Free Pearson Edexcel International A Level Physics lesson
Resultant force and acceleration
Find the net force. Predict how motion changes.
What you’ll learn
- Draw a force diagram for one chosen object and calculate its resultant force.
- Use resultant force = mass × acceleration with direction and consistent units.
- Explain balanced forces, changing speed and terminal velocity using one physical model.
Start with one object
A force describes an interaction, such as a rope pulling a trolley or a track pushing up on it. Force is a vector: both its size and direction matter. To analyse motion, first choose the object or system you are studying. A free-body diagram shows only forces exerted on that chosen object. Label each arrow by the force and the object causing it; make longer arrows represent larger forces when a scale is useful.
The resultant force is the vector sum of those forces. For a straight track, choose one positive direction and give forces in the opposite direction a negative sign. Add them algebraically. A 9 N pull to the right and 3 N resistance to the left give +6 N if right is positive. The individual forces still exist; the resultant is one equivalent force for predicting the acceleration.
Check perpendicular directions separately. If a trolley does not accelerate vertically, its upward normal contact force balances its weight. Neither belongs in the horizontal sum. For two perpendicular unbalanced forces, the resultant magnitude follows Pythagoras: 3 N east and 4 N north combine to 5 N, directed 53° north of east. Oblique vector resolution and extended-body equilibrium need further study beyond this introductory lesson.
Fᵣ = ΣF = +9 N − 3 N = +6 N
Symbols and units
- Σ means add the signed forces on the chosen object; Fᵣ is their resultant in newtons (N).
- A positive or negative sign records direction along a declared axis; it is not a different unit.
How the equation explains the idea
- Choose right as positive and include only forces acting on this object. A 9 N rightward force contributes +9 N and a 3 N leftward force contributes −3 N.
- Add the components: Fᵣ = +9 − 3 = +6 N. The positive result specifies rightward; the magnitudes must not simply be added.
Worked example
A crate on a level floor
- Choose the crate as the object and right as positive.
- A rope pulls with +18 N. Sliding friction acts with −7 N.
- Horizontal resultant = 18 − 7 = +11 N. Weight and the normal force cancel vertically.
The resultant is 11 N to the right.
Keep this idea. Choose one object, draw its external forces, then add signed forces along the direction of interest.
A resultant force changes velocity
Acceleration is the rate of change of velocity. A velocity change may involve a change in speed, direction, or both. For an object of constant mass viewed from a non-accelerating reference frame, Newton’s second law links its acceleration to the resultant external force: ΣF = ma. The acceleration points in the direction of that resultant. This is a relationship between net force and acceleration, not between a single applied force and speed.
In SI units, force is measured in newtons, mass in kilograms and acceleration in metres per second squared. One newton produces an acceleration of 1 m s⁻² in a mass of 1 kg. Convert grams to kilograms before substituting. Write the force sum before dividing by the mass so that friction, drag or tension are not silently omitted.
At fixed mass, doubling the resultant force doubles acceleration. At fixed resultant force, doubling mass halves acceleration. Greater mass means greater resistance to a change in velocity. Neither statement alone predicts final speed: speed also depends on initial velocity and how long the force acts.
ΣF = ma → a = ΣF / m
Symbols and units
- ΣF is resultant force in N; m is inertial mass in kg; a is acceleration in m s⁻².
- 1 N = 1 kg m s⁻². Acceleration has the direction of resultant force in this constant-mass model.
How the equation explains the idea
- Start from ΣF = ma and divide both sides by the nonzero mass: a = ΣF/m.
- At fixed mass, doubling resultant force doubles acceleration. At fixed resultant force, doubling mass halves acceleration; this does not say that velocity doubles.
Conditions. Computed from a = F/m for constant masses of 2 kg and 4 kg. F is the total signed force along one axis; these are nonrelativistic models.
Resultant force / N; Acceleration / m s⁻².
The gradient of acceleration against force is 1/m. At 4 N, acceleration is 2 m s⁻² for 2 kg and 1 m s⁻² for 4 kg. Both curves pass through zero because zero resultant force gives zero acceleration.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A moving delivery trolley
- Mass = 2.0 kg. Forward pull = 9.0 N. Backward resistance = 3.0 N.
- Take forward as positive: ΣF = 9.0 − 3.0 = +6.0 N.
- a = 6.0 / 2.0 = +3.0 m s⁻². Each second, velocity increases by 3.0 m s⁻¹ while these forces remain constant.
Acceleration is 3.0 m s⁻² forwards.
Keep this idea. Calculate acceleration from the resultant force and total mass of the chosen system.
Balanced forces do not mean stopped
If the resultant force is zero, acceleration is zero. Velocity therefore stays constant: the object remains at rest if it started at rest, or continues moving at the same speed in the same direction. This is Newton’s first law. A moving suitcase pulled at constant velocity may have a forward pull exactly balanced by resistance; a steady speed does not mean there are no forces.
A force opposite to velocity initially reduces speed. With right positive, a cart travelling at +4 m s⁻¹ and accelerating at −2 m s⁻² reaches rest after 2 s. If the same resultant continues, it then moves left and gains speed. “Negative acceleration” is a statement about the chosen direction, not a synonym for slowing down. Speed increases when velocity and acceleration have the same sign.
Balanced forces acting on one object are not a Newton’s third-law pair. For example, the upward support and downward weight of a resting book both act on the book. An interaction pair acts on two different objects: the desk pushes the book up and the book pushes the desk down. Do not cancel a force on the book with a force on the desk in the book’s force diagram.
ΣF = 0 → a = 0 → constant velocity
Symbols and units
- ΣF is the vector sum of external forces in N; a is acceleration in m s⁻².
- v is velocity in m s⁻¹; m is constant positive mass in kg.
How the equation explains the idea
- Substitute ΣF = 0 into a = ΣF/m to obtain a = 0.
- Since a measures change of velocity per unit time, zero acceleration preserves whatever velocity the object already has. Rest is just the special case v = 0.
Worked example
Braking with a sign convention
- A 500 kg cart moves right. A net braking force of 1,000 N acts left.
- Right is positive, so a = −1,000 / 500 = −2.0 m s⁻².
- While velocity is positive, this acceleration reduces its magnitude. After the cart stops, this constant-force model would reverse its motion.
A direction statement makes “−2.0 m s⁻²” meaningful.
Keep this idea. Zero net force preserves velocity; opposite-signed velocity and acceleration reduce speed.
Connect the force diagram to motion
The lab models a constant-mass cart moving on a straight, level track. The adjustable force is already the horizontal resultant: do not subtract friction again. Vertical forces balance. Each run begins from rest at position zero; the force and mass stay constant during that run. The track is idealised and position is displacement from the starting point.
Predict before running. If the force is doubled while mass stays fixed, acceleration doubles. Starting from rest over the same elapsed time, both velocity and displacement double too. If only the mass doubles, all three halve. A fair comparison uses the same initial conditions and elapsed time. Changing two variables at once makes the cause harder to identify.
Under constant acceleration, velocity is u + at and displacement is ut + ½at². With u = 0, velocity grows linearly with time, whereas displacement grows quadratically. The velocity–time graph has gradient a; its area gives displacement. The position drawing, arrows, graphs and numerical values should all tell this same story. Real carts may have variable resistance or a limited track, so the ideal model is a prediction under stated assumptions.
v = u + at s = ut + ½at² (constant a)
Symbols and units
- u and v are initial and final velocities in m s⁻¹; a is constant acceleration in m s⁻².
- t is elapsed time in s and s is signed displacement in m, not total distance if motion reverses.
How the equation explains the idea
- Constant acceleration means equal velocity changes in equal times: Δv = at, hence v = u + at.
- The straight velocity–time line has mean velocity (u + v)/2. Multiplying by t and substituting v = u + at gives s = ut + ½at².
- Changing the force while keeping mass fixed changes a = ΣF/m and therefore both the slope of v against t and the curvature of s against t.
Worked example
Predict two seconds of motion
- Set resultant force to 4.0 N and mass to 2.0 kg. Then a = 2.0 m s⁻².
- From rest at t = 2.0 s: v = 0 + 2.0 × 2.0 = 4.0 m s⁻¹.
- Displacement = ½ × 2.0 × (2.0)² = 4.0 m; this equals the triangular area under the velocity–time graph.
After 2.0 s: acceleration 2.0 m s⁻², velocity 4.0 m s⁻¹, displacement 4.0 m.
Keep this idea. A model is useful when its assumptions and its force, velocity and displacement predictions agree.
When the forces change: falling and drag
Weight is the gravitational force on a mass: W = mg. Close to Earth, use the value of g supplied by the question; otherwise 9.81 N kg⁻¹ is a useful approximation. Mass is measured in kilograms and is not itself a force. For vertical motion choose up or down as positive and keep that choice throughout the calculation.
Consider a falling object released from rest in air, with approximately constant weight and negligible upthrust. Initially drag is small, so weight produces a downward acceleration. As downward speed grows, upward drag grows. The downward resultant, and therefore acceleration, becomes smaller. At terminal velocity drag balances weight: the object is still falling, but its velocity is constant.
Terminal velocity is not a moment when gravity disappears. If the forces become unbalanced again, for example when an opening parachute increases drag above weight, acceleration points upward while the person may still move downward. This reduces downward speed. A new, smaller steady falling speed is approached as drag and weight balance again.
Down positive: ΣF = mg − D a = (mg − D) / m
Symbols and units
- m is mass in kg, g is gravitational acceleration in m s⁻², D is upward drag magnitude in N.
- a is downward acceleration in m s⁻²; v is downward speed in m s⁻¹.
How the equation explains the idea
- With down positive, weight contributes +mg and opposing drag contributes −D, so ma = mg − D.
- Divide by m: a = g − D/m. As drag rises with falling speed, acceleration decreases even though speed can still increase.
- At terminal speed D = mg, giving a = 0 and constant nonzero speed. The equality is a force balance, not disappearance of either force.
Conditions. Illustrative linear-drag model D = bv with m = 1 kg and g = 10 m s⁻², released from rest. Buoyancy is neglected. Linear drag is a stated model, not a universal law for falling objects.
Time / s; Downward speed / m s⁻¹.
The solution v = (mg/b)(1 − exp(−bt/m)) begins with slope g and flattens towards mg/b. Doubling b halves terminal speed. The curve never reaches its asymptote in finite time in this ideal model.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
An object approaching terminal speed
- Mass = 0.50 kg, g = 9.81 N kg⁻¹, upward drag = 3.905 N.
- Weight = 0.50 × 9.81 = 4.905 N. Net force = 4.905 − 3.905 = 1.000 N downward.
- Acceleration = 1.000 / 0.50 = 2.0 m s⁻² downward. It has not yet reached terminal velocity.
At terminal velocity the drag would be 4.905 N and acceleration zero.
Keep this idea. Explain changing motion by tracking the resultant force, not by assuming the applied forces stay constant.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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