Work and energy: the Year 3 test

In Year 3 physics goes from two to three hours a week and changes nature: it stops being descriptive and becomes quantitative. The test on work and energy, which almost always falls between November and mid-December, is the point where this leap turns into a grade — and it is, together with the one on vectors in Year 1, the physics test in which I see the most unexpected failing grades.

The reason is not the difficulty of the formulas, of which there are four. It is that this chapter, for the first time, asks for a choice: faced with a problem you have to decide whether to use dynamics, conservation of energy or momentum. Anyone who has studied only the formulas tries them all until one gives a number; anyone who has understood the criterion takes ten seconds.

Here you will find the criterion, the formulas you need, three worked exercises with the calculations in full and the five mistakes that cost the most marks.

The idea: energy is a shortcut, not one more formula

Everything that is solved with energy could be solved with $F = ma$ and kinematics. The point is that it often does not pay, and sometimes it cannot be done: if the force changes along the path, or the path is curved, dynamics becomes impractical at Year 3 level.

Energy sidesteps the problem because it ignores the route and looks only at the endpoints: where you start, where you arrive, and how much energy was exchanged in between. A body sliding down from a height $h$ along a curved slide reaches the bottom with the same speed it would have sliding down a smooth inclined plane of the same height, and with energy you see this in one line.

That is why the chapter is called “conservation”: it is not a new formula, it is a way of looking at the same problem that skips all the intermediate steps.

Work, and the sign everyone gets wrong

L = F \cdot s = F s cos θ

where $θ$ is the angle between the force and the displacement. The cosine is the whole physics of this formula:

Angle between $F$ and $s$	$cos θ$	Work	Typical example
$0°$	$+ 1$	driving, positive	weight going downhill
between $0°$ and $90°$	between 0 and 1	positive, reduced	towing with an inclined rope
$90°$	$0$	zero	normal reaction, centripetal force
$180°$	$- 1$	resisting, negative	friction, weight going uphill

The line to memorise is the third: a force perpendicular to the displacement does no work. The normal reaction of the plane never does work, nor does the centripetal force in uniform circular motion, nor does the tension of a string in a pendulum. Every year I see tests where they are added into the energy balance: they are marks given away.

The unit is the joule, $1 J = 1 N \cdot m$ . Power is work over time:

P = \frac{L}{Δ t} = F v

The work-energy theorem

L_{tot} = Δ K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}

The total work — of all forces, friction included — equals the change in kinetic energy. It is the most general theorem in the chapter: it always holds, with or without friction, with constant or variable forces.

Two useful consequences in a test: if the speed does not change, the total work is zero (not that there are no forces, but that their works cancel); and if the body starts from rest, $Δ K$ is simply $\frac{1}{2} m v_{f}^{2}$ , with no subtraction.

Conservation, with and without friction

Conservative forces (weight, elastic force) have an associated potential energy:

U_{g} = m g h, U_{e} = \frac{1}{2} k x^{2}

If only conservative forces act, the mechanical energy $E = K + U$ is conserved:

K_{i} + U_{i} = K_{f} + U_{f}

If there is friction — that is, almost always, in December problems — mechanical energy is not conserved, and the difference is exactly the work of friction, which is negative:

E_{f} - E_{i} = L_{friction} = - μ_{d} N d

On a horizontal plane $N = m g$ ; on an inclined plane of angle $θ$ it is $N = m g cos θ$ . The $cos θ$ is the second source of mistakes in the test after the sign of the work.

A note on method: you can write the balance in two equivalent ways — “initial energy minus dissipated energy equals final energy”, or “change in mechanical energy equals the work of friction”. Choose one and always use that one. Alternating between them is the quickest way to get a sign wrong.

Momentum and collisions

p = m v, J = F Δ t = Δ p

The rule that solves collision exercises fits on one line: in every collision the total momentum of the system is conserved; kinetic energy is not, except in elastic collisions.

Type of collision	Is $p$ conserved?	Is $K$ conserved?	What happens
Elastic	yes	yes	the bodies bounce off
Inelastic	yes	no	part of $K$ becomes heat and deformation
Perfectly inelastic	yes	no (maximum loss)	the bodies stick together

The perfectly inelastic case is the one that comes up almost every time, because it is solved in two lines: after the collision the bodies have the same velocity.

m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v_{f}

Energy or momentum? The criterion

This is the question that decides the test, and the answer is short:

there is a collision or an explosion → start from momentum, always;
there is a climb, a descent, a spring, a slide → start from energy;
you are asked for a force, an acceleration or a time → you need dynamics, $F = ma$ ;
in inelastic collisions, the kinetic energy lost is a legitimate question, but it is calculated after finding the final velocity with momentum. It is never the starting point.

The most reliable linguistic signal in the text of a problem is the word “collides”, “hits”, “embeds itself”, “sticks”: from there you always start from $p$ .

Three worked exercises

1. Inclined plane with friction

A body of mass $m = 2.0$ kg starts from rest and slides down an inclined plane of $30°$ for a length $d = 4.0$ m. The coefficient of kinetic friction is $μ_{d} = 0.20$ . What speed does it reach at the bottom? (Use $g = 9.8 m/ s^{2}$ .)

First thing, the height — not the length of the plane:

h = d sin 30° = 4.0 \cdot 0.50 = 2.0 m

Initial potential energy:

U_{i} = m g h = 2.0 \cdot 9.8 \cdot 2.0 = 39.2 J

Energy dissipated by friction, with $N = m g cos 30°$ :

∣ L_{friction} ∣ = μ_{d} m g cos 30° \cdot d = 0.20 \cdot 2.0 \cdot 9.8 \cdot 0.866 \cdot 4.0 \approx 13.6 J

The energy balance gives the final kinetic energy:

K_{f} = U_{i} - ∣ L_{friction} ∣ = 39.2 - 13.6 = 25.6 J

And finally the speed:

v_{f} = \frac{2 K _{f}}{m} = \frac{2 \cdot 25.6}{2.0} \approx 5.1 m/s

Sense check: without friction it would have been $2 g h = 6.3$ m/s. With friction it must come out less, and it does. If you had got more, the mistake would have been a sign.

2. Perfectly inelastic collision

A trolley of $3.0$ kg travelling at $4.0$ m/s hits a stationary trolley of $1.0$ kg; after the collision they stay hooked together. Find the final speed and the kinetic energy lost.

There is a collision: start from momentum.

p_{i} = 3.0 \cdot 4.0 + 1.0 \cdot 0 = 12 kg \cdot m/s

v_{f} = \frac{p _{i}}{m _{1} + m _{2}} = \frac{12}{4.0} = 3.0 m/s

Now, and only now, the energy:

K_{i} = \frac{1}{2} \cdot 3.0 \cdot 4. 0^{2} = 24 J, K_{f} = \frac{1}{2} \cdot 4.0 \cdot 3. 0^{2} = 18 J

$6$ J were lost, that is 25% of the initial energy, turned into heat and deformation. It is not a calculation error: it is what “inelastic” means.

3. Spring and maximum height

A spring of constant $k = 200$ N/m, compressed by $x = 0.10$ m, launches a block of $m = 0.50$ kg onto a smooth horizontal plane, which then continues into a smooth climb. Find the speed on the plane and the maximum height reached.

Stored elastic energy:

U_{e} = \frac{1}{2} k x^{2} = \frac{1}{2} \cdot 200 \cdot (0.10)^{2} = 1.0 J

Everything is smooth, so it is conserved: it becomes kinetic energy on the plane.

v = \frac{2 U _{e}}{m} = \frac{2 \cdot 1.0}{0.50} = 2.0 m/s

On the climb the same energy becomes gravitational potential energy:

h = \frac{U _{e}}{m g} = \frac{1.0}{0.50 \cdot 9.8} \approx 0.20 m

Note that the mass was not needed for the height if you reason in terms of speed ( $h = v^{2} /2 g$ ), and that the shape of the climb is irrelevant: it is exactly the advantage of the energy approach.

The five mistakes that cost the most marks

1. Using the length of the plane instead of the height. In $U = m g h$ , $h$ is the vertical height: on an inclined plane it is $d sin θ$ , not $d$ . It is by far the most frequent mistake in this test.

2. Conserving mechanical energy when there is friction. If the text gives a coefficient of friction, mechanical energy is not conserved. The coefficient is in the text precisely to check that you know this.

3. Getting the sign of the work wrong. Friction always does negative work; weight does positive work downhill and negative uphill. If an energy comes out negative, the sign is almost always the culprit.

4. Putting forces that do no work into the balance. Normal reaction, tension in the string of a pendulum, centripetal force: perpendicular to the displacement, zero work.

5. Applying conservation of kinetic energy in an inelastic collision. $p$ is conserved, not $K$ . It is the mistake that makes the problem impossible, because it leads to a system with no solution.

Then there is a sixth mistake that is not about physics but weighs a lot: units. Mass in kg, distances in m, speeds in m/s. There is almost always a problem with grams or km/h to convert, and it is put there on purpose.

The two weeks before the test

Every day, for a week, write on a blank sheet the four formulas ( $L$ , $K$ , $U_{g}$ , $U_{e}$ ) and the work-energy theorem. Two minutes.
Take ten problems and write down only which principle you would use, without solving them: energy, momentum or dynamics. It is the exercise that pays off most of all for this test.
Do five inclined-plane-with-friction problems in a row, which is the most likely type, and five perfectly inelastic collisions.
Check every result with common sense: the speed with friction must be less than the one without, the kinetic energy after an inelastic collision less than before. A plausibility check catches more mistakes than rereading.

The set-up mistakes that recur throughout high-school physics — units, signs, reading the text — I have collected in typical physics mistakes in high school.

The December test comes after the November one on vectors and kinematics: if that is not yet solid, start from the first physics test in Year 3. For energy as seen in other courses there are work and energy in Year 4 at the classical high school and work, power and energy at the art high school, while for collisions with more exercises there is momentum and collisions. The energy that friction dissipates becomes heat: the sequel is specific heat and thermal equilibrium.

How we work on this in lessons

On this chapter my first lesson contains almost no calculations. We take ten problem texts and for each we decide, in ten seconds, which principle to start from and why. Only afterwards do we solve. Anyone who arrives at the test knowing how to choose the principle finishes the hour with all the exercises done; anyone who tries them all hands in half, even knowing the formulas by heart.

The second thing I insist on is a diagram with the forces and heights marked before writing any equation: where you put the zero of potential energy is your choice, but it must be written down, otherwise the signs become random. Lessons are online and one-to-one: rates and packages are on the pricing page, and the 10-hour package is worth it if in Year 3 the jump in difficulty has left gaps on vectors and dynamics too.

The quickest way to see where you stand is to try the simulated test on work and energy: fifteen questions in fifty minutes, with a breakdown of weak points by topic at the end. The correct answer to every question is free and needs no registration; the guided step-by-step solution is unlocked with the first lesson. If after the test you see that energy holds but collisions do not (or vice versa), write to me for a free introductory call: we look at the syllabus of your test and I tell you honestly how many lessons you need. The full picture of how I set up the path is on the page for physics tutoring.

Work, energy and collisions: the Year 3 December test