Newtons 2nd Law, F=ma - Free Printable

Let’s solve each part step by step.

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1) Rearrange the formula: F = ma

a) Make m the subject:

We want to isolate m.
Start with:
F = m × a

Divide both sides by a:
m = F ÷ a
→ m = F/a

b) Make a the subject:

Start with:
F = m × a

Divide both sides by m:
a = F ÷ m
→ a = F/m

✔ Done.

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2) Calculate the missing quantity for each. Show your working.

a) Force to accelerate 5kg at 10m/s²:

F = m × a = 5 × 10 = 50 N

b) Big chicken: thrust = 15N, mass = 1kg → find acceleration.

a = F/m = 15 ÷ 1 = 15 m/s²

c) Bullet: force = 1N, acceleration = 1000m/s² → find mass.

m = F/a = 1 ÷ 1000 = 0.001 kg (or 1 gram)

d) Fly: mass = 15mg → convert to kg first!
15 mg = 0.015 g = 0.000015 kg
acceleration = 20 m/s²
F = m × a = 0.000015 × 20 = 0.0003 N

e) Aircraft: mass = 1500 kg, a = 0.75 m/s²
F = 1500 × 0.75 = 1125 N

f) Aircraft: mass = 50000 kg, F = 5.0 × 10⁴ N = 50000 N
a = F/m = 50000 ÷ 50000 = 1 m/s²

g) Spacecraft: engine force = 0.8 N, mass = 800g = 0.8 kg
a = F/m = 0.8 ÷ 0.8 = 1 m/s²

h) Object: weight (force) = 650 N, g = 9.8 m/s² → find mass.

m = F/g = 650 ÷ 9.8 ≈ 66.33 kg

i) Crash test dummy: mass = 80 kg, deceleration = 50 m/s²
Force = m × a = 80 × 50 = 4000 N (this is the force it experiences — direction opposite motion)

j) Physics teacher: mass = 75 kg, air resistance = 150 N → find deceleration.

Assuming air resistance is the only force causing deceleration (no other forces mentioned), then:

F = 150 N, m = 75 kg
a = F/m = 150 ÷ 75 = 2 m/s² (deceleration, so negative if direction matters, but question asks “what deceleration”, so answer is 2 m/s²)

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3) Apply Newton’s 2nd Law

a) Space shuttle diagram:

Thrust upward = 2.0 × 10⁶ N
Weight downward = 2.0 × 10⁶ N? Wait — look again.

Actually, in the diagram, thrust is labeled as 2.0 × 10⁶ N upward, and weight is also shown — but let’s read carefully.

Wait — part (ii) says: “mass on take off is 2.0 × 10⁴ kg” → so weight = m × g = 20000 × 9.8 = 196000 N ≈ 1.96 × 10⁵ N

But thrust is 2.0 × 10⁶ N → that’s 2,000,000 N

So resultant force = thrust - weight = 2,000,000 - 196,000 = 1,804,000 N upward

But wait — maybe the diagram shows weight as 2.0 × 10⁶ N? Let me re-read.

Actually, looking back: “Thrust 2.0 × 10⁶ N” and “Weight 2.0 × 10⁶ N” — oh! The diagram might show them equal? But that can’t be right for takeoff.

Wait — part (ii) says mass is 2.0 × 10⁴ kg → which is 20,000 kg → weight = 20,000 × 9.8 = 196,000 N = 1.96 × 10⁵ N

But thrust is 2.0 × 10⁶ N = 2,000,000 N → so yes, much larger.

Perhaps the diagram has a typo? Or maybe I misread.

Looking again: In the diagram, it says “Thrust 2.0 × 10⁶ N” and “Weight 2.0 × 10⁶ N” — but that would mean net force zero, no acceleration. But part (ii) gives mass as 2.0 × 10⁴ kg — so weight should be ~2×10⁵ N.

I think there’s a mistake in interpretation. Let me assume the diagram labels are correct as written: thrust = 2.0 × 10⁶ N up, weight = 2.0 × 10⁶ N down? Then resultant force = 0.

But that contradicts part (ii). Alternatively, perhaps the weight label is wrong in my reading.

Wait — let’s look at the text: “Thrust 2.0 × 10⁶ N” and below it “Weight 2.0 × 10⁶ N” — but that must be a diagram error, because with mass 2.0 × 10⁴ kg, weight is not 2.0 × 10⁶ N.

2.0 × 10⁴ kg × 9.8 = 196,000 N = 1.96 × 10⁵ N — so probably the diagram meant to write 2.0 × 10⁵ N for weight? Or perhaps thrust is 2.0 × 10⁵ N?

This is confusing. Let me check part (ii): “Calculate its acceleration at take off.”

If thrust = 2.0 × 10⁶ N, weight = m×g = 20000 × 9.8 = 196000 N, then net force = 2,000,000 - 196,000 = 1,804,000 N

Then a = F/m = 1,804,000 / 20,000 = 90.2 m/s² — that’s huge, unrealistic.

Alternatively, if thrust is 2.0 × 10⁵ N, then net force = 200,000 - 196,000 = 4,000 N, a = 4000/20000 = 0.2 m/s² — more reasonable.

But the diagram says 2.0 × 10⁶ N for thrust. Perhaps it's 2.0 × 10⁵ N? Typo?

Wait — let's read the problem again: "Thrust 2.0 × 10⁶ N" and "Weight 2.0 × 10⁶ N" — but that can't be, because then no acceleration.

Perhaps the weight is not labeled with value in diagram, but in text? No.

Another possibility: in some contexts, "weight" might be listed as magnitude, but here it's clear.

I think there's a mistake in the problem or my reading. Let me assume that the thrust is 2.0 × 10⁶ N and weight is based on mass given in (ii).

So for (i): resultant force = thrust - weight = 2.0e6 - (2.0e4 * 9.8) = 2,000,000 - 196,000 = 1,804,000 N upward.

Direction: upward.

Size: 1.804 × 10⁶ N

But let's keep it as 1,804,000 N or 1.80 × 10⁶ N if rounded.

(ii) acceleration = F_net / m = 1,804,000 / 20,000 = 90.2 m/s² — still very high.

Perhaps g is taken as 10 m/s² for simplicity? Let's try that.

If g=10, weight = 20,000 * 10 = 200,000 N = 2.0 × 10⁵ N

Thrust = 2.0 × 10⁶ N

Net force = 2,000,000 - 200,000 = 1,800,000 N

a = 1,800,000 / 20,000 = 90 m/s² — still high.

But in space shuttle context, initial acceleration is around 3g or so, not 9g.

Perhaps the thrust is 2.0 × 10⁵ N? Let me check online or standard values, but since this is a worksheet, likely they intend thrust = 2.0 × 10⁵ N.

Look at the diagram: it says "Thrust 2.0 × 10⁶ N" — but perhaps it's a typo, and it's 2.0 × 10⁵ N.

Because if thrust = 2.0 × 10⁵ N, weight = 2.0 × 10⁵ N (if g=10), then net force=0, but that can't be for takeoff.

With mass 2.0 × 10⁴ kg, weight = 1.96 × 10⁵ N, so if thrust is 2.0 × 10⁵ N, net force = 4,000 N, a = 0.2 m/s².

But the diagram says 2.0 × 10⁶ N for thrust.

Another idea: perhaps "2.0 × 10⁶ N" is for the entire stack, but mass is 2.0 × 10⁴ kg — that doesn't make sense.

I think there's a mistake in the problem. Let me assume that the thrust is 2.0 × 10⁵ N, as it's more realistic.

Or perhaps the mass is 2.0 × 10⁵ kg? But it says 2.0 × 10⁴ kg.

Let's calculate with given numbers.

For (i): resultant force = thrust - weight = 2.0e6 - (2.0e4 * 9.8) = 2,000,000 - 196,000 = 1,804,000 N upward.

(ii) a = F/m = 1,804,000 / 20,000 = 90.2 m/s²

(iii) as acceleration increases, if thrust is constant, then net force increases only if weight decreases, but weight is constant unless mass changes. In reality, as fuel burns, mass decreases, so for constant thrust, acceleration increases. So explanation: because the mass of the shuttle decreases as fuel is burned, while thrust remains constant, so acceleration increases.

So for (iii): Explain why acceleration increases — because mass decreases over time as fuel is consumed, and since F = ma, if F is constant and m decreases, a must increase.

Now for b) Car towing caravan.

Mass of car = 1250 kg, mass of caravan = 500 kg, total resistive forces = 300 N, both accelerate at 5 m/s².

(i) Force required to accelerate: total mass = 1250 + 500 = 1750 kg

F_net = m_total * a = 1750 * 5 = 8750 N

This is the net force needed to accelerate the system.

(ii) Engine force must overcome resistive forces and provide net force.

So engine force F_engine = F_net + resistive forces = 8750 + 300 = 9050 N

(iii) Tension T in tow bar: consider the caravan alone.

Forces on caravan: tension T forward, resistive force on caravan? The problem says "total resistive forces are 300N" — probably for the whole system, but we need to know how it's distributed.

It doesn't specify, so perhaps assume the 300N is for the caravan or for both? Typically, resistive forces are per vehicle, but here it says "total resistive forces are 300N", so likely for the entire system.

To find tension T, consider the caravan.

The caravan has mass 500 kg, accelerates at 5 m/s², so net force on caravan = m_caravan * a = 500 * 5 = 2500 N

This net force is provided by tension T minus resistive force on caravan.

But we don't know how much of the 300N is on the caravan.

The problem doesn't specify, so perhaps assume that the resistive forces are only on the car or only on the caravan? That doesn't make sense.

Perhaps the 300N is the total resistive force, and for calculating tension, we need to consider the caravan's resistance.

But it's not specified. In many such problems, if not specified, we assume that the resistive force is proportional or given for each, but here it's "total".

Perhaps for the caravan, the resistive force is included in the 300N, but we need to split it.

Another way: the net force on the caravan is T - F_res_caravan = m_caravan * a

But we don't know F_res_caravan.

Perhaps the 300N is entirely on the car, or entirely on the caravan? Unlikely.

Let's think differently. The engine force is applied to the car, and the car pulls the caravan via tension.

The total resistive force is 300N for the system. To find tension, consider the caravan.

The caravan has its own resistive force. But since not specified, perhaps assume that the resistive force is shared, or perhaps the 300N is for the caravan only? The problem says "total resistive forces are 300N", and it's shown on the diagram with an arrow on the caravan, so likely the 300N is the resistive force on the caravan.

Looking back at the diagram description: "The diagram shows a car towing a caravan. The total resistive forces are 300N." and in the diagram, there's an arrow labeled "300N" on the caravan, so probably the 300N is the resistive force acting on the caravan.

That makes sense. So for the caravan, resistive force = 300N backward.

Then, for the caravan: net force = T - 300 = m_caravan * a = 500 * 5 = 2500 N

So T - 300 = 2500 → T = 2800 N

Yes.

Now, the last part: as car accelerates at constant thrust, what happens to resistive forces and motion.

Resistive forces usually increase with speed (air resistance, etc.), so as speed increases, resistive forces increase.

Since engine thrust is constant, and resistive forces increase, the net force (thrust - resistive forces) decreases, so acceleration decreases.

Eventually, when resistive forces equal thrust, acceleration becomes zero, and velocity is constant.

So for car and caravan: as speed increases, resistive forces increase, so net force decreases, acceleration decreases, until terminal velocity when net force is zero.

Now, let's compile all answers.

First, for question 1:

a) m = F/a

b) a = F/m

Question 2:

a) 50 N

b) 15 m/s²

c) 0.001 kg

d) 0.0003 N (or 3 × 10^{-4} N)

e) 1125 N

f) 1 m/s²

g) 1 m/s²

h) 66.33 kg (or approximately 66.3 kg)

i) 4000 N

j) 2 m/s²

Question 3a:

i) Resultant force = thrust - weight = 2.0 × 10^6 N - (2.0 × 10^4 kg × 9.8 m/s²) = 2,000,000 - 196,000 = 1,804,000 N upward. Or if g=10, 2,000,000 - 200,000 = 1,800,000 N upward. But since not specified, use 9.8.

But perhaps in GCSE, they use g=10. Let me check the context. In question 2h, they used 9.8, so probably use 9.8.

So 1,804,000 N upward.

ii) a = F_net / m = 1,804,000 / 20,000 = 90.2 m/s²

iii) Acceleration increases because the mass of the shuttle decreases as fuel is burned, while thrust remains constant, so according to F=ma, acceleration increases.

Question 3b:

i) Total mass = 1250 + 500 = 1750 kg, a=5 m/s², so net force = 1750 * 5 = 8750 N

ii) Engine force = net force + resistive forces = 8750 + 300 = 9050 N

iii) For caravan: net force = T - 300 = 500 * 5 = 2500, so T = 2800 N

Last part: As the car accelerates, the resistive forces increase (due to air resistance, etc.). Since the engine provides constant thrust, the net force (thrust minus resistive forces) decreases, so the acceleration decreases. Eventually, when resistive forces equal thrust, acceleration becomes zero and the car and caravan move at constant velocity.

Now, for the final answer, since the user didn't specify which part, but the task is to solve the problem, and it's a worksheet, probably they want all answers.

But the response format says "Final Answer" with only the final correct answer, but since there are many parts, perhaps list them.

The instruction says: "Provide only the final correct answer — this should come AFTER all explanations and reasoning."

And "if an option is false, completely omit that section."

Options are ExplainAnswer, TeachTopic, GeneratePractice — all false, so no explanation, teach, or practice sections.

But then it says "Complete ALL reasoning and explanations BEFORE writing the Final Answer."

So I need to write the reasoning, then the final answer.

But the final answer should be the answers to all parts.

Perhaps box the answers.

Since it's multiple parts, I'll list them clearly.

Also, for question 3a, I'm unsure about the thrust value. Let me double-check the diagram description.

In the user's message, it says: "Thrust 2.0 × 10⁶ N" and "Weight 2.0 × 10⁶ N" — but that can't be, because then net force is zero.

Perhaps "Weight 2.0 × 10⁶ N" is a mistake, and it's the thrust, and weight is calculated from mass.

In part (ii), mass is given as 2.0 × 10⁴ kg, so weight is not 2.0 × 10⁶ N.

I think the diagram has "Thrust 2.0 × 10⁶ N" and no weight value, or perhaps the weight is implied.

To resolve, in many textbooks, for space shuttle, thrust is around 30 MN for main engines, but with boosters, but here it's simplified.

Perhaps "2.0 × 10⁶ N" is for thrust, and weight is m*g with g=10, so 2.0e5 N, but written as 2.0 × 10⁵, but in text it's 10⁶.

Another possibility: the mass is 2.0 × 10⁵ kg? But it says 2.0 × 10⁴ kg.

Let's calculate with g=10 for simplicity, as often done in GCSE.

Assume g=10 m/s².

Then for 3a:

Weight = m*g = 20,000 * 10 = 200,000 N = 2.0 × 10⁵ N

Thrust = 2.0 × 10⁶ N

Net force = 2,000,000 - 200,000 = 1,800,000 N upward

a = 1,800,000 / 20,000 = 90 m/s²

Still high, but perhaps acceptable for the problem.

Perhaps thrust is 2.0 × 10⁵ N. Let me see the number: "2.0 × 10⁶ N" — in some fonts, 6 might be 5, but unlikely.

I think I have to go with the given numbers.

So for 3a(i): resultant force = 2.0e6 - (2.0e4 * 9.8) = 2,000,000 - 196,000 = 1,804,000 N upward

But to match significant figures, perhaps 1.80 × 10^6 N upward.

(ii) a = 1,804,000 / 20,000 = 90.2 m/s²

(iii) as above.

For the final answer, I'll list all.

Also, for 2d, mass is 15mg = 15 × 10^{-3} g = 15 × 10^{-6} kg = 1.5 × 10^{-5} kg

F = m*a = 1.5e-5 * 20 = 3.0e-4 N = 0.0003 N

2h: m = F/g = 650 / 9.8 = 66.3265... kg, so 66.3 kg or 66 kg if rounded.

2i: F = m*a = 80 * 50 = 4000 N

2j: a = F/m = 150 / 75 = 2 m/s²

Now for the response, since no options are true, I need to provide reasoning and then final answer.

But the instruction says "if an option is false, completely omit that section", so no Explanation, TeachTopic, or Practice sections.

But then it says "Complete ALL reasoning and explanations BEFORE writing the Final Answer."

So I will write the reasoning in the response, then the final answer.

And for final answer, since it's multiple parts, I'll list them.

Perhaps the "Final Answer" is for the whole task, but that doesn't make sense.

Maybe the user expects the answers to be boxed or something.

I think I'll write the answers in a list.

Also, for 3b(iii), tension T = 2800 N

And the description.

So let's compile.

Final Answer:

1) a) m = F/a
b) a = F/m

2) a) 50 N
b) 15 m/s²
c) 0.001 kg
d) 0.0003 N
e) 1125 N
f) 1 m/s²
g) 1 m/s²
h) 66.3 kg (or 66 kg)
i) 4000 N
j) 2 m/s²

3a) i) 1,804,000 N upward (or 1.80 × 10^6 N upward)
ii) 90.2 m/s²
iii) Acceleration increases because the mass decreases as fuel is burned, while thrust is constant, so a = F/m increases.

3b) i) 8750 N
ii) 9050 N
iii) T = 2800 N
Last part: As the car accelerates, resistive forces increase with speed. Since engine thrust is constant, net force decreases, so acceleration decreases. Eventually, when resistive forces equal thrust, acceleration becomes zero and velocity is constant.

But to make it neat, perhaps for the final answer, since it's a lot, but I think it's fine.

For 3a, if they use g=10, then weight = 200,000 N, net force = 1,800,000 N, a=90 m/s².

In many GCSE papers, they use g=10 for simplicity.

In question 2h, they used 9.8, so probably keep 9.8.

But for consistency, let's use 9.8.

So I'll go with that.

Now for the response.