Energy and Fields

01

Account for energy stores and transfers

Start by naming the system: the objects you include in the account. Then identify which energy stores change and which transfers cross its boundary.

A moving object has kinetic energy. Interacting objects can have gravitational or electric potential energy, and a deformed material can store elastic potential energy. Chemical and internal stores also matter when explaining batteries, fuels, warming and friction.

Kinetic store

Associated with motion, such as a moving trolley. Its kinetic energy depends on its mass and speed.

Potential-energy stores

Associated with an interaction or configuration: the separation of a load and Earth, the positions of interacting charges, or the deformation of a spring.

Chemical store

A battery or fuel can supply energy through changes in its chemical system.

Internal store

Associated with microscopic motion and interactions within matter. A rise in temperature is one possible sign of an increase in internal energy.

A transfer is a process, not another store. Mechanical work transfers energy when a force acts through a displacement. Electrical transfer can carry energy from a battery to a motor. Radiation can transfer energy to an object and warm it. Heating is transfer associated with a temperature difference. Power describes how quickly energy transfers.

For example, a battery-driven lifting motor draws on the battery's chemical store. Energy transfers electrically to the motor and mechanically to the lifted load. The gravitational store belongs to the load-Earth system, because it depends on their separation.

Use conservation with a stated boundary

This account applies to a fixed collection of matter. If no net energy crosses its boundary, its total energy stays constant. Its kinetic-plus-potential energy need not stay constant: some energy may become internal energy.

Energy dispersed into a motor and its surroundings is still conserved. It is less useful for lifting the load. The intended task determines usefulness: warming water is useful for a heater, while unwanted warming reduces a lifting motor's useful output.

Changing the boundary can change which terms are transfers and which are internal changes. Keep one boundary throughout a calculation, and count each amount once. The power comparison uses this same lift to distinguish an energy total from a rate.

02

Work and kinetic energy

Work is a mechanical transfer of energy. Identify the force, the body it acts on and that body's displacement before calculating it.

For a constant force, work is the force multiplied by displacement in the force's direction. If the angle between force and displacement is θ:

Here F and s are the force and displacement magnitudes. Equivalently, multiply the force component along the displacement by s. The angle is between these two vectors, not automatically the angle to a vertical line or surface normal. Work is a scalar, in J = N m.

• A force component along the displacement does positive work.
• A force component opposite the displacement does negative work.
• A force perpendicular to the displacement does zero work, even when the force is nonzero.

A force holding an object stationary does no mechanical work on that object because its displacement is zero. This does not imply that the person holding it uses no metabolic energy.

If the force component varies with displacement, divide the motion into small intervals. Add force component × displacement for those intervals: the result is the signed area under a force-component/displacement graph. Final force × total displacement is not a general replacement for that area.

A force model identifies all external forces on the chosen body. Work by one named force and net work by all the forces are different quantities.

Derive kinetic energy

Consider a constant-mass particle, or a body translating without relevant rotation or internal changes. It moves in a straight line under a constant resultant force. Choose a positive direction and use signed F_resultant, a and s consistently.

1. Use work and Newton's second law:

   W_net = F_resultants = mas

2. Constant force and constant mass give constant acceleration. From the straight-line motion equations, v² - u² = 2as, so:

   W_net = ½m(v² - u²)

3. This is the change in kinetic energy. Taking kinetic energy to be zero at rest gives:

   E_k = ½mv²
   W_net = E_k,final - E_k,initial

The derivation uses constant acceleration at its middle step. It also explains why the resultant force matters. Substituting a pulling force while ignoring resistance will not give the kinetic-energy change.

Kinetic energy depends on speed squared and has no direction. Reversing velocity while keeping the same speed leaves kinetic energy unchanged, although momentum changes. Its units are kg m² s^-2 = J.

If the system is enlarged to include the crate and floor, the 12 J pulling input gives a 6 J kinetic increase and a 6 J internal increase under the stated friction model. This is the same energy account with a different boundary. Do not subtract friction's 6 J work and then subtract the corresponding 6 J internal increase again.

Estimate the scale: a body of roughly 1 kg moving at about 1 m/s has E_k ≈ ½(1)(1)² = 0.5 J, of order 1 J. This uses rough mass and speed assumptions; it is a magnitude check, not a precise measurement.

03

Field strength, patterns and force

A field describes how a suitable body may experience a force at each position. It belongs to the source arrangement and exists whether or not a test body is currently placed there.

Define field strength at a point

Gravitational field strength, g

The gravitational force per unit mass on a mass placed at that point. Its direction is the force direction, and its unit is N/kg.

Electric field strength, E

The electric force per unit charge on a positive test charge placed at that point. Its direction is the force direction on that positive charge, and its unit is N/C.

Use a sufficiently small test charge so that it does not substantially disturb the source arrangement. A smaller test mass or charge experiences a smaller force in the same field; that alone does not mean the field has become weaker.

When gravity is the only force, F = ma and gravitational force = mg give a = g. This connects g in N/kg with free-fall acceleration in m/s². Determining force per unit mass and timing a falling body's motion are different measurement methods, even when they give the same numerical value for g.

Electric field strength is also expressed in the equivalent unit V/m. Its connection with potential difference across a distance uses a uniform-field model. Keep the present definition, force per unit positive charge, clear. The letter E can mean field strength or energy; its definition and units distinguish them.

Interpret field lines

• Uniform gravitational field: in a small region near Earth's surface, lines are approximately parallel, equally spaced and downward.
• Radial gravitational field: outside a spherically symmetric mass, lines point towards its centre.
• Uniform electric field: between parallel plates, away from their edges, lines point from the positive plate to the negative plate.
• Radial electric fields: lines point outward from a positive point charge and inward towards a negative point charge.

The tangent to a line gives the local field direction. Closer lines indicate a stronger field under a consistent drawing convention. Lines cannot cross where the field has a defined direction, because one point cannot have two different resultant field directions. They are representations, not physical strings.

A positive mass experiences force along g. For a charge, F = qE with the charge's sign included: a positive charge has force along E; a negative charge has force opposite E. A field line is not a compulsory trajectory. The body's velocity may point in a different direction from the force.

04

Field work and potential energy

Potential energy belongs to an interaction or configuration. Determine the field force and displacement before deciding whether that energy increases or decreases.

Gravitational potential energy

Associated with the relative positions of masses, such as a load and Earth. It is not restricted to a height model in every gravitational situation.

Electric potential energy

Associated with the positions of interacting charges. The test charge's sign matters when relating a movement to its potential-energy change.

Elastic potential energy

Associated with deformation of a material, such as stretching or compressing a spring. Its change is related to the force-extension area.

Choose a reference when giving an absolute potential-energy value. Adding the same constant to the initial and final values does not change their difference. A change can therefore be calculated even when the chosen zero is elsewhere.

Work by a static field

For the gravitational and electric fields considered here:

Positive work by the field decreases the associated potential energy. Negative field work increases it. This statement applies to the field's work; it does not say that every force's work equals minus a potential-energy change.

For gravity, the force on a positive mass is along the gravitational field. In an electric field, F = qE: the force on a negative charge is opposite the field arrow. Calculate work using the actual force direction.

Move a mass upward

A 0.500 kg body is moved 0.800 m upward in a uniform gravitational field of 9.81 N/kg downward. Choose upward as positive. Gravitational force is 0.500 × 9.81 = 4.905 N downward, or -4.905 N in this coordinate.

Gravity opposes the upward displacement, so the load-Earth gravitational store increases. Reversing the displacement reverses the two signs: gravity then does positive work and the gravitational potential energy decreases.

Compare positive and negative charges

Choose right as positive. In a uniform electric field E = 2.0 × 10³ N/C right, move a charge 0.050 m right. Recall that 1 microC = 10^-6 C.

Distinguish field work from an agent's work

If an external agent moves the body slowly with unchanged kinetic energy and no other energy transfers, its work is +ΔE_p. Thus steadily lifting the mass requires +3.924 J from the agent, and steadily moving the negative charge right requires +3.0 × 10^-4 J.

If the body also gains kinetic energy, or energy dissipates, the agent's total work need not equal the potential-energy increase alone. Write the complete energy account rather than assigning every work term to one store.

05

Elastic energy from a force-extension graph

The stretching force usually changes as a material extends. Find the work from the area under its force-extension graph, then state when that work is stored as recoverable elastic energy.

Extension x is the increase from the unloaded reference length. For a small extra extension, work is approximately stretching force × extra extension. Adding these small areas gives the loading work.

Use a settled, slow-loading model with negligible kinetic-energy change, reversible deformation and negligible dissipative heating. Under those conditions:

The graph shows the applied stretching-force magnitude. The material's restoring force on the load points the opposite way. Do not change the stored energy's sign because these forces have different recipients and directions.

A linear spring

For a spring obeying Hooke's law, F = kx. Take k = 75 N/m and x = 0.040 m. The final stretching force is 75 × 0.040 = 3.0 N. The line begins at zero extension and zero applied load.

Substituting F = kx into the triangular area gives E_elastic = ½kx². This expression follows from the linear relation through the unloaded origin. It is not a rule that every force-extension curve encloses a triangle.

A force-extension model with changing gradient

The following supplied model joins each pair of neighbouring points with a straight segment. Assume the same slow, reversible, negligible-loss conditions. These are model values rather than recorded experimental observations.

For each segment, the trapezium area is its mean endpoint force multiplied by its extension interval:

1. 0 to 0.020 m: (0 + 1.0)/2 × 0.020 = 0.010 J.
2. 0.020 to 0.040 m: (1.0 + 3.0)/2 × 0.020 = 0.040 J.
3. 0.040 to 0.060 m: (3.0 + 6.0)/2 × 0.020 = 0.090 J.

Total energy stored from zero extension is 0.140 J. The additional energy from 0.020 m to 0.060 m is 0.040 + 0.090 = 0.130 J. Equivalently, subtract the energy already stored at 0.020 m from the final total.

Using the endpoint triangle would give ½(0.060)(6.0) = 0.180 J; using the endpoint rectangle would give 0.060 × 6.0 = 0.360 J. Neither follows the actual graph. The trapezium sum is exact for its stated straight segments. Joining sparse points sampled from a smooth curve would instead give an estimate.

Check the area units. With force in N and extension in m, the numerical area is in J. If the horizontal axis is in cm, convert the length scale before interpreting the area as joules.

What would the measurements establish?

Measure the unloaded reference, then use settled length readings to calculate extension as load increases. Keep the force and extension axes and their units explicit. The spring investigation's force readings and extension graph can then determine loading work from area.

Compare loading and unloading behaviour before claiming that all that work is recoverable. In a material that dissipates energy or remains permanently deformed, not all loading work is recovered as useful mechanical output. Reaching the same final extension is not enough evidence of an ideal spring model.

If the elastic energy is later transferred to a body's kinetic energy, an equality between the two requires negligible losses and no other changing stores. The presence of a spring alone does not establish those conditions.

06

Power, losses and efficiency

Energy measures how much transfers. Power measures the rate of transfer. Efficiency compares the useful output with the total input for a stated task.

Average and instantaneous power

The unit is W = J/s. Instantaneous power is the rate at a particular time. Always identify the transfer being described: total input, useful output or the work of a named force.

Over a short interval, a force's work is Fs cos θ. Divide by the interval and use displacement per unit time in the limit of a short interval:

Here θ is the angle between force and velocity. Mechanical power is the force multiplied by the velocity component in its direction. P = Fv is the aligned special case. An opposing force has negative power; a perpendicular force has zero power.

For the crate in the angled-pull example, consider an instant when its speed is 2.0 m/s right. The 8.0 N pull at 60 degrees supplies 8.0 × 2.0 × cos 60° = 8.0 W. The 2.0 N resistance contributes -4.0 W, giving net mechanical power +4.0 W, the rate of kinetic-energy increase. The perpendicular support contributes zero.

Compare the useful output and total input

Efficiency is unitless and is often expressed as a percentage. For powers averaged over the same interval, the same ratio can be calculated as useful output power / total input power.

If a second supplied run transfers the same 320 J input and 200 J useful output in 2.00 s, its input power is 160 W and useful output power 100 W. Both powers double; efficiency remains 62.5%. The equal energies are supplied conditions, not a general claim that changing a real motor's speed leaves efficiency unchanged.

The 120 J internal increase does not lift the load, so the motor needs more than 200 J input. Reducing unwanted mechanical friction can reduce that loss in an otherwise comparable lifting task. All energy is still accounted for; the loss is a reduction in useful output for this task.

A quoted efficiency alone does not give an energy amount: the input is also needed. For a complete account with the stated useful output and total input, efficiency cannot exceed 100%.

Kilowatt-hours measure energy

A kilowatt-hour is the energy transferred at 1 kW for 1 hour. Its symbol is kWh; multiplying power by time gives energy.

For a supplied constant power of 2.0 kW over 3.0 minutes:

The 2.0 kW value is a rate. The 0.10 kWh value is the amount transferred during the stated interval.

Investigate a lift with matching measurements

To determine a lifting device's work, power and efficiency, measure quantities for the same defined interval:

1. Load force: use a suitable force reading, or measure mass and multiply by the stated local g. Check zero and choose a range that covers the load.
2. Vertical rise: measure the difference between start and finish heights with a suitable rule or tape. A sloping string length is not the vertical rise. View a pointer perpendicular to the scale to avoid parallax, and retain the same reference.
3. Elapsed time: identify the same start and finish events used for the displacement and energy readings. Use a steady interval, or account for kinetic-energy changes if the speed changes.
4. Input energy: use the change in a suitable calibrated energy reading over that interval. A motor's rated power is not a measurement of the energy taken in during a particular run.
5. Analysis: calculate useful work from force and vertical rise, then divide each measured energy by the same time. Use useful energy / input energy for efficiency.

Choose ranges and resolutions that suit the force, length and time scales. Record raw values with units, repeat comparable runs to assess scatter, and retain guard digits in calculations. An underestimated rise makes the calculated useful work and efficiency too low if the other readings are correct. Repeating the same incorrect height reference will not remove that error.

Load force × speed gives useful mechanical output power here, not electrical input power. An input-energy reading supplies the latter account without assuming that every input joule lifts the load. Friction and warming may explain a real energy difference; they are not automatically faults in the instruments.

Estimate before calculating: a roughly 1 kg load has weight about 10 N near Earth. Raising it about 1 m in about 1 s at steady speed takes roughly 10 J and needs useful power of order 10 W. An inefficient device needs greater input power. The mass, rise, time and local field are the stated assumptions behind this estimate.