Mastering Energy Stores and Transfers: Your Exam Success Guide

What the examiner is testing

The examiner is assessing your ability to identify and quantify energy changes within a system, often requiring you to apply the principle of conservation of energy. Marks are typically awarded for correctly naming energy stores, identifying transfer pathways, and performing calculations involving specific energy equations.

The method

1. Identify the system: Clearly define what is being studied (e.g., a falling ball, a heating kettle, a car braking).
2. Name the initial energy store(s): Before any change occurs, what type of energy does the system possess? Common stores include kinetic (movement), gravitational potential (height), elastic potential (stretch/compression), chemical (bonds), thermal (temperature), and nuclear.
3. Name the final energy store(s): After the change, what type of energy does the system possess?
4. Identify the energy transfer pathway(s): How did energy move from the initial store(s) to the final store(s)? Common pathways are mechanical work (force over a distance), electrical work (current flow), heating (temperature difference), and radiation (waves).
5. Apply the conservation of energy: State that energy cannot be created or destroyed, only transferred from one store to another or from one system to another. This means the total energy at the start equals the total energy at the end.
6. Select and apply relevant equations: Based on the identified stores and transfers, choose the appropriate formula (e.g., \(E_k = \frac{1}{2}mv^2\), \(E_p = mgh\), \(W = Fd\), \(Q = mc\Delta\theta\)).
7. Substitute values and calculate: Plug in the given numbers, ensuring all units are consistent (e.g., mass in kg, height in m, velocity in m/s).
8. State the final answer with correct units: Include the appropriate unit (e.g., Joules for energy, Watts for power).

Worked example

A 2.0 kg object is lifted vertically by 3.0 m. Calculate the gravitational potential energy gained by the object. (Take \(g = 9.8 \text{ N/kg}\)).

$$ \begin{aligned} \text{Mass } (m) &= 2.0 \text{ kg} \\ \text{Height } (h) &= 3.0 \text{ m} \\ \text{Gravitational field strength } (g) &= 9.8 \text{ N/kg} \\ \text{Gravitational potential energy } (E_p) &= mgh \\ E_p &= (2.0 \text{ kg}) \times (9.8 \text{ N/kg}) \times (3.0 \text{ m}) \\ E_p &= 58.8 \text{ J} \end{aligned} $$
Sanity check: Lifting an object increases its potential energy. A 2 kg object lifted 3 m gains a reasonable amount of energy, and the value of 58.8 J is consistent with the magnitudes of the input values.

Worked example: a harder one

A car with a mass of 1200 kg is travelling at 20 m/s. The brakes are applied, and the car comes to a stop over a distance of 50 m. Calculate the average braking force exerted on the car.

Initial thought process: The car has kinetic energy. When it stops, this kinetic energy is transferred to thermal energy in the brakes and surroundings due to the work done by friction. We need to find the kinetic energy and then relate it to the work done.

Why the obvious first move fails (or leads to a longer path): You might immediately think of \(F = ma\). However, we don't know the acceleration directly. We have initial velocity, final velocity (0 m/s), and distance. We could calculate acceleration first using SUVAT equations, then use \(F=ma\). This is a valid two-step approach but there's a more direct energy method.

Energy method:
The initial kinetic energy of the car is transferred as work done by the braking force.
$$ \begin{aligned} \text{Mass } (m) &= 1200 \text{ kg} \\ \text{Initial velocity } (v) &= 20 \text{ m/s} \\ \text{Final velocity } (u) &= 0 \text{ m/s} \\ \text{Distance } (d) &= 50 \text{ m} \\ \\ \text{Initial kinetic energy } (E_k) &= \frac{1}{2}mv^2 \\ E_k &= \frac{1}{2} \times (1200 \text{ kg}) \times (20 \text{ m/s})^2 \\ E_k &= \frac{1}{2} \times 1200 \text{ kg} \times 400 \text{ m}^2\text{/s}^2 \\ E_k &= 240000 \text{ J} \\ \\ \text{Work done by braking force } (W) &= \text{Force } (F) \times \text{distance } (d) \\ \text{By conservation of energy: } \text{Work done} &= \text{change in kinetic energy} \\ W &= E_k \\ F \times d &= E_k \\ F \times (50 \text{ m}) &= 240000 \text{ J} \\ F &= \frac{240000 \text{ J}}{50 \text{ m}} \\ F &= 4800 \text{ N} \end{aligned} $$
The average braking force is 4800 N.

Practice

1. A kettle contains 1.5 kg of water. How much thermal energy is needed to raise its temperature by \(60^\circ\text{C}\)? (Specific heat capacity of water = \(4200 \text{ J/kg}^\circ\text{C}\)).
2. A student pulls a block with a force of 15 N over a distance of 4.0 m. How much work is done by the student?
3. A 0.5 kg ball is dropped from a height of 10 m. Assuming no air resistance, what is its kinetic energy just before it hits the ground? (Take \(g = 9.8 \text{ N/kg}\)).
4. A cyclist and their bike have a combined mass of 80 kg. They are travelling at 10 m/s. They then apply the brakes and slow down to 2 m/s over a distance of 20 m. Calculate the average braking force.

Answers:
1. \(Q = mc\Delta\theta = (1.5 \text{ kg}) \times (4200 \text{ J/kg}^\circ\text{C}) \times (60^\circ\text{C}) = 378000 \text{ J}\)
2. \(W = Fd = (15 \text{ N}) \times (4.0 \text{ m}) = 60 \text{ J}\)
3. \(E_p = mgh = (0.5 \text{ kg}) \times (9.8 \text{ N/kg}) \times (10 \text{ m}) = 49 \text{ J}\). By conservation of energy, \(E_k = E_p = 49 \text{ J}\).
4. $$ \begin{aligned} \text{Initial kinetic energy } (E_{k, \text{initial}}) &= \frac{1}{2}mv_{\text{initial}}^2 \\ E_{k, \text{initial}} &= \frac{1}{2} \times (80 \text{ kg}) \times (10 \text{ m/s})^2 \\ E_{k, \text{initial}} &= \frac{1}{2} \times 80 \text{ kg} \times 100 \text{ m}^2\text{/s}^2 \\ E_{k, \text{initial}} &= 4000 \text{ J} \\ \\ \text{Final kinetic energy } (E_{k, \text{final}}) &= \frac{1}{2}mv_{\text{final}}^2 \\ E_{k, \text{final}} &= \frac{1}{2} \times (80 \text{ kg}) \times (2 \text{ m/s})^2 \\ E_{k, \text{final}} &= \frac{1}{2} \times 80 \text{ kg} \times 4 \text{ m}^2\text{/s}^2 \\ E_{k, \text{final}} &= 160 \text{ J} \\ \\ \text{Work done by braking force } (W) &= \text{Change in kinetic energy} \\ W &= E_{k, \text{initial}} - E_{k, \text{final}} \\ W &= 4000 \text{ J} - 160 \text{ J} \\ W &= 3840 \text{ J} \\ \\ \text{Also, } W &= Fd \\ F \times (20 \text{ m}) &= 3840 \text{ J} \\ F &= \frac{3840 \text{ J}}{20 \text{ m}} \\ F &= 192 \text{ N} \end{aligned} $$

The three mistakes that lose marks

1. Confusing energy stores with energy transfers: Students sometimes say "heat energy" when they mean "thermal energy store" or "heating by radiation" when they mean "radiation". Wrong answer example: "The car lost heat energy." (Should be "The car's kinetic energy was transferred by heating to the surroundings, increasing their thermal energy store.")
2. Incorrectly applying the conservation of energy: Assuming that all initial energy is converted into a useful final energy store, ignoring dissipation. Wrong answer example: "When a light bulb is switched on, all the electrical energy becomes light energy." (Ignores the significant transfer to thermal energy in the filament and surroundings).
3. Using inconsistent units in calculations: Mixing grams with kilograms, or cm with meters, leading to incorrect numerical answers. Wrong answer example: Calculating \(E_p\) with mass in grams and height in meters, getting an answer 1000 times too large or too small.

30-second recap

Energy is stored in various forms (kinetic, potential, thermal, chemical, etc.) and can be transferred between these stores or between systems by work (mechanical, electrical), heating, or radiation. The total energy in a closed system remains constant, meaning energy is conserved. Calculations involve identifying stores, transfers, and applying specific formulae with correct units.