Lesson 3 of 4
Resultant force and acceleration
I can apply resultant force = mass x acceleration to new situations.Syllabus
Syllabus K223 / K224, 4(c). Apply resultant force = mass x acceleration to new situations.
Make a guess. You are not marked.
A box is pushed along a rough floor and moves at a constant velocity of 2 m/s. What is the resultant force on the box?
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Zero, because the push and friction are balanced.
Constant velocity means zero acceleration, and F = ma gives zero resultant force. The push is exactly balanced by friction, so the forces cancel.
In short
- Resultant force = mass x acceleration, F = ma.
- Zero resultant force: the object stays at rest or keeps moving at a constant velocity.
- A larger resultant force gives a larger acceleration; a larger mass gives a smaller one.
The resultant force determines the acceleration. Calculate it from all the forces on the body before using F = ma.
Acceleration is the change in velocity per unit time. Use a free-body diagram to find the resultant and choose a positive direction.
Fresultant = maFor a body of constant mass: force in N, mass in kg and acceleration in m/s2. Acceleration is in the direction of the resultant force.
This is Newton's second law. A mark scheme may write "by Newton's second law, F = ma".
A resultant force of 1 N gives a 1 kg mass an acceleration of 1 m/s2. The equation concerns acceleration, not the velocity at that instant. An object may be moving quickly while its resultant force and acceleration are both zero.
Unbalanced forces on a moving trolley
Selected body: 2.0 kg trolley
The trolley is moving right along a horizontal track.
Horizontal resultant: 7 - 3 = 4 N right. The vertical forces balance.
Worked example
Subtract the opposing force first
The trolley shown is moving forwards. Find its acceleration. Take forwards as positive.
- Find the horizontal resultant: Fresultant = 7 - 3 = +4 N. The forces oppose each other, so subtract their magnitudes.
- Rearrange: a = Fresultant / m.
- Substitute: a = 4 / 2.0 = 2.0 m/s2 forwards.
The balanced vertical forces give no vertical acceleration. Using 7 / 2.0 would treat the pull as the entire resultant and ignore the track's resistance.
Try it step by step
Work it out in steps: Resultant force in N, then Acceleration a = F / m in m/s2.
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Problem 1.
Resultant force in N: F = 20 - 8.0 = 12 N forwards
Acceleration a = F / m in m/s2: a = 12 / 4.0 = 3.0 m/s2 (2 sf)
Problem 2.
Resultant force in N: F = 50 - 20 = 30 N forwards
Acceleration a = F / m in m/s2: a = 30 / 12 = 2.5 m/s2 (2 sf)
Problem 3.
Resultant force in N: F = 9.0 - 3.0 = 6.0 N forwards
Acceleration a = F / m in m/s2: a = 6.0 / 1.5 = 4.0 m/s2 (2 sf)
A forward-moving object can accelerate backwards
In a different interval, suppose the trolley is still moving forwards but its string pull is 3 N and track resistance is 7 N. Keeping forwards positive gives:
a = -4 / 2.0 = -2.0 m/s2The acceleration is backwards while the velocity is forwards, so the trolley slows during this interval.
The negative sign specifies direction. It does not mean the trolley is already moving backwards. If the trolley reaches a stop, reassess the forces before predicting what happens next; the same resistance direction cannot simply be assumed throughout a reversal.
How mass affects acceleration
With the same resultant force, a larger mass has a smaller acceleration. A 4 N resultant gives a 2 kg trolley an acceleration of 2 m/s2, but a 4 kg trolley an acceleration of 1 m/s2. Doubling mass halves acceleration when the resultant is unchanged.
Investigating the relationship
The following idealised data describe a trolley of fixed mass 0.50 kg. The forces listed are resultant forces, after accounting for resistance.
| Resultant force / N | Acceleration / (m/s2) |
|---|---|
| 0.20 | 0.40 |
| 0.40 | 0.80 |
| 0.60 | 1.20 |
Doubling the resultant from 0.20 to 0.40 N doubles acceleration from 0.40 to 0.80 m/s2. The ratio Fresultant / a is 0.50 kg in every row, consistent with the fixed mass. An acceleration-against-resultant-force graph would be a straight line through the origin for this model.
For a real investigation, keep the trolley's total mass unchanged and use a level track. Change the horizontal pull, estimate acceleration from motion readings, and account for resistance when finding the resultant. A motion sensor or timed velocity readings can provide the changes needed for a = (v - u) / t.
More detail (optional)
Repeat readings to judge variation. A slightly sloping track adds a component of weight along the track, while changing resistance can make a measured pull different from the assumed resultant. Address those causes instead of calling every disagreement "human error". Real data need not fall exactly on the model values.
Watch out for this
A student says: "A ball thrown upwards keeps rising because the force of the throw is still acting on it." What is wrong with this?
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The throwing force stops the moment the ball leaves the hand. The ball keeps rising because it already has an upward velocity, and its weight, now the resultant force, slows it down.
Check your understanding
A 4.0 kg trolley has a horizontal pull of 14 N to the right and a resistance of 6 N to the left. Its vertical forces balance. What is its acceleration?
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2.0 m/s2 to the right.
The horizontal resultant is 14 - 6 = 8 N to the right. Using resultant force = mass x acceleration gives a = 8 / 4.0 = 2.0 m/s2 to the right.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
A car of mass 1200 kg moves along a straight, level road. The driving force on it is 3000 N and the total resistive force is 600 N. Calculate the acceleration of the car. [3]
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