A. Space, time and motion

Choose a reference

Position $\vec r$ specifies where an object is relative to a chosen origin. A position is not meaningful without a reference frame and coordinate direction.

Track change in position

Velocity $\vec v$ describes how position changes with time. It is a vector: its direction is the direction of motion at that instant, and on a curved path the velocity arrow is tangent to the path.

Track change in velocity

Acceleration $\vec a$ describes how velocity changes with time. A change in speed, direction, or both is acceleration; an object can accelerate even while its instantaneous speed is zero.

Common trap

Do not use “velocity” as a synonym for speed. Speed gives only magnitude; velocity also requires direction relative to the chosen coordinate system.

Velocity is a rate

Velocity is the rate of change of position:

$$\vec v=\frac{d\vec r}{dt}$$

Over a finite interval, average velocity is displacement divided by elapsed time.

Acceleration changes velocity

Acceleration is the rate of change of velocity:

$$\vec a=\frac{d\vec v}{dt}$$

A constant acceleration gives equal changes in velocity during equal time intervals.

Read the gradient

On a position–time graph, the gradient represents velocity. On a velocity–time graph, the gradient represents acceleration. The graph’s slope, not its height alone, carries the rate-of-change meaning.

Common trap

Distance divided by time gives average speed, not instantaneous velocity. Likewise, a large velocity does not imply a large acceleration unless the velocity is changing rapidly.

Displacement is a vector

Displacement is the change in position:

$$\Delta\vec r=\vec r_{final}-\vec r_{initial}$$

It has a magnitude and a direction from the initial position to the final position.

Ignore the route for displacement

The path taken between the two positions does not determine displacement. A curved or complicated journey can still have a straight-line displacement between its endpoints.

Use components when needed

For perpendicular changes, resolve displacement into components and combine them vectorially. A signed one-dimensional displacement is positive or negative according to the chosen axis.

Common trap

A return to the starting point gives zero displacement even though the distance travelled is non-zero.

Distance

Distance is the total path length travelled. It is a scalar, so it has magnitude only and cannot be negative.

Displacement

Displacement is the vector change in position from start to finish. Its magnitude is the shortest endpoint-to-endpoint separation, not generally the length of the route.

Match the average quantity

Average speed uses total distance divided by total time. Average velocity uses displacement divided by total time. A route with turns can therefore have average speed greater than the magnitude of average velocity.

Common trap

For a complete oscillation, the displacement is zero but the distance is four times the amplitude. Choose the quantity named in the question before substituting.

Average values use an interval

Average speed is total distance divided by total time. Average velocity is displacement divided by elapsed time. Average acceleration is change in velocity divided by elapsed time.

Instantaneous values use one moment

Instantaneous velocity is the tangent gradient on a position–time graph; instantaneous speed is its magnitude and is what an ideal speedometer reports. Instantaneous acceleration is the tangent gradient on a velocity–time graph. Average values instead use a finite interval.

Connect graph quantities

The gradient of a velocity–time graph is acceleration, and the area under it is displacement. A constant acceleration therefore produces a straight-line velocity–time graph.

Common trap

Do not use the average gradient when a question asks for an instantaneous value. Use the tangent at the specified time.

Uniform-acceleration model

SUVAT equations apply when acceleration is constant along the chosen one-dimensional axis:

$$v=u+at,\quad s=ut+\frac12at^2,\quad v^2=u^2+2as,\quad s=\frac{u+v}{2}t$$

Choose an equation

List the known and unknown quantities $s,u,v,a,t$. Select an equation containing the required unknown and only known quantities; keep signs consistent with the positive direction.

Check the model

A constant acceleration means equal changes in velocity in equal time intervals. Use separate horizontal and vertical equations only when the motion has been resolved into independent components.

Worked example from local Question Bank row 22687

A glider accelerates uniformly from rest to $27.0\,\mathrm{m\,s^{-1}}$ in $11.0\,\mathrm{s}$. Use $s=\frac{u+v}{2}t$:

$$s=\frac{0+27.0}{2}\times 11.0=148.5\,\mathrm{m}\approx149\,\mathrm{m}$$

The result is the launch-run displacement; the constant-acceleration assumption is essential.

Common trap

Do not use SUVAT when acceleration varies significantly with time or position. A formula can produce a neat number while still violating the model’s constant-acceleration assumption.

Uniform acceleration

Acceleration is uniform when the velocity changes by equal amounts in equal time intervals. The velocity–time graph is a straight line with constant gradient.

Non-uniform acceleration

Acceleration is non-uniform when its magnitude or direction changes. The velocity–time graph then has a changing gradient, and a single SUVAT value cannot describe the entire interval.

Model versus reality

A constant-acceleration model can be useful over a limited interval even when real forces vary. State the approximation and identify the neglected force or changing condition.

Common trap

A curved trajectory does not by itself prove that acceleration is non-uniform: projectile motion without drag has constant downward acceleration while its velocity direction changes.

Separate the axes

With negligible fluid resistance, projectile motion is independent horizontal and vertical motion. Resolve the launch velocity:

$$u_x=u\cos\theta,\qquad u_y=u\sin\theta$$

Horizontal motion

There is no horizontal acceleration in the ideal model, so $v_x=u_x$ and $x=u_xt$. Use the horizontal displacement to find time or horizontal speed.

Vertical motion

Use one-dimensional constant-acceleration equations vertically, usually with $a_y=-g$ if upward is positive. The horizontal and vertical equations share the same time $t$.

Worked example from local Question Bank row 31723

A tennis ball travels $11.9\,\mathrm{m}$ horizontally after launch at $64.0\,\mathrm{m\,s^{-1}}$ and $7^\circ$ to the horizontal.

$$u_x=64.0\cos7^\circ=63.52\,\mathrm{m\,s^{-1}}$$
$$t=\frac{x}{u_x}=\frac{11.9}{63.52}=0.187\,\mathrm{s}$$

The same $0.187\,\mathrm{s}$ must then be used in the vertical equation.

Common trap

Do not use the launch speed as the horizontal speed. Resolve it first, and do not use the horizontal time independently of the vertical motion.

Drag opposes instantaneous velocity

Fluid resistance acts opposite the projectile's velocity and usually grows with speed. Its direction changes through the flight, so the resultant acceleration is not the constant downward $g$ of the ideal model.

Use the force direction

Before the peak, drag has horizontal and downward components; after the peak, it has horizontal and upward components. Therefore acceleration is not determined by velocity alone and changes continuously.

Terminal-speed condition

For vertical descent, terminal speed is reached when upward drag (and any buoyancy included in the model) balances weight. The object then continues at constant downward velocity.

Common trap

Zero acceleration at terminal speed does not mean zero velocity. At the top of a projectile path, vertical velocity may be zero while acceleration remains non-zero.

Three linked laws

1. If the resultant force is zero, velocity is constant.
2. A resultant force changes momentum; for constant mass, $\vec F_{net}=m\vec a$.
3. Forces between two bodies are equal in magnitude and opposite in direction, acting on different bodies.

Choose the system

Draw forces acting on the chosen object, then use the resultant force to predict its acceleration. For action–reaction pairs, identify the two different bodies before applying the third law.

Use interactions to explain motion

A rocket pushes gas backward; the gas exerts an equal and opposite force on the rocket. The rocket can therefore accelerate even in the absence of a supporting surface.

Common trap

The forces in a third-law pair do not cancel in one free-body diagram because they act on different objects.

A force needs an interaction

A force is an interaction between bodies. One body exerts the force and another body experiences it. Contact, gravitational, electric and magnetic interactions can all change momentum.

Name both bodies

When explaining a force, state the interacting pair and the direction of the force on the chosen body. The reaction force acts on the other body, not back on the same free-body diagram.

Fields can mediate interaction

Bodies do not need to touch for gravitational, electric or magnetic forces. For example, current-carrying coils interact through their magnetic fields, producing attraction or repulsion depending on the field arrangement.

Common trap

Do not describe a force as a property that an isolated object “has” without naming the other body or field involved.

Isolate one body

A free-body diagram shows only the chosen body and the external forces acting on it. Replace the body with a point or simple shape and choose useful axes.

Draw actual forces

Use arrows from the body, label each interaction and draw the direction physically. Typical labels include weight $mg$, normal force $N$, tension $T$, friction and drag.

Resolve only when needed

If a force is angled, resolve it into the chosen axes. Then apply $\sum F_x=ma_x$ and $\sum F_y=ma_y$ to the same body.

Common trap

Do not draw velocity, acceleration or a force exerted by the chosen body on its surroundings as forces acting on the chosen body.

Add force components

The resultant force is the vector sum of all forces on the chosen body:

$$\vec F_{net}=\sum\vec F$$

Resolve angled forces into perpendicular components before adding.

Connect to acceleration

For constant mass, apply Newton’s second law along each axis:

$$\sum F_x=ma_x,\qquad \sum F_y=ma_y$$

Use equilibrium correctly

If the resultant force is zero, acceleration is zero, but the object may still have constant non-zero velocity. A balanced vertical component does not imply every force is absent.

Common trap

Do not add force magnitudes without their directions. A component that balances another contributes zero only along the same axis.

Contact-force family

Contact forces arise when bodies or a body and fluid interact: normal force, friction, tension, elastic restoring force, viscous drag and buoyancy.

Use the interaction geometry

Normal force is perpendicular to the surface; friction acts along the surface opposing relative motion or attempted motion; tension acts along a taut string; drag opposes motion through a fluid; buoyancy acts upward due to fluid pressure differences.

Check the condition

Friction can be static or kinetic, drag depends on speed and shape, and buoyancy depends on displaced fluid. The magnitudes are determined by the interaction and constraints, not by a memorized universal value.

Common trap

Do not include every possible contact force. Include only interactions actually present in the described situation.

Normal means perpendicular

The normal force $N$ is the contact force exerted by a surface perpendicular to that surface. Its direction follows the local surface normal, not necessarily the vertical direction.

Find it from the force balance

Use the component of Newton’s second law perpendicular to the surface. In a curved path, the normal force may combine with a component of weight to provide the required centripetal resultant.

Do not assume $N=mg$

$N=mg$ applies only in situations where the perpendicular acceleration and other perpendicular force components make that balance valid. Inclines, lifts, loops and vertical acceleration change the normal force.

Friction follows the contact

Friction acts parallel to the contact surface and opposes relative motion or the tendency of surfaces to move relative to each other.

Static friction adapts

Before slipping, static friction has whatever value is needed up to a maximum:

$$F_f\leq\mu_sN$$

It is not automatically equal to $\mu_sN$; that value occurs at impending motion.

Dynamic friction during sliding

Once surfaces slide, the model gives

$$F_f=\mu_dN$$

Use the normal force for the actual contact and combine friction with the other forces along the surface.

Common trap

Do not use the dynamic coefficient before motion begins, or assume static friction is always at its maximum.

Tension is a pull

Tension is the force exerted by a taut string, cable or rope on an attached body. It acts along the string and pulls away from the body.

Use the ideal-string model carefully

For a light, inextensible string over a frictionless pulley, tension has the same magnitude throughout. If the string, pulley or contact is non-ideal, tension can vary and must be found from each body’s force balance.

Connect tension to motion

Draw tension in the string direction, then use $\sum F=ma$. A body can have non-zero tension while at rest if other forces balance it.

Common trap

A string can pull but not push. Do not draw tension toward the string’s far end through the body or assume its value equals the weight without a force balance.

Restoring force

For an ideal elastic element within its proportional range,

$$\vec F_H=-k\vec x$$

The minus sign means the force acts opposite the displacement from equilibrium.

Use extension correctly

For a spring, $x$ is extension or compression relative to its natural length. In a vertical equilibrium, the spring tension can balance weight, but the extension is not the total spring length.

Respect the model boundary

Hooke’s law is a linear approximation. Beyond the limit of proportionality, the force–extension graph is no longer linear and the same $k$ cannot be used.

Stokes drag

For a small sphere moving slowly through a viscous fluid,

$$F_d=6\pi\eta r v$$

where $\eta$ is viscosity, $r$ is sphere radius and $v$ is speed relative to the fluid.

Drag opposes motion

The drag force points opposite the sphere’s velocity. As speed increases, drag increases linearly in this model, reducing the resultant force when the driving force is fixed.

Approach to terminal speed

For a falling sphere, weight drives the motion and viscous drag grows with speed. When drag balances the effective weight, acceleration becomes zero and terminal speed is reached.

Common trap

Do not treat viscosity $\eta$ as the same quantity as drag force, and do not forget that the formula applies to the stated small-sphere, viscous-flow model.

Buoyancy from pressure difference

A fluid exerts a net upward buoyant force on an immersed object because pressure is greater at greater depth. In the IB model,

$$F_b=\rho_fV_{disp}g$$

where $V_{disp}$ is the displaced fluid volume.

Separate buoyancy from net force

The buoyant force is one force in the free-body diagram. The net force is found after combining it with weight, tension, drag or other forces.

Floating condition

For an object at rest on the fluid, buoyancy balances its weight. This gives a useful density or submerged-volume relationship, but only after the equilibrium assumption is stated.

Common trap

Use the density of the displaced fluid and the displaced volume, not automatically the object’s total volume or density.

Three field interactions

Gravitational, electric and magnetic forces are field forces: bodies can interact without contact. Identify the source of the field, the object acted on and the force direction.

Keep the mechanisms distinct

Gravity acts on mass, electric force acts on charge, and magnetic force acts on moving charges or currents in a magnetic field. Their equations and direction rules are not interchangeable.

Use the force relevant to the system

A free-body diagram may contain more than one field force. Add them as vectors and apply Newton’s second law to the selected body.

Common trap

Do not call every non-contact force “electromagnetic”; gravitational attraction is a separate interaction.

Weight is a force

Weight is the gravitational force on a mass:

$$F_g=mg$$

The direction is toward the local gravitational field source.

Use local $g$

The value of $g$ depends on location. Use the value stated or the local field strength appropriate to the body’s position; mass does not change when the object is moved.

Common trap

Mass is measured in kilograms and is not a force. Weight is measured in newtons and can change when $g$ changes.

Electric interaction

Electric force acts between charged bodies. Its direction depends on the signs of the charges: like charges repel and unlike charges attract.

Use the electric field

A positive test charge is pushed in the electric-field direction; a negative charge experiences force opposite to the field. Keep field direction and force direction separate when the charge sign matters.

Common trap

Do not reverse the force direction for a positive charge, and do not treat electric force as a contact force.

Magnetic interaction

A magnetic force acts on a moving charge or current in a magnetic field. Its direction is perpendicular to the relevant velocity/current and magnetic-field directions.

Apply the direction rule

Use the stated right-hand rule or vector relationship, then reverse the result for a negative charge. Parallel motion and field give zero magnetic force in the ideal model.

Common trap

A magnetic field can change the direction of velocity without doing work on an ideal moving charge; do not automatically infer a speed change from a magnetic force.

Momentum

Linear momentum is

$$\vec p=m\vec v$$

It is a vector. For an isolated system, total momentum is conserved before and after an interaction.

Check the system

Momentum is conserved when the resultant external impulse on the chosen system is negligible. Internal forces can change individual momenta while leaving the vector total unchanged.

Use signs or components

Choose a positive direction and conserve momentum component-by-component. A negative final velocity means motion opposite to the chosen positive direction.

Common trap

Do not conserve kinetic energy automatically. Momentum conservation and kinetic-energy conservation are separate claims.

Impulse changes momentum

Impulse is the integral of resultant force over time. For a constant average force,

$$\vec J=\vec F_{net}\Delta t=\Delta\vec p$$

Use the momentum change

Calculate $\Delta\vec p=\vec p_f-\vec p_i$, including direction. A rebound reverses the velocity component and can make the momentum change larger than either momentum magnitude alone.

Average force

If the force varies, $F\Delta t$ represents average resultant force over the contact interval. Use consistent units for impulse in N s or kg m s⁻¹.

Common trap

Do not use the initial momentum alone when the object rebounds or ends with a non-zero final velocity.

Impulse is external to the system

For a chosen system, the net external impulse equals the system’s change in total momentum:

$$\vec J_{ext}=\Delta\vec p_{system}$$

Same momentum change, different force

If an object must undergo the same $\Delta p$, increasing the stopping time reduces the average resultant force:

$$F_{avg}=\frac{\Delta p}{\Delta t}$$

Apply to safety systems

A flexible safety net, airbag or crumple zone extends the interaction time while producing the required momentum change, reducing the average force on the person or vehicle.

Common trap

Extending the stopping time does not make the momentum change disappear; it changes the rate at which that change occurs.

Constant mass

For a body of constant mass, Newton’s second law becomes

$$\vec F_{net}=m\vec a$$

Use the resultant force, not one arbitrarily selected force.

General momentum form

The broader statement is

$$\vec F_{net}=\frac{\Delta\vec p}{\Delta t}$$

or its instantaneous form. This is the safer form when mass changes or when momentum is the quantity given.

Check what changes

If mass is constant, $\Delta p=m\Delta v$, so the two forms agree. If mass enters or leaves the system, include the momentum carried by that mass and define the system carefully.

Common trap

Do not double the acceleration simply because an applied force doubles when a fixed resistive force remains; calculate the new resultant force first.

Momentum first

In an isolated collision, total linear momentum is conserved for both elastic and inelastic collisions.

Kinetic energy distinguishes them

In an elastic collision, total kinetic energy is also conserved. In an inelastic collision, some kinetic energy is transferred to internal energy, sound or deformation; in a perfectly inelastic collision the bodies move together afterward.

Use the right conservation law

Apply momentum conservation to find final velocities, then compare initial and final kinetic energy if the collision type is required.

Common trap

“Inelastic” does not mean momentum is lost. It means kinetic energy is not conserved.

Explosion model

An explosion is an interaction in which an initially combined system separates into parts. If the external impulse is negligible, total momentum before and after is equal.

Use a sign convention

For an object initially at rest, the vector momenta after the explosion sum to zero. In one dimension, equal and opposite momenta can give different speeds when the masses differ.

Energy is separate

The chemical, elastic or other internal energy released can increase total kinetic energy while momentum remains conserved.

Common trap

Do not assume the fragments have equal speeds. Momentum magnitudes are equal and opposite only when the initial total momentum is zero.

Track the energy store

Total energy is conserved, but kinetic energy may be transferred to internal energy, sound, deformation or chemical energy during an interaction.

Collision comparison

Elastic collisions conserve total kinetic energy as well as momentum. Inelastic collisions conserve momentum but have a lower final total kinetic energy.

Explosion comparison

An explosion can convert internal energy into kinetic energy, so final kinetic energy can exceed the initial kinetic energy while total momentum remains conserved.

Common trap

“Kinetic energy is lost” is shorthand for transferred to other stores; it is not destroyed.

Radial acceleration

For uniform circular motion, the centripetal acceleration is directed toward the centre:

$$a_c=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}$$

Velocity can be constant in magnitude

Even when speed is constant, the velocity direction changes continuously. That directional change produces inward acceleration.

Choose the matching data

Use $v^2/r$ when speed and radius are given, $\omega^2r$ when angular speed is given, or $4\pi^2r/T^2$ when period is given.

Common trap

Centripetal acceleration is not tangential and does not point along the instantaneous velocity.

Centripetal force is a resultant

Centripetal force is the name for the net inward force required for circular motion:

$$F_c=ma_c=\frac{mv^2}{r}$$

Identify its physical source

Centripetal force is not an extra force. It may be supplied by tension, gravity, friction, normal force, electric force or a combination of forces.

Keep the direction clear

The required resultant points toward the centre and is perpendicular to instantaneous velocity in uniform circular motion.

Common trap

Do not add a separate “centripetal force” arrow to a free-body diagram unless the question explicitly uses it as a shorthand for the inward resultant.

Velocity direction changes

In circular motion, the inward centripetal acceleration changes the direction of the velocity. If speed is constant, the magnitude of velocity stays constant while its direction changes.

What happens if the inward force disappears

If the centripetal interaction is removed, the object continues along the tangent at the release point, consistent with Newton’s first law.

Maintain contact

In a vertical loop, the inward resultant must be sufficient to maintain the required radial acceleration. At the limiting contact condition, the normal force can fall to zero.

Common trap

The released object does not move along the radius; its instantaneous path is tangent to the circle.

Connect the descriptions

For uniform circular motion,

$$v=\frac{2\pi r}{T}=\omega r$$

Angular speed $\omega$ is the same for all points on a rigid rotating body, while linear speed increases with distance from the axis.

Use the period

One revolution takes period $T$, so $\omega=2\pi/T$. Keep radians and seconds consistent.

Compare points on one disk

If one point is twice as far from the centre, its linear speed is twice as large at the same angular speed; its centripetal acceleration is also twice as large.

Common trap

Do not assume equal linear speeds for all points on a rotating disk. Equal angular speed does not mean equal tangential speed.

Energy is conserved

Energy cannot be created or destroyed. In a defined system, energy is transferred between stores or across the system boundary, so the total energy accounting remains balanced.

Define the system first

Name the objects included and identify transfers by work, heating, radiation or electrical means. A falling object may transfer gravitational potential energy to kinetic energy, internal energy or sound.

Follow the chain

Write the initial store, the useful output store and any dissipated or transferred energy. A Sankey diagram or energy-flow statement should account for all significant branches.

Common trap

Energy “lost” from a useful store has been transferred elsewhere; it has not disappeared.

Work transfers energy

Work done by a force is the energy transferred by that force. For a constant force,

$$W=Fs\cos\theta$$

where $\theta$ is the angle between force and displacement.

Use the sign

Positive work transfers energy into the object’s relevant store; negative work transfers energy out of it. A force perpendicular to displacement does zero work.

Follow the physical process

Wind can transfer kinetic energy to a turbine through work, while resistive forces can transfer mechanical energy to internal energy of the surroundings.

Common trap

Do not call every force an energy transfer. Check whether the force has a component along the displacement.

Read the width as energy

A Sankey diagram shows an input energy flowing into useful output and other transfers. Arrow width is proportional to energy, so the branches must account for the whole input.

Identify useful output

Label the useful branch before calculating efficiency. Other branches may represent heating, sound or unwanted mechanical transfers.

Connect to efficiency

The useful fraction of the input is

$$\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}$$

Common trap

Do not compare branch widths without checking whether the diagram uses the same scale and whether the requested quantity is energy or power.

Constant-force work

For a force $F$ acting through displacement $s$,

$$W=Fs\cos\theta$$

Only the component parallel to displacement transfers energy by work.

Area under a force–distance graph

For a variable force, the area under an $F$-against-$s$ graph gives work. A negative area represents work against the chosen displacement direction.

Check the angle

Use the angle between force and displacement, not the angle between the force and an unrelated axis unless the component has first been resolved.

Worked example from local Question Bank row 39177

A kite pulls a ship with force $2.50\times10^5\,\mathrm{N}$ at $39^\circ$ to its $1.00\,\mathrm{km}$ displacement. Convert $1.00\,\mathrm{km}=1.00\times10^3\,\mathrm{m}$, then

$$W=Fs\cos\theta=(2.50\times10^5)(1.00\times10^3)\cos39^\circ=1.94\times10^8\,\mathrm{J}\approx1.9\times10^8\,\mathrm{J}$$

Only the force component along the ship's displacement transfers energy.

Work–energy theorem

The net work done by the resultant force on a system equals its change in kinetic energy:

$$W_{net}=\Delta E_k$$

Use force–distance area

For a variable resultant force, the signed area under the force–distance graph gives the work and therefore the kinetic-energy change.

Include all resultant forces

Friction, applied forces and gravity may each do work. Add their signed contributions before relating the result to the final kinetic energy.

Worked example from local Question Bank row 31356

A constant net force of $100\,\mathrm{N}$ moves an object from rest through $2.0\,\mathrm{m}$ until its speed is $10\,\mathrm{m\,s^{-1}}$.

$$W_{net}=Fs=(100)(2.0)=200\,\mathrm{J}$$
$$200=\Delta E_k=\frac12m(10)^2-0$$
$$m=4.0\,\mathrm{kg}$$

The positive net work is exactly the object's kinetic-energy gain.

Common trap

Do not use the work of one force as the net work unless all other force contributions are zero or already included.

Mechanical energy stores

Mechanical energy is the sum of translational kinetic energy, gravitational potential energy and elastic potential energy:

$$E_{mech}=E_k+E_{p,g}+E_{p,elastic}$$

Use the chosen system

Mechanical energy describes these stores within the system. Internal energy, chemical energy and sound may also be present in the full energy account but are not mechanical energy.

Common trap

Do not call all conserved energy mechanical energy; classify the store before applying a mechanical-energy equation.

Condition for conservation

Mechanical energy is conserved when only conservative forces transfer energy within the system and friction or other resistive transfers are absent or negligible.

Write the balance

$$E_{k,i}+E_{p,i}=E_{k,f}+E_{p,f}$$

Choose a convenient zero for potential energy and keep the same reference throughout.

When it is not conserved

Friction, drag or deformation transfer mechanical energy to internal energy. Total energy is still conserved, but the mechanical-energy equation needs an additional transfer term.

Conservative transformations

When mechanical energy is conserved, energy can move between translational kinetic, gravitational potential and elastic potential stores without changing their sum.

Use the endpoints

For a car descending a frictionless track, gravitational potential energy decreases while kinetic energy increases. For a spring system, elastic potential energy can become kinetic energy and then return.

Add non-conservative transfers

If friction or drag acts, part of the mechanical energy transfers to internal energy. The endpoint equation must include that loss from the mechanical stores.

Kinetic-energy forms

Translational kinetic energy is

$$E_k=\frac12mv^2=\frac{p^2}{2m}$$

Choose the known quantity

Use $\frac12mv^2$ when mass and speed are given, or $p^2/(2m)$ when momentum is given. Kinetic energy is scalar and cannot be negative.

Worked example from local Question Bank row 35674

For $m=0.14\,\mathrm{g}=1.4\times10^{-4}\,\mathrm{kg}$ and $v=3.1\,\mathrm{m\,s^{-1}}$,

$$E_k=\frac12(1.4\times10^{-4})(3.1)^2=6.7\times10^{-4}\,\mathrm{J}=0.67\,\mathrm{mJ}$$

Converting grams to kilograms before substitution keeps the energy unit in joules.

Common trap

Doubling speed quadruples kinetic energy; do not scale it linearly with speed.

Near-Earth gravitational potential energy

For a height change $\Delta h$ in a uniform gravitational field,

$$\Delta E_{p,g}=mg\Delta h$$

Use the height change

Raising an object gives positive change in gravitational potential energy; lowering it gives negative change relative to the chosen reference.

Link to power

If height changes at constant speed, the rate of gravitational potential-energy gain is $mgv$, before accounting for efficiency or other transfers.

Worked example from local Question Bank row 37039

An object's weight is $6.10\times10^2\,\mathrm{N}$ and it rises vertically by $8.0\,\mathrm{m}$. Since $mg$ is its weight,

$$\Delta E_{p,g}=(6.10\times10^2)(8.0)=4.88\times10^3\,\mathrm{J}\approx4.9\,\mathrm{kJ}$$

The positive result means the gravitational potential-energy store increases.

Common trap

Use the local value of $g$ and the vertical height change, not the distance along a slope.

Elastic store

For a spring within its linear range,

$$E_{p,elastic}=\frac12k(\Delta x)^2$$

where $\Delta x$ is extension or compression from the natural length.

Area under the graph

The elastic potential energy equals the work done in stretching or compressing the spring. On a force–extension graph it is the area under the graph.

Worked example from local Question Bank row 31357

A spring with $k=100\,\mathrm{N\,m^{-1}}$ is compressed by $0.10\,\mathrm{m}$.

$$E_{p,elastic}=\frac12(100)(0.10)^2=0.50\,\mathrm{J}$$

This is the energy available for transfer when the ideal spring is released.

Common trap

Do not use the total spring length as $\Delta x$, and remember that doubling extension quadruples the stored energy in the ideal model.

Power is rate

Power is the rate of work or energy transfer:

$$P=\frac{\Delta W}{\Delta t}=\frac{\Delta E}{\Delta t}$$

Mechanical shortcut

For a constant force parallel to velocity,

$$P=Fv$$

Keep energy and power distinct

Energy is measured in joules; power is measured in watts, or joules per second. Multiply power by time to recover transferred energy.

Worked example from local Question Bank row 29322

A student of weight $600\,\mathrm{N}$ climbs $6.0\,\mathrm{m}$ vertically in $8.0\,\mathrm{s}$.

$$\Delta W=(600)(6.0)=3.6\times10^3\,\mathrm{J}$$
$$P=\frac{3.6\times10^3}{8.0}=4.5\times10^2\,\mathrm{W}=450\,\mathrm{W}$$

The result is the average rate of energy transfer against gravity.

Useful fraction

Efficiency is the ratio of useful output to total input:

$$\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}$$

Choose matching quantities

Use energy ratios for the same process and time interval, or power ratios when input and output are rates. Efficiency is dimensionless and is often reported as a percentage.

Worked example from local Question Bank row 29709

Solar intensity is $240\,\mathrm{W\,m^{-2}}$ over $2.50\times10^4\,\mathrm{m^2}$, so input power is

$$P_{in}=(240)(2.50\times10^4)=6.0\times10^6\,\mathrm{W}=6.0\,\mathrm{MW}$$