VCE Physics (Year 12) · Motion and Forces · 20m read

Energy, Work and Power

Ever wonder why pushing a box up a hill is harder than rolling it down, even though the same forces are involved? Energy and work are your secret weapon for solving tricky mechanics problems without getting bogged down in forces and acceleration. Instead of tracking every detail of motion, you'll use the simple idea that energy doesn't disappear—it just transforms—letting you leap straight to answers on springs, slopes, and friction scenarios.

Learning objective: Use the work–energy theorem and conservation of energy to solve mechanical problems including springs and friction.

By the end of this lesson you can:

I calculate work as $W = Fs\cos\theta$ and identify when work is zero.
I apply $KE = \frac{1}{2}mv^2$, $PE_{grav} = mgh$, $PE_{elastic} = \frac{1}{2}kx^2$.
I apply conservation of mechanical energy with and without friction.
I calculate power as $P = W/t = Fv$.

Energy is money moving between accounts. Work transfers energy; power is how fast you spend it; total energy is always conserved.

Work, energy types and the work–energy theorem

### Work

$$W = Fs\cos\theta \quad [\text{J} = \text{N}\cdot\text{m}]$$

Only the component of force parallel to displacement does work.
- $\theta = 0°$: maximum positive work (force and displacement in same direction)
- $\theta = 90°$: zero work (force perpendicular to displacement — normal force, centripetal force)
- $\theta = 180°$: negative work (friction opposing motion)

### Types of mechanical energy

Kinetic energy:
$$KE = \frac{1}{2}mv^2$$

Gravitational potential energy:
$$PE_{grav} = mgh \qquad \text{(h measured from chosen reference level)}$$

Elastic potential energy (spring):
$$PE_{spring} = \frac{1}{2}kx^2 \qquad \text{(x = extension/compression from natural length)}$$

### Work–energy theorem

$$W_{net} = \Delta KE = KE_f - KE_i$$

The net work done on an object equals the change in its kinetic energy. Useful when forces vary (like a spring, or varying friction).

### Conservation of mechanical energy

If only conservative forces act (no friction, no air resistance):

$$KE_1 + PE_1 = KE_2 + PE_2$$

With friction (non-conservative force):

$$KE_1 + PE_1 = KE_2 + PE_2 + W_{friction}$$

where $W_{friction} = f \cdot d$ (always positive, this is the energy dissipated as heat).

### Power

$$P = \frac{W}{t} = Fv \quad [\text{W} = \text{J/s}]$$

$P = Fv$ only when force and velocity are parallel (or use the component of F parallel to v).

Energy and Power Formulas

Work: W = Fs\cos\theta ($\theta$ between force and displacement)
Kinetic energy: KE = \tfrac{1}{2}mv^2 (Scalar, always $\geq 0$)
Gravitational PE: PE_{grav} = mgh (Choose your own $h = 0$ reference)
Spring PE: PE_{spring} = \tfrac{1}{2}kx^2 ($x$ from natural length, not from wall)
Energy conservation: KE_1 + PE_1 = KE_2 + PE_2 (No friction only)
With friction: E_{lost} = f \cdot d (Friction converts to heat)
Power: P = W/t = Fv (Force parallel to velocity)

Energy conservation strategy

1. Choose PE reference: Set $h = 0$ at the lowest point (or wherever convenient).
2. List all energies: At each position: $KE$, $PE_{grav}$, $PE_{spring}$. Write expressions.
3. Write energy equation: $E_1 = E_2$ (frictionless) or $E_1 - f \cdot d = E_2$ (friction).
4. Mass cancels: If no spring or friction, $m$ cancels — speed at a height is independent of mass.
5. Verify: Check: energy at bottom must be $\geq$ energy at top (energy cannot be created).

Quick reference

| Formula | Meaning |
|---|---|
| W = Fs cos θ | Work |
| Ek = ½mv² | Kinetic energy |
| Eg = mgΔh | Gravitational PE |
| Es = ½kx² | Spring PE |
| P = W/t = Fv | Power |

Worked example (Easy)

A 2.0 kg block slides down a frictionless 1.5 m high ramp from rest. Find its speed at the bottom.

Set reference. $h = 0$ at the bottom.
Write conservation. mgh = \tfrac{1}{2}mv^2
Cancel mass, solve. v^2 = 2gh = 2(9.8)(1.5) = 29.4 \implies v = 5.4 \ \text{m/s}

Answer: $v = 5.4 \ \text{m/s}$ — note mass cancelled, so any block on this frictionless ramp has the same speed.

Worked example (Hard)

A 1.5 kg mass slides from rest down a 2.0 m frictionless slope (height 0.80 m), then compresses a spring ($k = 600 \ \text{N/m}$) by $x$ before momentarily stopping. Find the maximum compression $x$.

Energy at start. Only gravitational PE ($h = 0.80$ m, $v = 0$): E_i = mgh = 1.5(9.8)(0.80) = 11.76 \ \text{J}
Energy at maximum compression. Only spring PE ($v = 0$ at max compression, $h = 0$): E_f = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(600)x^2 = 300x^2
Apply conservation. 11.76 = 300x^2 \implies x^2 = 0.0392 \implies x = 0.198 \ \text{m}

Answer: Maximum compression $x \approx 0.20 \ \text{m}$. All gravitational PE has been stored as elastic PE at the moment of stopping.

Common mistakes

Using $x$ as the total spring compression, not the compression from natural length $PE_{spring} = \frac{1}{2}kx^2$ requires $x$ to be measured from the natural (uncompressed) length. If a spring is pre-compressed, you must use the additional compression, not the total position. Fix: Identify the natural length position. Measure $x$ as the displacement from that point. Sketch the spring at natural length and at the compressed/stretched position.
Forgetting that friction does negative work (removes energy) Students often set $E_1 = E_2$ even when friction is present, giving an answer that violates energy conservation (final KE comes out larger than it should be). Fix: With friction: $E_1 - f \cdot d = E_2$. The friction term $f \cdot d$ is always positive — energy is removed, so final energy is always less than initial.
Using $P = Fv$ when the force isn't parallel to velocity Power is $P = \vec{F} \cdot \vec{v} = Fv\cos\theta$. If a car is being pushed up a slope, the engine force may not be parallel to the velocity direction used. Fix: Always use the component of force parallel to motion, or use $P = W/t$ directly.

Conservative vs non-conservative forces — why it matters for energy

Conservative forces (gravity, springs): the work done by them depends only on start and end position, not the path taken. This is why we can define a "potential energy" function — the PE at any point uniquely represents the energy stored regardless of how you got there.

Non-conservative forces (friction, air resistance): work done depends on the path. A long, winding path with friction does more work than a short path. These forces cannot be represented as potential energy.

Why this matters for problem-solving: For conservative-only systems, you can choose any two points and set their total energies equal — $E_1 = E_2$. For systems with friction, energy is being transferred to thermal energy: $E_1 - W_{friction} = E_2$. The thermal energy is "gone" from the mechanical system — you cannot recover it as kinetic energy later.

Interesting consequence: If all forces were conservative, we could create perpetual motion machines — running a pendulum forever. The existence of friction and dissipation means this is physically impossible. The irreversibility of friction is deeply connected to the Second Law of Thermodynamics.

Practice questions

A force of 50 N acts perpendicular to the direction of motion over 10 m. The work done is:
1. $500 \ \text{J}$
2. $0 \ \text{J}$
3. $50 \ \text{J}$
4. $5 \ \text{J}$
(1 marks)
A spring with $k = 200 \ \text{N/m}$ is compressed $0.050 \ \text{m}$. Elastic PE stored is:
1. $0.25 \ \text{J}$
2. $5.0 \ \text{J}$
3. $10 \ \text{J}$
4. $50 \ \text{J}$
(2 marks)
A 1500 kg car accelerates from 0 to 20 m/s in 8.0 s. Find the average useful power delivered.
1. $37.5 \ \text{kW}$
2. $18.75 \ \text{kW}$
3. $75 \ \text{kW}$
4. $300 \ \text{kW}$
(3 marks)
A 0.50 kg trolley rolls down a frictionless track from a height of 0.80 m, then compresses a spring ($k = 400 \ \text{N/m}$) before stopping. Find the maximum compression.
(4 marks)
A 3.0 kg block slides 4.0 m down a 30° slope with friction coefficient $\mu = 0.20$. Find its speed at the bottom if it starts from rest. ($g = 9.8 \ \text{m/s}^2$)
(4 marks)

Summary

$W = Fs\cos\theta$. Perpendicular force does zero work.
$KE = \frac{1}{2}mv^2$; $PE_{grav} = mgh$; $PE_{spring} = \frac{1}{2}kx^2$.
No friction: $KE_1 + PE_1 = KE_2 + PE_2$. Mass cancels for gravity-only problems.
Friction: $E_1 - f\cdot d = E_2$. Energy removed becomes heat.
$P = W/t = Fv$ (parallel components). Watch for perpendicular components.