a vector quantity, so direction must always be considered when combining several forces acting on the same object.
Worked example
Question: A resultant force of 24 N is applied to a trolley of mass 6 kg, initially at rest on a smooth horizontal surface. Calculate the acceleration produced and the velocity of the trolley after 3 s.
Given: F = 24 N, m = 6 kg. Using F = ma: a = F / m = 24 N ÷ 6 kg = 4 m s⁻².
To find the velocity after 3 s, use v = u + at with u = 0 m s⁻¹: v = 0 m s⁻¹ + (4 m s⁻² × 3 s) = 12 m s⁻¹. The trolley accelerates at 4 m s⁻² and reaches a velocity of 12 m s⁻¹ after 3 s.
This example shows how F = ma links directly to the equations of motion once acceleration has been found, allowing further quantities such as velocity or displacement to be calculated.
How it is examined
Paper 1 objective items commonly ask candidates to calculate force, mass or acceleration using F = ma, or to identify the SI unit of force and its definition in terms of base units. Paper 2 structured and essay questions frequently require candidates to calculate a resultant force from several forces acting on an object before applying F = ma, or to explain how Newton's second law relates to a described scenario, such as why a heavier loaded lorry accelerates more slowly than an empty one under the same driving force.
Paper 3 practical work may involve trolleys with varying mass or applied force to investigate how acceleration depends on each.
Common mistakes include using the total applied force instead of the net force when other forces such as friction are present, mixing up mass and weight when substituting into F = ma, and giving an answer for force without the unit newton. Since SPM Physics provides no formula sheet, F = ma must be memorised precisely, along with the definition of the newton as 1 kg m s⁻², so that unit derivation questions can be answered without hesitation.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)