Work, Energy and Conservation

Spec mapping: OCR H556 Module 3.3 — Work, energy and power (definition of work; kinetic and gravitational potential energy; conservation of energy; energy transfers in mechanical systems). Refer to the official OCR H556 specification document for exact wording.

Energy is arguably the single most powerful unifying concept in physics. Forces and velocities are tied to particular bodies at particular moments; energy is a bookkeeping quantity that transfers between bodies and between forms but is never created or destroyed. The general principle of conservation of energy was crystallised across the nineteenth century through the experiments of Joule, Helmholtz, Mayer and others — and it remains, in modified relativistic form, exact today. This lesson develops the mechanical instances of that principle — work, kinetic energy $\tfrac{1}{2}mv^2$ , gravitational potential energy $mgh$ — and the link back to Newton's second law via the work-energy theorem.

Key Definition: The work done by a constant force $\vec{F}$ on a body that undergoes displacement $\vec{s}$ is $W = \vec{F}\cdot\vec{s} = Fs\cos\theta$ , where $\theta$ is the angle between force and displacement. Work is scalar: positive when the force has a component along the motion, negative when it opposes motion, zero when it is perpendicular.

Work Done by a Constant Force

For a constant force $\vec{F}$ acting through displacement $\vec{s}$ :

$W = F\,s\cos\theta$

where:

$F$ is the magnitude of the force, in newtons (N).
$s$ is the magnitude of the displacement, in metres (m).
$\theta$ is the angle between $\vec{F}$ and $\vec{s}$ (so $0 \leq \theta \leq 180°$ ).

The SI unit of work is the joule (J), defined as one newton acting over one metre in the direction of the force. Equivalently, $1\,\text{J} = 1\,\text{N m} = 1\,\text{kg m}^2\,\text{s}^{-2}$ .

Three special cases are worth committing to memory:

$\theta = 0°$ (force along motion): $W = Fs$ , maximum positive. Energy is transferred to the body — its KE rises.
$\theta = 90°$ (force perpendicular to motion): $W = 0$ . The force can change the direction of motion but does not change the body's KE. Two crucial examples: (i) the normal contact force on a block sliding on a horizontal surface; (ii) the centripetal tension on a ball on a string moving in a horizontal circle at constant speed.
$\theta = 180°$ (force opposite to motion): $W = -Fs$ , negative. The body's KE falls. Friction on a decelerating block does negative work.

Common Exam Mistake: Forgetting the $\cos\theta$ factor when the force is not aligned with the motion. The full definition $W = Fs\cos\theta$ should be the starting point for every work calculation.

Worked Example 1 — Sledge Pulled by an Inclined Rope

A sledge is pulled $12$ m along horizontal snow by a rope inclined at $30°$ above the horizontal. The tension in the rope is $50$ N. Calculate the work done on the sledge by the tension.

$W = F\,s\cos\theta = 50 \times 12 \times \cos 30° = 50 \times 12 \times 0.8660 = 520\;\text{J}$

Only the horizontal component of the tension ( $50\cos 30° = 43.3$ N) does work on the sledge along its direction of motion; the vertical component ( $50\sin 30° = 25$ N) is perpendicular to the displacement and does no work — though it does slightly reduce the normal force on the snow, which in turn reduces the kinetic friction. The full energetic picture of a sledge therefore involves three forces (tension, weight, friction) and only two of them — tension and friction — actually do work.

Kinetic Energy

The kinetic energy of a body of mass $m$ moving at speed $v$ is:

$E_k = \tfrac{1}{2}mv^2$

This formula falls naturally from Newton's second law: consider a body of mass $m$ initially at rest, accelerated by a constant net force $F$ over distance $s$ . The body reaches speed $v$ given by SUVAT $v^2 = u^2 + 2as$ with $u = 0$ :

$s = \frac{v^2}{2a}$

The work done by the net force is:

$W_{\text{net}} = F\,s = ma \times \frac{v^2}{2a} = \tfrac{1}{2}mv^2$

So the work done by the net force exactly equals the kinetic energy gained.

This generalises to the work-energy theorem:

$W_{\text{net}} = \Delta E_k = E_{k,\text{final}} - E_{k,\text{initial}}$

Stated cleanly: the net work done on a body equals the change in its kinetic energy. This theorem is exact for any constant or varying force in classical mechanics — derived directly from $F = ma$ — and it is the single most useful identity in problem-solving.

Worked Example 2 — Braking Car

A $1500$ kg car travelling at $25$ m s $^{-1}$ brakes to rest. Find (a) its initial KE, (b) the work done by the brakes, (c) the average braking force if the car stops over $40$ m.

(a) $E_k = \tfrac{1}{2} \times 1500 \times 25^2 = \tfrac{1}{2} \times 1500 \times 625 = 4.69 \times 10^5$ J $\approx 469$ kJ.

(b) $W_{\text{brakes}} = \Delta E_k = 0 - 469\,000 = -469$ kJ. The negative sign means energy is taken from the car — it ends up as heat in the brake pads and discs (the discs glow visibly red after sustained braking), plus a small amount of acoustic energy and tyre-rubber dissipation.

(c) $W = -F\,s$ where $F$ is the magnitude of the braking force. So $F = 469\,000 / 40 = 11.7$ kN. Compare with the car's weight $mg = 14.7$ kN — the deceleration is about $0.8g$ , near the limit of dry-road friction.

Gravitational Potential Energy

Near the Earth's surface (where the gravitational field strength $g$ is approximately uniform at $9.81$ N kg $^{-1}$ ), the gravitational potential energy of a mass $m$ at height $h$ above a chosen reference level is:

$E_p = mgh$

This formula derives directly from work: to lift a mass $m$ at constant velocity through height $h$ requires an upward force of $mg$ (to balance gravity), doing work $W = mgh$ . By the work-energy theorem the KE is unchanged (constant velocity), so the energy must have gone somewhere: it is stored as gravitational PE in the Earth-mass system. When the mass is later released, gravity does positive work returning that $mgh$ to the body as kinetic energy.

The choice of reference level is arbitrary — only changes $\Delta E_p$ are physically meaningful. Set "zero PE" at the floor, the table top, sea level — whatever makes the algebra simplest. A common convention is to set PE = 0 at the lowest point of motion.

Common Exam Mistake: Applying $E_p = mgh$ to a problem with large height changes — e.g. a satellite orbit, or a rocket reaching low Earth orbit at $\sim 200$ km altitude. The formula assumes $g$ constant; for the full variation, $E_p = -GMm/r$ is required, introduced in the gravitational fields topic later in the course.

Worked Example 3 — Bag onto a Shelf

A $5.0$ kg bag is lifted at constant velocity onto a $1.5$ m high shelf. Find the work done by the lifter and the PE gained.

$W_{\text{lift}} = m\,g\,h = 5.0 \times 9.81 \times 1.5 = 73.6\;\text{J}$

This is the PE gained by the bag-Earth system. If the bag is then released, $73.6$ J of GPE converts back into KE during the fall. By the work-energy theorem and conservation:

$\tfrac{1}{2}mv^2 = mgh \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.5} = 5.42\;\text{m s}^{-1}$

This is the same answer obtained by SUVAT in free fall ( $v^2 = u^2 + 2as$ , $u = 0$ , $a = g$ , $s = h$ ). Energy methods and force methods always agree — but energy methods are usually quicker because intermediate quantities (acceleration, time) are bypassed.

Conservation of Energy

The full principle of conservation of energy states:

Energy cannot be created or destroyed, only transferred from one form to another. The total energy of an isolated system is constant.

This is a law of physics in the strongest sense: it has been confirmed in every experiment ever performed, and any apparent violation has so far been traced to a previously unrecognised form of energy. (The discovery of the neutrino in 1956 was driven by exactly this reasoning: $\beta$ -decay seemed to violate energy conservation until Pauli postulated the missing particle in 1930.)

For mechanical systems where only conservative forces act (gravity, ideal springs, electrostatic), the principle reduces to:

$E_k + E_p = \text{constant}\quad\Leftrightarrow\quad \Delta E_k + \Delta E_p = 0$

In the presence of non-conservative forces (friction, drag, air resistance, inelastic-collision losses), some mechanical energy is dissipated to thermal / acoustic forms. We then write:

$E_{k,i} + E_{p,i} = E_{k,f} + E_{p,f} + W_{\text{dissipated}}$

where $W_{\text{dissipated}}$ is the energy lost to non-conservative work. The total energy (mechanical + thermal + acoustic + ...) is still conserved; it is only the mechanical sub-total that decreases.

flowchart TD
  A[Initial state: mass at height h, at rest] --> B[Gravitational PE only]
  B --> C[Mass released, accelerates under gravity]
  C --> D[GPE converts continuously to KE]
  D --> E{Friction or drag present}
  E -->|No: smooth, vacuum| F[KE at base = initial GPE]
  E -->|Yes: rough or air| G[KE at base = initial GPE minus W_dissipated]
  G --> H[W_dissipated appears as heat in surface or air]
  F --> I[Mechanical energy conserved]
  H --> J[Total energy conserved overall]

Worked Example 4 — Frictionless Slide

A child of mass $30$ kg slides from rest down a smooth slide whose top is $3.0$ m above the bottom. Find the speed at the bottom.

Conservation of mechanical energy: $mgh = \tfrac{1}{2}mv^2$ . Mass cancels:

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 3.0} = \sqrt{58.86} = 7.67\;\text{m s}^{-1}$

This exit speed is independent of the child's mass and independent of the slide's shape — only the vertical drop $h$ matters. A heavy child and a light child both leave a smooth $3$ m slide at $7.67$ m s $^{-1}$ . This shape-independence is the deepest physical content of energy conservation; in classical mechanics, the work done by gravity between two points depends only on the height difference, not the path. (Gravity is a conservative force.)

Worked Example 5 — Slide With Friction

The same slide as Worked Example 4, but friction now does $400$ J of work against the child. Find her exit speed.

Initial PE: $E_p = mgh = 30 \times 9.81 \times 3.0 = 882.9$ J.

Energy balance: $E_{k,f} = E_{p,i} - W_{\text{friction}} = 882.9 - 400 = 482.9$ J.

So $\tfrac{1}{2} \times 30 \times v^2 = 482.9 \Rightarrow v^2 = 32.19 \Rightarrow v = 5.67$ m s $^{-1}$ .

Friction took $400/883 = 45\%$ of the available energy. The exit speed dropped from $7.67$ to $5.67$ m s $^{-1}$ — a $26\%$ decrease, smaller than the energy fraction because KE goes as $v^2$ .

Worked Example 6 — Pendulum

A simple pendulum of length $1.2$ m is pulled sideways so the bob rises $h = 0.20$ m above its lowest point. Find the speed at the lowest point (frictionless).

$mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.20} = 1.98$ m s $^{-1}$ .

Notice we did not need to integrate the tension force around the arc, or solve the differential equation of pendulum motion, or even know the length! Only the vertical drop $h$ matters because tension does no work (always perpendicular to motion of the bob). Energy methods make this a one-line answer where Newton's second law would require calculus.

Worked Example 7 — Block-on-Spring Energy Transfer

A $0.50$ kg block, initially at rest, is pushed horizontally against a spring ( $k = 800$ N m $^{-1}$ ) compressing it by $0.10$ m. The block is released on a smooth surface. Find the block's speed when the spring returns to its natural length.

All elastic PE converts to KE (smooth surface, no friction):

$\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2 \Rightarrow v = x\sqrt{k/m} = 0.10 \times \sqrt{800/0.50} = 0.10 \times 40 = 4.0\;\text{m s}^{-1}$

If the surface beyond the spring is rough with coefficient of friction $\mu = 0.30$ , the block decelerates: friction force is $\mu mg = 0.30 \times 0.50 \times 9.81 = 1.47$ N, and the block stops after $\tfrac{1}{2}mv^2 / (\mu mg) = 4.0 / 2.94 = 1.36$ m.

Synoptic Links

Newton's second law (Module 3.2) — multiplying $F = ma$ by displacement $s$ along the line of action gives $Fs = mas$ , and SUVAT then gives $mas = \tfrac{1}{2}m(v^2 - u^2)$ . So $W_{\text{net}} = \Delta E_k$ is equivalent to Newton II for one-dimensional motion. Energy methods are not "a separate theory" but a different bookkeeping of the same physics.
Power (next lesson, Module 3.3) — the rate at which work is done is power: $P = W/t$ or $P = Fv$ . Same quantity, different time-derivative.
Hooke's law and springs (Module 3.4) — elastic PE $\tfrac{1}{2}kx^2$ is the spring analogue of GPE $mgh$ . The same energy-balance methods apply.
Momentum (Module 4.1) — in collisions both momentum and energy are useful: momentum is always conserved (vector), KE only in elastic collisions (scalar).
Practical work (PAG 6, Newtonian mechanics) — measuring kinetic energy at the foot of a ramp and comparing with initial GPE quantifies the energy "lost" to friction. A trolley-and-ramp setup is standard.

Work, Energy and Conservation

Work, Energy and Conservation

Work Done by a Constant Force

Worked Example 1 — Sledge Pulled by an Inclined Rope

Kinetic Energy

Worked Example 2 — Braking Car

Gravitational Potential Energy

Worked Example 3 — Bag onto a Shelf

Conservation of Energy

Worked Example 4 — Frictionless Slide

Worked Example 5 — Slide With Friction

Worked Example 6 — Pendulum

Worked Example 7 — Block-on-Spring Energy Transfer

Synoptic Links

Specimen question modelled on the OCR H556 paper format

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