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In this level
Use forces on one constant-mass object to calculate and explain horizontal acceleration.
Learning objectives:
Combine forces with a stated direction convention.
Select and rearrange F = ma for the required quantity.
Find a required push when an opposing force acts.
Explain balanced forces and the direction of acceleration.
Two carts can have different pushes but the same acceleration. Compare their net forces and masses.
Prerequisite review
Equality and inverse operations — Rightwards can be positive and leftwards negative. Add signed forces. An equation stays equal when the same valid operation is performed on both sides.
net force: The combined effect of all forces acting on the same object.
acceleration: Change in velocity per second; its direction follows net force.
Watch the prerequisite 1
Solve 2x − 10 = 6.
Step 1
Before: 2x − 10 = 6
Action: Add ten to both sides.
After: 2x − 10 + 10 = 6 + 10; 2x = 16
Why: The opposite terms sum to zero.
Apply +10 to both sides
Step 2
Before: 2x = 16
Action: Divide both sides by two.
After: 2x/2 = 16/2; x = 8
Why: The matching factor two reduces to one.
Apply /2 to both sides
Step 3
Before: x = 8
Action: Substitute into the original equation.
After: 2 × 8 − 10 = 16 − 10 = 6
Why: The original equation is satisfied.
x = 8. Both operations preserve equality.
Complete the missing step 1
Complete: 3y − 6 = 9. Add the same quantity to both sides, then divide both sides.
Use equal operations to isolate the unknown.
Add six to both sides: 3y − 6 + 6 = 9 + 6.
Now 3y = 15. Divide both sides by three; simplify the unknown.
3y − 6 + 6 = 9 + 6; 3y = 15. Divide both sides by 3: y = 5. Check: 3 × 5 − 6 = 9.
Try a fresh question 1
Solve 4z + 3 = 19 and check.
Use the same operation on both sides.
Subtract three from both sides.
4z = 16. Divide both sides by four, then check in the original equation.
Subtract 3 from both sides: 4z + 3 − 3 = 19 − 3; 4z = 16. Divide both sides by 4: z = 4. Check: 4 × 4 + 3 = 19.
Signed quantities and opposing forces — A chosen direction gives positive and negative contributions. Same-direction forces add; opposing directions subtract in magnitude. Keep the direction of the larger contribution.
positive direction: The direction chosen to give positive signed components; the opposite direction is negative.
Watch the prerequisite 1
Combine 9 N rightwards and 5 N leftwards.
Step 1
Before: The forces oppose each other.
Action: Choose rightwards positive.
After: Rightward force +9 N; leftward force −5 N
Why: The signs describe directions, not different kinds of force.
Step 2
Before: F = (+9) + (−5)
Action: Add the signed contributions.
After: F = 9 − 5 = +4 N
Why: Five opposing units balance; four rightward units remain.
Combine opposite directions
Step 3
Before: Net effect 4 N rightwards.
Action: Compare with the original directions.
After: The larger original force points right.
Why: The result direction agrees. Both original forces still act.
4 N rightwards.
Complete the missing step 1
Complete the signed sum for 7 N right and 3 N left: F = (+__) + (−__).
Use direction signs before arithmetic.
The rightward force is positive and leftward force negative.
Write 7 − 3 and subtract; the sign gives direction.
Right positive: F = (+7) + (−3) = 7 − 3 = +4 N; 4 N rightwards.
Try a fresh question 1
Combine 2 N right and 8 N left using rightwards positive.
Use the signed direction convention.
Write the sum (+2) + (−8).
The leftward magnitude is greater; subtract and state the direction.
F = (+2) + (−8) = 2 − 8 = −6 N; the net effect is 6 N leftwards.
Division, mass, force and acceleration units — Mass is measured in kg, force in N and acceleration in m s⁻². One newton is kg m s⁻². Dividing net force by mass gives acceleration, not speed.
mass: A measure of inertia, measured in kilograms.
force: A push or pull, measured in newtons.
velocity: Speed together with direction, measured in metres per second.
Watch the prerequisite 1
Explain and calculate 12 N divided by 3 kg.
Step 1
Before: 12 divided among three equal groups.
Action: Divide twelve by three.
After: 12/3 = 4
Why: Three groups of four reconstruct twelve.
Division into equal groups
Step 2
Before: One newton is one kg m s⁻².
Action: Replace the newton unit with its definition.
After: 12 N/3 kg = (12 kg m s⁻²)/(3 kg)
Why: Numerical division and unit division are separate.
Step 3
Before: (12 kg m s⁻²)/(3 kg)
Action: Divide numbers and reduce matching kilogram factors.
After: (12/3)(kg/kg) m s⁻² = 4 m s⁻²
Why: Kilogram divided by kilogram is one. The acceleration unit remains.
4 m s⁻².
Complete the missing step 1
Complete (10 N)/(2 kg) = (10/2)(kg/kg) m s⁻² and name the quantity.
Separate numerical division from unit division.
Replace N with kg m s⁻².
Divide ten by two and reduce the matching kilogram factors.
10/2 = 5; kg/kg = 1. Therefore 10 N/2 kg = 5 m s⁻². This is acceleration, a change of velocity each second.
Try a fresh question 1
Calculate (6 N)/(4 kg). Explain its unit and whether the result must be a whole number.
Use numerical and unit division.
Write (6/4)(kg/kg) m s⁻².
Reduce the kilogram ratio to one and complete the numerical quotient.
6/4 = 1.5 and kg/kg = 1; result 1.5 m s⁻². Acceleration need not be a whole number. This is not speed, which has unit m s⁻¹.
Readiness check
Watch — I do
Newton’s second law: net force and acceleration
Find the force that belongs in the equation
Study one trolley. Choose rightwards positive. Include every horizontal force. The upward support force balances weight in this level-surface model. The diagram is schematic.
F = sum of signed horizontal forces
Horizontal forces on one trolley
Use Newton’s second law
For constant mass in this horizontal model, F denotes the signed horizontal net force. Acceleration has the same direction as F. Mass is in kilograms; force in newtons; acceleration in metres per second squared.
F = ma
a = F/m
Divide both sides by m; matching factors reduce to one
Select the equation for the unknown
Find net force first. If a push P acts rightwards against resistance R, P − R = ma. Add R to both sides to find P. This arrangement determines the signs.
P − R = ma
P − R + R = ma + R
P = ma + R
Apply +3 to both sides
Acceleration is a change in velocity
A negative acceleration points left under our convention. Whether speed rises or falls also depends on velocity direction. Zero net force gives zero acceleration and unchanged velocity, including a possible nonzero velocity.
F = 0 ⇒ a = 0; velocity remains unchanged
Compare velocity and acceleration
Worked examples
I do
A 2.0 kg trolley has 12.0 N rightwards and 4.0 N leftwards. Find acceleration.
Step 1
Before: One trolley on a level surface.
Action: Draw and label the forces.
After: Mass 2.0 kg; horizontal forces 12.0 N and 4.0 N.
Why: Support balances weight; only horizontal motion is considered.
Horizontal forces on one trolley
Step 2
Before: 12.0 N right; 4.0 N left.
Action: Choose rightwards positive.
After: F = +12.0 + (−4.0) = 8.0 N
Why: Signs record direction. Net force is a combined effect, not an extra force.
Step 3
Before: F = ma
Action: Divide both sides by mass.
After: F/m = ma/m; a = F/m
Why: Mass divided by itself is one; acceleration remains.
Divide both sides by m; matching factors reduce to one
Step 4
Before: a = F/m
Action: Substitute net force and mass.
After: a = 8.0/2.0 = 4.0 m s⁻²
Why: The acceleration is 4.0 m s⁻² rightwards.
Step 5
Before: a = 4.0 m s⁻²
Action: Check force and units.
After: ma = 2.0 × (4.0) = 8.0 N
Why: The force agrees; N/kg equals metres per second squared.
4.0 m s⁻² rightwards.
I do
A 2.0 kg trolley has 4.0 N rightwards and 10.0 N leftwards. Find acceleration.
Step 1
Before: One trolley on a level surface.
Action: Draw and label the forces.
After: Mass 2.0 kg; horizontal forces 4.0 N and 10.0 N.
Why: Support balances weight; only horizontal motion is considered.
Horizontal forces on one trolley
Step 2
Before: 4.0 N right; 10.0 N left.
Action: Choose rightwards positive.
After: F = +4.0 + (−10.0) = −6.0 N
Why: Signs record direction. Net force is a combined effect, not an extra force.
Step 3
Before: F = ma
Action: Divide both sides by mass.
After: F/m = ma/m; a = F/m
Why: Mass divided by itself is one; acceleration remains.
Divide both sides by m; matching factors reduce to one
Step 4
Before: a = F/m
Action: Substitute net force and mass.
After: a = −6.0/2.0 = −3.0 m s⁻²
Why: The acceleration is 3.0 m s⁻² leftwards.
Step 5
Before: a = −3.0 m s⁻²
Action: Check force and units.
After: ma = 2.0 × (−3.0) = −6.0 N
Why: The force agrees; N/kg equals metres per second squared.
3.0 m s⁻² leftwards.
I do
A 4.0 kg trolley needs 2.0 m s⁻² rightwards acceleration against 3.0 N leftwards resistance. Find the rightward push.
Step 1
Before: Unknown push P; known mass and acceleration.
Action: Find the required net force first.
After: F = ma = 4.0 × 2.0 = 8.0 N
Why: The required net force differs from the push because resistance also acts.
Step 2
Before: P rightwards; 3.0 N leftwards.
Action: Write the signed force equation.
After: P − 3.0 = 8.0
Why: Rightwards is positive.
Step 3
Before: P − 3.0 = 8.0
Action: Add 3.0 to both sides.
After: P − 3.0 + 3.0 = 8.0 + 3.0; P = 11.0 N
Why: The opposite resistance terms sum to zero.
Apply +3 to both sides
Step 4
Before: P = 11.0 N
Action: Subtract resistance and divide by mass.
After: (11.0 − 3.0)/4.0 = 2.0 m s⁻²
Why: The requested acceleration is recovered.
Push 11.0 N rightwards.
I do
A 3.0 kg trolley needs 2.0 m s⁻² rightwards acceleration against 1.0 N leftwards resistance. Find the rightward push.
Step 1
Before: Unknown push P; known mass and acceleration.
Action: Find the required net force first.
After: F = ma = 3.0 × 2.0 = 6.0 N
Why: The required net force differs from the push because resistance also acts.
Step 2
Before: P rightwards; 1.0 N leftwards.
Action: Write the signed force equation.
After: P − 1.0 = 6.0
Why: Rightwards is positive.
Step 3
Before: P − 1.0 = 6.0
Action: Add 1.0 to both sides.
After: P − 1.0 + 1.0 = 6.0 + 1.0; P = 7.0 N
Why: The opposite resistance terms sum to zero.
Apply +1 to both sides
Step 4
Before: P = 7.0 N
Action: Subtract resistance and divide by mass.
After: (7.0 − 1.0)/3.0 = 2.0 m s⁻²
Why: The requested acceleration is recovered.
Push 7.0 N rightwards.
I do
A net force of 9.0 N rightwards gives acceleration 3.0 m s⁻² rightwards. Find mass.
Step 1
Before: F = ma
Action: Divide both sides by nonzero acceleration.
After: F/a = ma/a; m = F/a
Why: The factor a/a reduces to one.
Divide both sides by a; matching factors reduce to one
Step 2
Before: m = F/a
Action: Substitute.
After: m = 9.0/3.0 = 3.0 kg
Why: Matching force and acceleration directions give positive mass.
Step 3
Before: m = 3.0 kg
Action: Multiply mass by acceleration.
After: 3.0 × 3.0 = 9.0 N
Why: This matches net force. If acceleration and net force are both zero, they cannot determine mass.
3.0 kg.
I do
A net force of 12 N leftwards produces acceleration 4.0 m s⁻² leftwards. Find mass.
Step 1
Before: Force and acceleration point left.
Action: Choose rightwards positive.
After: F = −12 N; a = −4.0 m s⁻²
Why: Both quantities use the same sign convention.
Step 2
Before: F = ma
Action: Divide both sides by nonzero a.
After: F/a = ma/a; m = F/a
Why: The matching acceleration factor reduces to one.
Divide both sides by a; matching factors reduce to one
Step 3
Before: m = F/a
Action: Substitute both signed values.
After: m = (−12)/(−4.0) = 3.0 kg
Why: A negative divided by a negative is positive; direction does not create negative mass.
Step 4
Before: m = 3.0 kg
Action: Multiply by signed acceleration.
After: ma = 3.0 × (−4.0) = −12 N
Why: The result reproduces the leftward net force.
3.0 kg.
I do
A trolley already moves rightwards at 3.0 m s⁻¹. Opposing forces are each 6.0 N. Describe motion.
Step 1
Before: +6.0 N and −6.0 N
Action: Add signed forces.
After: F = 6.0 − 6.0 = 0 N
Why: Forces balance.
Step 2
Before: a = F/m
Action: Use zero net force.
After: a = 0/m = 0 m s⁻²
Why: Any positive mass gives zero acceleration.
Step 3
Before: Initial velocity 3.0 m s⁻¹ rightwards.
Action: Keep velocity unchanged.
After: Velocity remains 3.0 m s⁻¹ rightwards.
Why: Zero acceleration does not imply zero velocity.
Constant velocity 3.0 m s⁻¹ rightwards while forces remain balanced.
I do
A stationary trolley has opposing horizontal forces each 8.0 N. Describe its motion while they stay balanced.
Step 1
Before: Initial velocity is zero; forces +8.0 N and −8.0 N.
Action: Add the signed forces.
After: F = 8.0 − 8.0 = 0 N
Why: The two forces balance.
Step 2
Before: a = F/m for positive mass.
Action: Use the net force zero.
After: a = 0/m = 0 m s⁻²
Why: There is no change in velocity.
Step 3
Before: Initial velocity zero; acceleration zero.
Action: Keep the initial velocity.
After: The trolley remains stationary.
Why: Contrast the moving balanced-force example: the same zero acceleration preserves whichever velocity the object already had.
It remains stationary while the forces stay balanced.
Together — We do
Worked examples
We do
Together: a 3 kg trolley has a 13 N push right and 4 N resistance left. Complete F = __ − __, then 3a = __, and explain each operation.
Step 1
Before: Forces act in opposite horizontal directions.
Action: Choose rightwards as positive.
After: Push = +13 N; resistance = −4 N
Why: The signs identify direction on one axis.
Step 2
Before: F = __ − __
Action: Include both forces and subtract their magnitudes.
After: F = 13 − 4 = 9 N rightwards
Why: Newton’s second law uses the resultant.
Step 3
Before: F = ma; F = 9 N; m = 3 kg
Action: Substitute the resultant and mass.
After: 9 = 3a
Why: Acceleration is the unknown; the same relation can be written 3a = 9.
Step 4
Before: 3a = 9
Action: Divide both sides by three.
After: (3a)/3 = 9/3; a = 3 m s⁻² rightwards
Why: Matching factors reduce to one; N/kg = m s⁻².
Apply ÷ 3 to both sides
Step 5
Before: a = 3 m s⁻² rightwards
Action: Multiply back and compare the direction.
After: ma = 3 × 3 = 9 N = 13 − 4
Why: The stronger rightward force and acceleration point right.
Net force is 9 N rightwards; acceleration is 3 m s⁻² rightwards.
Complete the method: a 3.0 kg trolley has 15 N rightwards and 3.0 N leftwards. Fill F = 15 − __, then a = F/__.
Horizontal forces on one trolleyHint 1
Find net force before acceleration.
Hint 2
Write F = 15 − 3.0.
Hint 3
The net force is 12 N. Divide it by the stated mass.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 15 − 3.0 = 12 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/3 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 3 × (4) = 12 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 15 − 3.0 = 12 N rightwards. a = 12/3.0 = 4.0 m s⁻² rightwards. Check: 3.0 × 4.0 = 12 N.
F = 15 − 3.0 = 12 N rightwards. a = 12/3.0 = 4.0 m s⁻² rightwards.
Check: 3.0 × 4.0 = 12 N.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Two forces 4.0 N and 2.0 N both act rightwards on a 2.0 kg object. Choose addition or subtraction, then find acceleration.
Same-direction forcesHint 1
Use directions to combine forces.
Hint 2
Both signed forces are positive.
Hint 3
Add the forces, then divide their sum by 2.0 kg.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 4.0 + 2.0 = 6.0 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (6)/2 = 3 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 3 m s⁻²
Action: Check force and direction.
After: ma = 2 × (3) = 6 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Same direction: F = 4.0 + 2.0 = 6.0 N rightwards. a = 6.0/2.0 = 3.0 m s⁻² rightwards.
Same direction: F = 4.0 + 2.0 = 6.0 N rightwards. a = 6.0/2.0 = 3.0 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Complete P − 1.0 = ma for a 3.0 kg trolley needing 2.0 m s⁻² rightward acceleration.
Apply +1 to both sidesHint 1
Find required net force first.
Hint 2
Calculate ma = 3.0 × 2.0.
Hint 3
P − 1.0 = 6.0. Add the resistance to both sides.
Show the complete working
Step 1
Before: Unknown push P; known mass and acceleration.
Action: Find the required net force first.
After: F = ma = 3.0 × 2.0 = 6.0 N
Why: The required net force differs from the push because resistance also acts.
Step 2
Before: P rightwards; 1.0 N leftwards.
Action: Write the signed force equation.
After: P − 1.0 = 6.0
Why: Rightwards is positive.
Step 3
Before: P − 1.0 = 6.0
Action: Add 1.0 to both sides.
After: P − 1.0 + 1.0 = 6.0 + 1.0; P = 7.0 N
Why: The opposite resistance terms sum to zero.
Apply +1 to both sides
Step 4
Before: P = 7.0 N
Action: Subtract resistance and divide by mass.
After: (7.0 − 1.0)/3.0 = 2.0 m s⁻²
Why: The requested acceleration is recovered.
ma = 3.0 × 2.0 = 6.0 N. P − 1.0 = 6.0. Add 1.0 to both sides: P = 7.0 N rightwards. Check: (7.0 − 1.0)/3.0 = 2.0 m s⁻².
ma = 3.0 × 2.0 = 6.0 N.
P − 1.0 = 6.0.
Add 1.0 to both sides: P = 7.0 N rightwards.
Check: (7.0 − 1.0)/3.0 = 2.0 m s⁻².
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Complete the mass method for F = 18 N rightwards and a = 6.0 m s⁻² rightwards: 18 = m × 6.0. Show division beneath both sides.
Hint 1
Identify mass as the unknown.
Hint 2
Divide both sides of 18 = m × 6.0 by the nonzero acceleration.
Hint 3
The matching acceleration factors reduce to one; finish 18/6.0 and use kilograms.
Show the complete working
Step 1
Before: F = 18 N; a = 6 m s⁻²
Action: Identify mass as the unknown.
After: F = ma
Why: Use matching directions for force and acceleration.
Step 2
Before: F = ma
Action: Divide both sides by nonzero a.
After: F/a = ma/a; m = F/a
Why: The matching a factors reduce to one.
Divide both sides by a; matching factors reduce to one
Step 3
Before: m = F/a
Action: Substitute both signed values.
After: m = (18)/(6) = 3 kg
Why: Matching signs give positive mass. Force divided by acceleration leaves kilograms.
Step 4
Before: m = 3 kg
Action: Multiply by the given acceleration.
After: ma = 3 × (6) = 18 N
Why: The force matches the original data.
Divide both sides by 6.0: 18/6.0 = (m × 6.0)/6.0. Matching 6.0 factors reduce to one, so m = 18/6.0 = 3.0 kg. Check: 3.0 × 6.0 = 18 N.
Divide both sides by 6.0: 18/6.0 = (m × 6.0)/6.0.
Matching 6.0 factors reduce to one, so m = 18/6.0 = 3.0 kg.
Check: 3.0 × 6.0 = 18 N.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A trolley moves right at 1.5 m s⁻¹. Forces of 4.0 N act right and left. Complete the net force, acceleration and subsequent velocity.
Hint 1
Distinguish velocity from a change in velocity.
Hint 2
Calculate F = 4.0 − 4.0.
Hint 3
Net force and acceleration are zero; preserve the velocity it already had.
Show the complete working
Step 1
Before: Opposing forces are both 4 N.
Action: Add their signed components.
After: F = 4 − 4 = 0 N
Why: Equal opposing forces balance.
Step 2
Before: F = 0 N; mass is positive.
Action: Use Newton’s second law.
After: a = F/m = 0/m = 0 m s⁻²
Why: Zero acceleration means no change in velocity.
Step 3
Before: The velocity before the balanced forces is 1.5 m s⁻¹ rightwards.
Action: Preserve that initial velocity.
After: Velocity remains 1.5 m s⁻¹ rightwards.
Why: Acceleration changes velocity; it does not specify the initial velocity.
F = 4.0 − 4.0 = 0 N. For positive mass, a = 0/m = 0 m s⁻². Its velocity remains 1.5 m s⁻¹ rightwards while those forces remain balanced.
F = 4.0 − 4.0 = 0 N.
For positive mass, a = 0/m = 0 m s⁻².
Its velocity remains 1.5 m s⁻¹ rightwards while those forces remain balanced.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
With help
Worked examples
You do
Solve a trolley problem with less help: mass 2 kg, 11 N right and 5 N left.
Step 1
Before: Choose a direction and identify all forces.
Action: Write net force, then isolate acceleration.
After: F = 11 − 5 = 6 N; a = 6/2 = 3 m s⁻² rightwards
Why: Check 2 × 3 = 6 N.
3 m s⁻² rightwards.
A 2.0 kg object has 10 N rightwards and 16 N leftwards. Find acceleration.
Horizontal forces on one trolleyHint 1
Combine signed forces on the same object.
Hint 2
Choose rightwards positive: F = 10 − 16.
Hint 3
F = −6.0 N; divide by 2.0 kg and interpret the sign.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 10 − 16 = −6.0 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (−6)/2 = −3 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = −3 m s⁻²
Action: Check force and direction.
After: ma = 2 × (−3) = −6 N; leftwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 10 − 16 = −6.0 N. a = −6.0/2.0 = −3.0 m s⁻², or 3.0 m s⁻² leftwards.
F = 10 − 16 = −6.0 N. a = −6.0/2.0 = −3.0 m s⁻², or 3.0 m s⁻² leftwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A net force of 16 N rightwards produces 4.0 m s⁻² rightwards acceleration. Find mass.
Hint 1
Identify the unknown as mass.
Hint 2
Divide F = ma by nonzero acceleration on both sides.
Hint 3
m = 16/4.0. Calculate and use kilograms.
Show the complete working
Step 1
Before: F = 16 N; a = 4 m s⁻²
Action: Identify mass as the unknown.
After: F = ma
Why: Use matching directions for force and acceleration.
Step 2
Before: F = ma
Action: Divide both sides by nonzero a.
After: F/a = ma/a; m = F/a
Why: The matching a factors reduce to one.
Divide both sides by a; matching factors reduce to one
Step 3
Before: m = F/a
Action: Substitute both signed values.
After: m = (16)/(4) = 4 kg
Why: Matching signs give positive mass. Force divided by acceleration leaves kilograms.
Step 4
Before: m = 4 kg
Action: Multiply by the given acceleration.
After: ma = 4 × (4) = 16 N
Why: The force matches the original data.
F = ma. Divide both sides by a: m = F/a = 16/4.0 = 4.0 kg. Check: 4.0 × 4.0 = 16 N.
F = ma.
Divide both sides by a: m = F/a = 16/4.0 = 4.0 kg.
Check: 4.0 × 4.0 = 16 N.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 5.0 kg trolley needs 2.0 m s⁻² rightwards against 4.0 N resistance. Find push.
Hint 1
Net force and pushing force differ.
Hint 2
Find F = ma = 5.0 × 2.0.
Hint 3
P − 4.0 = 10. Add 4.0 to both sides.
Show the complete working
Step 1
Before: Unknown push P; known mass and acceleration.
Action: Find the required net force first.
After: F = ma = 5.0 × 2.0 = 10.0 N
Why: The required net force differs from the push because resistance also acts.
Step 2
Before: P rightwards; 4.0 N leftwards.
Action: Write the signed force equation.
After: P − 4.0 = 10.0
Why: Rightwards is positive.
Step 3
Before: P − 4.0 = 10.0
Action: Add 4.0 to both sides.
After: P − 4.0 + 4.0 = 10.0 + 4.0; P = 14.0 N
Why: The opposite resistance terms sum to zero.
Apply +4 to both sides
Step 4
Before: P = 14.0 N
Action: Subtract resistance and divide by mass.
After: (14.0 − 4.0)/5.0 = 2.0 m s⁻²
Why: The requested acceleration is recovered.
F = 5.0 × 2.0 = 10 N. P − 4.0 = 10. Add 4.0 to both sides: P = 14 N rightwards. Check: (14 − 4.0)/5.0 = 2.0 m s⁻².
F = 5.0 × 2.0 = 10 N.
P − 4.0 = 10.
Add 4.0 to both sides: P = 14 N rightwards.
Check: (14 − 4.0)/5.0 = 2.0 m s⁻².
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Solve 4z + 3 = 19 and check.
Hint 1
Use the same operation on both sides.
Hint 2
Subtract three from both sides.
Hint 3
4z = 16. Divide both sides by four, then check in the original equation.
Show the complete working
Step 1
Before: 4z + 3 = 19
Action: Subtract three from both sides.
After: 4z + 3 − 3 = 19 − 3; 4z = 16
Why: The opposite terms sum to zero.
Apply −3 to both sides
Step 2
Before: 4z = 16
Action: Divide both sides by four.
After: 4z/4 = 16/4; z = 4
Why: The matching factor four reduces to one.
Apply /4 to both sides
Step 3
Before: z = 4
Action: Substitute into the original equation.
After: 4 × 4 + 3 = 16 + 3 = 19
Why: The result satisfies the original equality.
Subtract 3 from both sides: 4z + 3 − 3 = 19 − 3; 4z = 16. Divide both sides by 4: z = 4. Check: 4 × 4 + 3 = 19.
Subtract 3 from both sides: 4z + 3 − 3 = 19 − 3; 4z = 16.
Divide both sides by 4: z = 4.
Check: 4 × 4 + 3 = 19.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
On my own
A 2.0 kg object has 14 N rightwards and 6.0 N leftwards. Find net force and acceleration.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 14 − 6.0 = 8.0 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (8)/2 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 2 × (4) = 8 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 14 − 6.0 = 8.0 N rightwards. a = 8.0/2.0 = 4.0 m s⁻² rightwards.
F = 14 − 6.0 = 8.0 N rightwards. a = 8.0/2.0 = 4.0 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 3.0 kg object has 9.0 N rightwards and 15 N leftwards. Find acceleration.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 9.0 − 15 = −6.0 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (−6)/3 = −2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = −2 m s⁻²
Action: Check force and direction.
After: ma = 3 × (−2) = −6 N; leftwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 9.0 − 15 = −6.0 N. a = −6.0/3.0 = −2.0 m s⁻²; 2.0 m s⁻² leftwards.
F = 9.0 − 15 = −6.0 N. a = −6.0/3.0 = −2.0 m s⁻²; 2.0 m s⁻² leftwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Forces 6.0 N and 4.0 N both act rightwards on a 5.0 kg object. Find acceleration.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 6.0 + 4.0 = 10 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (10)/5 = 2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 2 m s⁻²
Action: Check force and direction.
After: ma = 5 × (2) = 10 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 6.0 + 4.0 = 10 N rightwards. a = 10/5.0 = 2.0 m s⁻² rightwards.
F = 6.0 + 4.0 = 10 N rightwards. a = 10/5.0 = 2.0 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 4.0 kg object accelerates at 1.5 m s⁻² rightwards. Find net force.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: m = 4 kg; a = 1.5 m s⁻²
Action: Find net force directly.
After: F = ma
Why: Mass and acceleration are both supplied.
Step 2
Before: F = ma
Action: Substitute and multiply.
After: F = 4 × (1.5) = 6 N
Why: kg m s⁻² is the newton unit; force follows the acceleration direction.
Step 3
Before: F = 6 N
Action: Divide by the original mass.
After: F/m = (6)/4 = 1.5 m s⁻²
Why: The original acceleration is recovered.
F = ma = 4.0 × 1.5 = 6.0 N rightwards.
F = ma = 4.0 × 1.5 = 6.0 N rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Net force 12 N rightwards produces acceleration 3.0 m s⁻² rightwards. Find mass.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: F = 12 N; a = 3 m s⁻²
Action: Identify mass as the unknown.
After: F = ma
Why: Use matching directions for force and acceleration.
Step 2
Before: F = ma
Action: Divide both sides by nonzero a.
After: F/a = ma/a; m = F/a
Why: The matching a factors reduce to one.
Divide both sides by a; matching factors reduce to one
Step 3
Before: m = F/a
Action: Substitute both signed values.
After: m = (12)/(3) = 4 kg
Why: Matching signs give positive mass. Force divided by acceleration leaves kilograms.
Step 4
Before: m = 4 kg
Action: Multiply by the given acceleration.
After: ma = 4 × (3) = 12 N
Why: The force matches the original data.
F = ma. Divide both sides by a: m = F/a = 12/3.0 = 4.0 kg.
F = ma.
Divide both sides by a: m = F/a = 12/3.0 = 4.0 kg.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 3.0 kg trolley needs 2.0 m s⁻² rightwards acceleration against 4.0 N resistance. Find push.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: Unknown push P; known mass and acceleration.
Action: Find the required net force first.
After: F = ma = 3.0 × 2.0 = 6.0 N
Why: The required net force differs from the push because resistance also acts.
Step 2
Before: P rightwards; 4.0 N leftwards.
Action: Write the signed force equation.
After: P − 4.0 = 6.0
Why: Rightwards is positive.
Step 3
Before: P − 4.0 = 6.0
Action: Add 4.0 to both sides.
After: P − 4.0 + 4.0 = 6.0 + 4.0; P = 10.0 N
Why: The opposite resistance terms sum to zero.
Apply +4 to both sides
Step 4
Before: P = 10.0 N
Action: Subtract resistance and divide by mass.
After: (10.0 − 4.0)/3.0 = 2.0 m s⁻²
Why: The requested acceleration is recovered.
Required F = 3.0 × 2.0 = 6.0 N. P − 4.0 = 6.0. Add 4.0 to both sides: P = 10 N rightwards. Check: (10 − 4.0)/3.0 = 2.0 m s⁻².
Required F = 3.0 × 2.0 = 6.0 N.
P − 4.0 = 6.0.
Add 4.0 to both sides: P = 10 N rightwards.
Check: (10 − 4.0)/3.0 = 2.0 m s⁻².
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A trolley moves right at 2.0 m s⁻¹. Opposing forces are each 6.0 N. Describe subsequent motion.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: Opposing forces are both 6 N.
Action: Add their signed components.
After: F = 6 − 6 = 0 N
Why: Equal opposing forces balance.
Step 2
Before: F = 0 N; mass is positive.
Action: Use Newton’s second law.
After: a = F/m = 0/m = 0 m s⁻²
Why: Zero acceleration means no change in velocity.
Step 3
Before: The velocity before the balanced forces is 2 m s⁻¹ rightwards.
Action: Preserve that initial velocity.
After: Velocity remains 2 m s⁻¹ rightwards.
Why: Acceleration changes velocity; it does not specify the initial velocity.
F = 6.0 − 6.0 = 0 N, so a = 0. It continues at 2.0 m s⁻¹ rightwards while forces remain unchanged.
F = 6.0 − 6.0 = 0 N, so a = 0.
It continues at 2.0 m s⁻¹ rightwards while forces remain unchanged.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
An 8.0 N net force acts on a 2.0 kg cart, then on a 4.0 kg cart. Compare accelerations.
Hint 1
Identify the unknown and the forces on one object.
Hint 2
Choose rightwards positive and write the signed net force where needed.
Hint 3
Choose F = ma, a = F/m or m = F/a; substitute the given values and interpret direction.
Show the complete working
Step 1
Before: Given net force 8 N; mass 2 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (8)/2 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 2 × (4) = 8 N; in the direction of the net force
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Step 5
Before: Given net force 8 N; mass 4 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 6
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 7
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (8)/4 = 2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 8
Before: a = 2 m s⁻²
Action: Check force and direction.
After: ma = 4 × (2) = 8 N; in the direction of the net force
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Step 9
Before: Accelerations 4 and 2 m s⁻².
Action: Compare the mass change with the acceleration change.
After: Doubling mass halves acceleration at unchanged net force.
Why: The force was held constant.
a = 8.0/2.0 = 4.0 m s⁻²; a = 8.0/4.0 = 2.0 m s⁻². Both follow the force direction. Doubling mass halves acceleration at unchanged net force.
a = 8.0/2.0 = 4.0 m s⁻²; a = 8.0/4.0 = 2.0 m s⁻².
Both follow the force direction.
Doubling mass halves acceleration at unchanged net force.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Which conclusion follows from zero net force?
Velocity remains unchanged.
The object must be stationary.
No forces can act on the object.
Hint 1
Identify the relationship being tested.
Hint 2
Use the definitions before comparing the options.
Hint 3
Reject options that contradict the relationship; select the remaining option.
Show the complete working
Step 1
Before: The supplied net force is zero.
Action: Use the given resultant directly.
After: F = 0 N
Why: Individual forces may be absent or may cancel; their magnitudes were not supplied.
Step 2
Before: F = 0 N; mass is positive.
Action: Use Newton’s second law.
After: a = F/m = 0/m = 0 m s⁻²
Why: Zero acceleration means no change in velocity.
Step 3
Before: The velocity before the balanced forces is not specified.
Action: Preserve that initial velocity.
After: Velocity remains unchanged; it need not be zero.
Why: Acceleration changes velocity; it does not specify the initial velocity.
F = ma with positive mass gives a = 0; velocity stays unchanged.
F = ma with positive mass gives a = 0; velocity stays unchanged.
Come back
Return on another day. Try these fresh questions before revealing a hint or answer. Explain what changed in your approach.
A 2 kg object has net force 8 N rightwards. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Given net force 8 N; mass 2 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (8)/2 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 2 × (4) = 8 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
a = F/m = 8/2 = 4 m s⁻² rightwards.
a = F/m = 8/2 = 4 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 3 kg object has net force 9 N leftwards. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Given net force −9 N; mass 3 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (−9)/3 = −3 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = −3 m s⁻²
Action: Check force and direction.
After: ma = 3 × (−3) = −9 N; leftwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
With rightwards positive: a = −9/3 = −3 m s⁻², or 3 m s⁻² leftwards.
With rightwards positive: a = −9/3 = −3 m s⁻², or 3 m s⁻² leftwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 5 kg object has net force 20 N rightwards. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Given net force 20 N; mass 5 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (20)/5 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 5 × (4) = 20 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
a = 20/5 = 4 m s⁻² rightwards.
a = 20/5 = 4 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 4 kg object has net force 6 N rightwards. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Given net force 6 N; mass 4 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (6)/4 = 1.5 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 1.5 m s⁻²
Action: Check force and direction.
After: ma = 4 × (1.5) = 6 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
a = 6/4 = 1.5 m s⁻² rightwards. Acceleration need not be an integer.
a = 6/4 = 1.5 m s⁻² rightwards.
Acceleration need not be an integer.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 3 kg object has zero net force. Find acceleration. Can its speed be determined?
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Given net force 0 N; mass 3 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (0)/3 = 0 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 0 m s⁻²
Action: Check force and direction.
After: ma = 3 × (0) = 0 N; no acceleration
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
a = 0/3 = 0 m s⁻². Initial velocity is needed to determine speed; it may already be moving.
a = 0/3 = 0 m s⁻².
Initial velocity is needed to determine speed; it may already be moving.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 4 kg object has 18 N rightwards and 6 N leftwards. Find net force and acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 18 − 6 = 12 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/4 = 3 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 3 m s⁻²
Action: Check force and direction.
After: ma = 4 × (3) = 12 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 18 − 6 = 12 N rightwards; a = 12/4 = 3 m s⁻² rightwards.
F = 18 − 6 = 12 N rightwards; a = 12/4 = 3 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Forces 6 N and 4 N both act rightwards on a 2 kg object. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 6 + 4 = 10 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (10)/2 = 5 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 5 m s⁻²
Action: Check force and direction.
After: ma = 2 × (5) = 10 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 6 + 4 = 10 N rightwards; a = 10/2 = 5 m s⁻² rightwards.
F = 6 + 4 = 10 N rightwards; a = 10/2 = 5 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 3 kg object has 4 N rightwards and 10 N leftwards. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 4 − 10 = −6 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (−6)/3 = −2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = −2 m s⁻²
Action: Check force and direction.
After: ma = 3 × (−2) = −6 N; leftwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 4 − 10 = −6 N; a = −6/3 = −2 m s⁻², or 2 m s⁻² leftwards.
F = 4 − 10 = −6 N; a = −6/3 = −2 m s⁻², or 2 m s⁻² leftwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 5 kg object has 8 N rightwards, 5 N rightwards and 3 N leftwards. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 8 + 5 − 3 = 10 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (10)/5 = 2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 2 m s⁻²
Action: Check force and direction.
After: ma = 5 × (2) = 10 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 8 + 5 − 3 = 10 N rightwards; a = 10/5 = 2 m s⁻² rightwards.
F = 8 + 5 − 3 = 10 N rightwards; a = 10/5 = 2 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 3 kg object accelerates at 2 m s⁻² leftwards. Find net force.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: m = 3 kg; a = −2 m s⁻²
Action: Find net force directly.
After: F = ma
Why: Mass and acceleration are both supplied.
Step 2
Before: F = ma
Action: Substitute and multiply.
After: F = 3 × (−2) = −6 N
Why: kg m s⁻² is the newton unit; force follows the acceleration direction.
Step 3
Before: F = −6 N
Action: Divide by the original mass.
After: F/m = (−6)/3 = −2 m s⁻²
Why: The original acceleration is recovered.
F = ma = 3 × (−2) = −6 N, or 6 N leftwards.
F = ma = 3 × (−2) = −6 N, or 6 N leftwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Net force 15 N rightwards gives acceleration 3 m s⁻² rightwards. Find mass.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: F = 15 N; a = 3 m s⁻²
Action: Identify mass as the unknown.
After: F = ma
Why: Use matching directions for force and acceleration.
Step 2
Before: F = ma
Action: Divide both sides by nonzero a.
After: F/a = ma/a; m = F/a
Why: The matching a factors reduce to one.
Divide both sides by a; matching factors reduce to one
Step 3
Before: m = F/a
Action: Substitute both signed values.
After: m = (15)/(3) = 5 kg
Why: Matching signs give positive mass. Force divided by acceleration leaves kilograms.
Step 4
Before: m = 5 kg
Action: Multiply by the given acceleration.
After: ma = 5 × (3) = 15 N
Why: The force matches the original data.
Divide F = ma by a: m = F/a = 15/3 = 5 kg. Check: 5 × 3 = 15 N.
Divide F = ma by a: m = F/a = 15/3 = 5 kg.
Check: 5 × 3 = 15 N.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 2 kg object has 10 N rightwards and 4 N leftwards. A learner calculates a = 10/2. Repair the method.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 10 − 4 = 6 N; the lone rightward force is not the net force.
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (6)/2 = 3 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 3 m s⁻²
Action: Check force and direction.
After: ma = 2 × (3) = 6 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
They used one force. F = 10 − 4 = 6 N rightwards; a = 6/2 = 3 m s⁻² rightwards.
They used one force.
F = 10 − 4 = 6 N rightwards; a = 6/2 = 3 m s⁻² rightwards.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 10 kg shopping trolley moves east. Push is 18 N east and resistance 8 N west. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 18 − 8 = 10 N eastwards
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (10)/10 = 1 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 1 m s⁻²
Action: Check force and direction.
After: ma = 10 × (1) = 10 N; eastwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
East positive: F = 18 − 8 = 10 N east; a = 10/10 = 1 m s⁻² east.
East positive: F = 18 − 8 = 10 N east; a = 10/10 = 1 m s⁻² east.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 0.5 kg model train has driving force 1.5 N east and resistance 0.5 N west. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 1.5 − 0.5 = 1.0 N eastwards
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (1)/0.5 = 2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 2 m s⁻²
Action: Check force and direction.
After: ma = 0.5 × (2) = 1 N; eastwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 1.5 − 0.5 = 1.0 N east; a = 1.0/0.5 = 2 m s⁻² east.
F = 1.5 − 0.5 = 1.0 N east; a = 1.0/0.5 = 2 m s⁻² east.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 20 kg crate moves right against 4 N resistance. Find the push for 0.5 m s⁻² rightward acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Unknown push P; known mass and acceleration.
Action: Find the required net force first.
After: F = ma = 20.0 × 0.5 = 10.0 N
Why: The required net force differs from the push because resistance also acts.
Step 2
Before: P rightwards; 4.0 N leftwards.
Action: Write the signed force equation.
After: P − 4.0 = 10.0
Why: Rightwards is positive.
Step 3
Before: P − 4.0 = 10.0
Action: Add 4.0 to both sides.
After: P − 4.0 + 4.0 = 10.0 + 4.0; P = 14.0 N
Why: The opposite resistance terms sum to zero.
Apply +4 to both sides
Step 4
Before: P = 14.0 N
Action: Subtract resistance and divide by mass.
After: (14.0 − 4.0)/20.0 = 0.5 m s⁻²
Why: The requested acceleration is recovered.
Required F = 20 × 0.5 = 10 N rightwards. P − 4 = 10. Add 4 to both sides: P = 14 N rightwards. Check: (14 − 4)/20 = 0.5 m s⁻².
Required F = 20 × 0.5 = 10 N rightwards.
P − 4 = 10.
Add 4 to both sides: P = 14 N rightwards.
Check: (14 − 4)/20 = 0.5 m s⁻².
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Two helpers push a 4 kg trolley east with 6 N and 9 N. Resistance is 3 N west. Find acceleration.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: F = 6 + 9 − 3 = 12 N eastwards
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/4 = 3 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 3 m s⁻²
Action: Check force and direction.
After: ma = 4 × (3) = 12 N; eastwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
F = 6 + 9 − 3 = 12 N east; a = 12/4 = 3 m s⁻² east.
F = 6 + 9 − 3 = 12 N east; a = 12/4 = 3 m s⁻² east.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A delivery robot accelerates at 2 m s⁻² leftwards under net force 24 N leftwards. Find mass.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: F = −24 N; a = −2 m s⁻²
Action: Identify mass as the unknown.
After: F = ma
Why: Use matching directions for force and acceleration.
Step 2
Before: F = ma
Action: Divide both sides by nonzero a.
After: F/a = ma/a; m = F/a
Why: The matching a factors reduce to one.
Divide both sides by a; matching factors reduce to one
Step 3
Before: m = F/a
Action: Substitute both signed values.
After: m = (−24)/(−2) = 12 kg
Why: Matching signs give positive mass. Force divided by acceleration leaves kilograms.
Step 4
Before: m = 12 kg
Action: Multiply by the given acceleration.
After: ma = 12 × (−2) = −24 N
Why: The force matches the original data.
Matching signed directions: m = F/a = (−24)/(−2) = 12 kg. Check magnitude: 12 × 2 = 24 N.
Matching signed directions: m = F/a = (−24)/(−2) = 12 kg.
Check magnitude: 12 × 2 = 24 N.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
A 6 kg trolley moves right at 2 m s⁻¹ with forces 13 N right and 13 N left. Describe motion.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Opposing forces are both 13 N.
Action: Add their signed components.
After: F = 13 − 13 = 0 N
Why: Equal opposing forces balance.
Step 2
Before: F = 0 N; mass is positive.
Action: Use Newton’s second law.
After: a = F/m = 0/m = 0 m s⁻²
Why: Zero acceleration means no change in velocity.
Step 3
Before: The velocity before the balanced forces is 2 m s⁻¹ rightwards.
Action: Preserve that initial velocity.
After: Velocity remains 2 m s⁻¹ rightwards.
Why: Acceleration changes velocity; it does not specify the initial velocity.
F = 13 − 13 = 0 N; a = 0/6 = 0. It continues at 2 m s⁻¹ rightwards while forces remain balanced.
F = 13 − 13 = 0 N; a = 0/6 = 0.
It continues at 2 m s⁻¹ rightwards while forces remain balanced.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Net force 12 N rightwards acts on carts of 3 kg and 6 kg. Calculate and compare accelerations.
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Given net force 12 N; mass 3 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/3 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 3 × (4) = 12 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Step 5
Before: Given net force 12 N; mass 6 kg.
Action: Identify acceleration as the unknown.
After: Use F = ma.
Why: Net force is already supplied; do not add it as an extra force.
Step 6
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 7
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/6 = 2 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 8
Before: a = 2 m s⁻²
Action: Check force and direction.
After: ma = 6 × (2) = 12 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Step 9
Before: Masses 3 and 6 kg; accelerations 4 and 2 m s⁻².
Action: Compare both ratios.
After: Mass doubled; acceleration halved.
Why: The same net force acted in both cases.
a = 12/3 = 4 m s⁻² and a = 12/6 = 2 m s⁻², both rightwards. Twice the mass gives half the acceleration at unchanged net force.
a = 12/3 = 4 m s⁻² and a = 12/6 = 2 m s⁻², both rightwards.
Twice the mass gives half the acceleration at unchanged net force.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Cart A: mass 3 kg, forces 18 N right and 6 N left. Cart B: mass 3 kg, forces 15 N right and 3 N left. Does A accelerate more?
Hint 1
Identify the unknown and use forces on the same object.
Hint 2
Write the signed net force or the required net force ma.
Hint 3
Isolate the unknown with equal operations; substitute values, then check units and direction.
Show the complete working
Step 1
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: Cart A: F = 18 − 6 = 12 N
Why: Opposing force contributions have opposite signs.
Step 2
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 3
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/3 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 4
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 3 × (4) = 12 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Step 5
Before: Rightwards or eastwards is positive.
Action: Combine every signed force on the object.
After: Cart B: F = 15 − 3 = 12 N
Why: Opposing force contributions have opposite signs.
Step 6
Before: F = ma
Action: Divide both sides by m.
After: F/m = ma/m; a = F/m
Why: The factor m/m reduces to one.
Divide both sides by m; matching factors reduce to one
Step 7
Before: a = F/m
Action: Substitute the signed net force and positive mass.
After: a = (12)/3 = 4 m s⁻²
Why: N/kg gives kg m s⁻²/kg; the matching kg factors reduce to one.
Step 8
Before: a = 4 m s⁻²
Action: Check force and direction.
After: ma = 3 × (4) = 12 N; rightwards
Why: The signed acceleration follows the net force; zero acceleration preserves the initial velocity.
Step 9
Before: Both net forces 12 N; both masses 3 kg.
Action: Compare their calculated accelerations.
After: Both accelerate at 4 m s⁻² rightwards.
Why: Different individual forces can produce the same net force.
A: F = 18 − 6 = 12 N. B: F = 15 − 3 = 12 N. Both: a = 12/3 = 4 m s⁻² rightwards. Applied force alone is insufficient; their net forces and masses match.
A: F = 18 − 6 = 12 N.
B: F = 15 − 3 = 12 N.
Both: a = 12/3 = 4 m s⁻² rightwards.
Applied force alone is insufficient; their net forces and masses match.
Review your response against these criteria
Uses the stated physical or chemical relationship.
Shows the relevant working and reasoning.
Checks the result against the original conditions.
Extension
Predict the change in acceleration when mass doubles while net force doubles. Justify.
Compare both changes
Optional extension task. It does not change lesson access.
A 4 kg object has 20 N right and 8 N left. Find acceleration.
Compare your explanation
F = 20 − 8 = 12 N right; a = 12/4 = 3 m s⁻² right.
Net force 10 N produces acceleration 2 m s⁻² in the same direction. Find mass.
Compare your explanation
m = F/a = 10/2 = 5 kg.
A 2 kg trolley needs 3 m s⁻² rightwards against 2 N resistance. Find push.
Compare your explanation
F = 2 × 3 = 6 N; P − 2 = 6, so P = 8 N rightwards.
An object moves left while accelerating right. Explain its immediate speed change and what balanced forces would mean.
Compare your explanation
Its speed decreases initially because acceleration opposes velocity. With balanced forces acceleration is zero and its existing velocity remains unchanged.
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