🧠 Work, Energy & Power: Efficiency and Resistance & Real Exam Questions

So, today we’re talking energy — not just plugging numbers into $W = Fs$ or $E_k = \tfrac12 mv^2$ , but thinking like someone who sees a moving system as an energy balance rather than a timeline of forces. SUVAT is brilliant, but examiners love when candidates switch view and use energy instead — because often the working becomes shorter, cleaner, calmer.

Imagine pushing a block across a rough floor. You could chase forces and accelerations, or you could say: work done in pushing = gain in kinetic energy + energy lost to friction. And suddenly three lines appear instead of twelve. That’s the goal today — building a way of working that feels instinctive. Something that strengthens your A Level Maths understanding rather than handing you another method to memorise — the kind of shift you get when you see A Level Maths problem-solving explained properly.

🔙 Previous topic:

Our lesson on Newton’s Laws: Lifts, Contact Forces, Blocks & Tension Problems leads neatly into this one because energy methods make far more sense once you’ve already seen how forces drive motion in those classic setups.

📘 Where These Questions Appear in Exams

Work–energy questions appear across mechanics papers. They look innocent — slopes, resistive forces, cars braking, power outputs — but the trick is whether you choose energy or Newton’s second law. Often energy solves what forces make messy.

📏 Setting Up the Energy Model

A particle of mass m is pulled along a rough horizontal surface by a constant force F for a distance s. The resistive force is R. The particle starts from rest. Find its final speed.

🖼️ Visual Setup for the Scenario

🧠 Breaking Down the Core Energy Ideas

🔨 Work Done — Force and Distance Working Together

When a force pushes something through distance s, energy transfers.

Work done by the applied force:

We write ( $W = Fs$ )

If the surface pushes back with resistance R, then work is done against that force too:

We might form ( $W_{\text{resist}} = Rs$ )

Already two pieces of an energy story.

🏎️ Kinetic Energy — How the Motion Shows Itself

Starting from rest means initial kinetic energy is zero.

Final kinetic energy is:

We write ( $E_k = \tfrac12 mv^2$ )

So energy input goes into kinetic energy unless something steals it — which is exactly what resistance does.

We balance it:

We might form
$Fs – Rs = \tfrac12 mv^2$

Solve gently:

( $v = \sqrt{\frac{2s(F – R)}{m}}$ )

And that’s the whole thing — no forces, no accelerations, no time.

⚙️ Power — How Fast Energy Is Being Used

Students often treat power like a spare formula. It isn’t — it’s energy per second.

We write ( $P = \frac{W}{t} = Fv$ )

That second relation surprises people: power is force × speed.
So if a car moves twice as fast, the engine must supply twice the power to maintain the same forward thrust — which explains why fuel economy collapses at high speeds.

Energy thinking unlocks understanding.

🪜 Lifting Work — Turning Effort Into Height Gain

Raise a mass m through height h, and you increase gravitational potential energy:

We form ( $mgh$ )

If no resistive loss:

Work done by the lifter = $mgh$

If there is a resistive loss — friction on pulleys, inefficiency in machinery:

We write
( $W_{\text{input}} = mgh + \text{energy lost}$ )

Examiners love these because students instinctively reach for Newton, even when energy is cleaner.

🎯 Efficiency — Tracking Useful Output vs Losses

Efficiency is about what portion of energy input ends up as useful output.

We write ( $\eta = \frac{\text{useful output}}{\text{input}}$ )

Multiply by 100% if you want a tidy percentage.

Real-world machines lose energy to sound, heat, vibration — energy doesn’t disappear, it spreads into less helpful forms.

Train braking, bike gears, engines, lifting cranes — all exam gold.

🧲 Resistance Problems — The Clean Energy Route

A car of mass m travels with velocity u; engine is turned off, resistive force R acts, car stops after distance d.

Energy view:

Initial kinetic energy = ( $\tfrac12 mu^2$ )

Energy lost to resistance = ( $Rd$ )

So balance:

We write
( $\tfrac12 mu^2 = Rd$ )

No accelerations needed — final velocity zero gives everything.

Exam writers absolutely love this, and it’s the kind of detail where A Level Maths revision during exam season pays you back almost instantly.

🚚 Mixed Problems — Power, Drag and Steady Motion

A lift moves upward at constant speed v, lifting mass M with power P. Resistive drag = D. Find tension T in the cable.

Energy per second used to lift:

We form ( $Mg v$ )

Energy lost to drag each second:

We form ( $Dv$ ]

Total power must supply both:

We write
( $P = v(T)$ )

But tension must support both weight and drag:

( $T = Mg + D$ )

Then combining gives:

We might form
( $P = v(Mg + D)$ )

Solve for whichever variable the exam target chooses.

This is exactly the type of modelling question where energy replaces brute force Newton’s law grind.

⚠️ Mistakes Students Make (and How to Dodge Them)

Mixing work done by vs work done against forces
Forgetting resistive forces reduce net energy gain
Using power incorrectly in non-steady motion
Trying to force SUVAT when energy is shorter
Not identifying what counts as “useful output”

One clear formulation:

Energy input − losses = energy gained

That single sentence solves whole pages.

🌍 Why Energy Methods Appear Everywhere

Cycling uphill burns legs. Cycling on flat into strong wind does too. One is gravitational energy, the other resistive work — but both feel identical in your quads. Cars slow when engines cut because resistive energy drains kinetic energy like a leak in a tank.

Energy thinking is reality-thinking.

🚀 Ready to Progress?

If you want energy questions — lifts, power, resistive motion, braking-distance modelling, engine output — to feel natural rather than uncertain, the complete online A Level Maths Revision Course develops energy fluency through guided problems, shortcut reasoning and real exam-style modelling.

📏 Recap Table

• Draw each block separately
• Tension is consistent only in ideal strings
• Normal reaction reveals lift behaviour
• Contact forces follow Newton’s 3rd Law
• Direction choice must be fixed
• Use one equation per block

Author Bio – S. Mahandru

I’m a mechanics teacher who thinks of moments like tiny stories about balance — shift a weight here, adjust a support there, and the whole system reveals what it wants to do. Once students see that equilibrium is more about intuition than algebra, the topic suddenly feels far less mysterious and far more satisfying.

🧭 Next topic:

Having explored how energy is transferred, lost and measured through efficiency and resistance — and seen these ideas in real exam questions — we’re now ready to step into Impulse and Momentum, where those same principles drive collisions, rebounds and conservation laws.

❓ Quick FAQs

Why do we bother drawing each block separately? I mean… can’t we just treat the whole system as one thing?

Honestly, yes — sometimes treating it as one thing is quicker, and I even encourage it… but only after you’ve drawn the separate diagrams. Each block has its own little world of forces acting on it, and the moment you skip that step, weird things start happening. You mix signs. You forget where tension goes. You treat friction like it magically applies to both. And then you’re four lines in thinking “hang on— why doesn’t this make sense?”

The separate diagrams force you to slow down long enough to see the structure: who pulls whom, who resists what, where acceleration actually points. Once you’ve got that, combining them is easy, almost obvious. But if you jump straight to the combined system, you lose the internal story — and that “internal story” is usually where the marks live. Treat them separately first, even if you don’t feel like it. Future-you will be grateful.

Why does the normal reaction get bigger or smaller in lifts? It feels weird — shouldn’t my weight just be my weight?

Yes — and that’s the exact misconception that tangles everyone. Your actual weight never changes; it’s $mg$ whether you’re in a lift or on the Moon. What changes is the push from the floor. And that push — the normal reaction — is just the lift trying to keep you moving the way it’s moving.

If the lift shoots upward, it has to shove harder on you to give you the same upward acceleration. If it slows down while going up, the shove relaxes a bit. And if it drops suddenly, the floor barely touches you — that odd flutter in your stomach? That’s just R dipping below mg for a moment.

Students often overthink this, but really, R is a mood ring for the lift’s motion: hard shove = accelerating up, soft shove = accelerating down, no shove = free fall (which hopefully never appears outside the exam paper). Once you separate R from “how heavy I feel,” lift problems calm right down.

Tension keeps confusing me. Why does it change when the setup changes? Isn’t a string just… a string?

I wish it were that simple. Think of tension less like a fixed quantity and more like the string doing whatever it must to keep itself un-stretched. That’s its entire personality. Nothing else.

If two blocks accelerate, tension adjusts to make that possible while keeping the string length constant. If one block is heavier, the system shifts around until the acceleration and tension fit the constraints. And here’s the messy bit students rarely say out loud: tension isn’t “decided by” the string — it’s decided by the equations of motion on each block.

Different masses? Tension settles somewhere between their weights. Add friction? Tension rebalances again. Put the system on an incline? It changes once more. It’s reactive, not constant. Honestly, once you stop expecting tension to behave nicely — like a polite houseguest — and start treating it as a force that negotiates between both ends of the string, everything clicks. The diagrams begin telling the story, and tension stops feeling mystical.

🧠 Work, Energy & Power: Efficiency and Resistance & Real Exam Questions