Kinetic & Potential Energy — Print · Grade 12 Science
Chapter: Work, Energy & MomentumKinetic & Potential Energy
Learning goals
- I can calculate kinetic energy with KE = ½mv² and gravitational potential energy with PE = mgh.
- I can calculate elastic potential energy stored in a spring using PE = ½kx².
Readings for this unit
Before this unit
Squared speed, height reference, and spring energy.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Where does a falling object’s kinetic energy come from?
Illustrative evidence case. A 2 kg object falls from a higher gravitational position and speeds up.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Where does a falling object’s kinetic energy come from?
Your first model
Define object–Earth boundary, reference frame and height zero.
Build the language from the idea
- kinetic energy
- The energy an object has because of its motion: KE = ½mv², measured in joules (J).
- gravitational potential energy
- Stored energy an object has because of its height in a gravitational field: PE = mgh.
- elastic potential energy
- Energy stored in a stretched or compressed elastic object such as a spring: PE = ½kx².
- spring constant (k)
- A measure of a spring's stiffness, in newtons per metre (N/m); it appears in Hooke's law F = kx.
- reference level
- The chosen height where gravitational potential energy is defined as zero; only changes in PE have physical meaning.
The energy of a moving object is its kinetic energy, KE = ½mv². The most important feature of this formula is that speed is *squared*, so kinetic energy is exquisitely sensitive to how fast something moves. Double an object's speed and its kinetic energy does not double — it quadruples; triple the speed and it grows nine-fold. This single fact explains why a highway crash releases so much more energy than a parking-lot fender-bender, and why even a light object, like a hailstone or a fast-pitched ball, can strike with surprising force when its speed is high.
Gravitational potential energy is the energy an object stores simply by being up high, PE = mgh, where m is mass, g = 9.8 m/s² near Earth's surface, and h is height above some chosen reference level. Raise a mass higher, or choose a heavier mass, and you bank more energy that gravity can later cash in as motion. A subtle but important point: the value of h depends on where you decide to call 'zero.' Only *changes* in height — and therefore changes in potential energy — carry physical meaning, so you are free to place the reference wherever makes a problem simplest.
Springs and other elastic objects store a third kind: elastic potential energy, PE = ½kx², where k is the spring constant (the spring's stiffness) and x is how far it is stretched or compressed from its natural length. This comes straight from Hooke's law, F = kx: because the restoring force grows steadily as you stretch a spring, the energy stored is the area under that force-versus-stretch line — a triangle, which is where the factor of ½ comes from. Compress a spring twice as far and you store four times the energy, ready to launch a dart, a pinball, or a valve.
These forms are not isolated curiosities; they are constantly converting into one another. A ball at the top of its arc holds mostly gravitational potential energy and almost no kinetic energy, then trades one for the other as it falls. A drawn bow stores elastic potential energy that becomes the arrow's kinetic energy at release. Tracking how much energy lives in each form at each moment — motion energy versus stored positional energy — is exactly the modelling skill this unit is building toward, and it sets up the conservation law of the next lesson.
Because all three are measured in the same unit, the joule, you can add them, compare them, and follow them as a single running total. That is what makes energy such a powerful accounting tool: an engineer can look at a roller-coaster car, a compressed spring in a mattress, or water held behind a dam and put a single number — in joules — on the energy each holds, then predict what that energy will do when it is released.
Model check: Gravitational potential energy needs a chosen reference; negative values can be valid. U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Identify the reference frame for kinetic energy
Kinetic energy is K=½mv² in ordinary nonrelativistic mechanics. Speed is measured in a specified frame, so an object can have zero kinetic energy relative to a moving vehicle but positive kinetic energy relative to the road. Squaring speed makes K nonnegative; direction still matters for momentum and force calculations. The graph’s quadratic shape means doubling speed quadruples kinetic energy for unchanged mass.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
If mass and speed both double, how does K change?
Visual reasoning — after your attempt
It grows by 2×2²=8 times in the same frame.
For ordinary low-speed motion, K=½mv² measures kinetic energy in a specified frame. It is nonnegative and depends on speed squared. Changing the velocity sign alone does not change K, but changing reference frame can change the speed and energy.
A 2 kg object at either +3 or −3 m/s has 9 J in that frame. Its momentum changes sign, so energy cannot replace a directional momentum account.
Pause and explain
Why can equal kinetic energies describe different directions?
Check your reasoning after answering
K uses speed squared and contains no direction sign; the two velocities can therefore have equal energy and opposite momentum.
Part 2 · Choose a potential-energy reference
Near Earth, gravitational potential energy of the object–Earth system can be written Ug=mgh relative to a chosen zero. Changing the zero shifts all potential values by a constant but leaves energy differences and predictions unchanged. The simple formula assumes g is approximately constant over the height interval. Potential energy belongs to an interaction/system account, rather than being a substance hidden inside the lifted object.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
If the zero moves 1 m upward, what happens to this 3 kg object’s Ug?
Visual reasoning — after your attempt
Its new h is 1 m, so Ug=29.4 J; the physical state and energy differences between fixed events are unchanged.
Near Earth with nearly uniform g, gravitational potential-energy change is mgΔh for the object-Earth interaction. The zero level is a convenient reference; changes matter for mechanical predictions. Negative potential energy relative to that choice does not imply an impossible physical state.
Choosing the floor or tabletop as zero changes U values but leaves the change for the same lift unchanged.
Pause and explain
Does changing the zero of potential energy change required lifting work?
Check your reasoning after answering
No. It changes the numerical reference, not the potential-energy difference between the same initial and final states.
Part 3 · Distinguish force and elastic energy
For an ideal linear spring, elastic potential energy relative to its unstrained state is Us=½kx². This is the area under the balancing force–extension curve, not the endpoint force itself. A spring stretched twice as far stores four times the energy within the same elastic range. Compression can also store energy because x² is positive, while the restoring force reverses direction according to −kx.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
For k=100 N/m and x=0.10 m, find both energy and force magnitude.
Visual reasoning — after your attempt
Us=0.50 J; force magnitude=10 N. Their units and meanings differ.
An ideal spring stores elastic potential energy ½kx². The energy depends on displacement squared, while restoring force is proportional to signed displacement. The formulas answer different questions and have different units. Use them only within the assumed elastic range.
A spring compressed or stretched by the same magnitude can have equal ideal stored energy, with restoring force in opposite directions.
Pause and explain
Why is elastic energy not simply kx?
Check your reasoning after answering
kx has force units. Energy requires integrating the varying spring force over displacement, giving ½kx² in the linear model.
Part 4 · Track transfers between compatible stores
An energy account uses one boundary, frame and reference convention throughout. For a 2 kg object falling from rest through a 3 m height change with negligible resistance, gravitational potential decreases by 58.8 J while kinetic energy increases by the same amount. The selected object–Earth mechanical total stays 98 J with ground as zero. Changing the height reference midway or excluding an important interaction would make the bookkeeping appear inconsistent.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What speed corresponds to K=58.8 J for 2 kg?
Visual reasoning — after your attempt
v=√(2K/m)=√58.8≈7.67 m/s.
An energy model must identify interactions and boundaries. A falling object can gain kinetic energy while object-Earth gravitational potential energy decreases. Drag or friction can transfer energy to thermal stores; an ideal prediction that omits them should be labeled accordingly.
A frictionless spring launcher gives a useful reference speed, but measured launch speed may be smaller if other stores gain energy.
Pause and explain
What does an energy shortfall suggest?
Check your reasoning after answering
It suggests omitted transfers, changed assumptions, or measurement error; total energy has not necessarily failed to be conserved.
Apply the whole unit · Predict a spring launch speed
Illustrative 0.500 kg cart starts at rest against a spring with k=200 N/m compressed 0.100 m. Assume a horizontal loss-free launch and negligible spring mass.
Your task
Calculate spring energy and cart speed after release, compare a measured 1.80 m/s, and identify the energy difference and plausible next measurements.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
Initial spring energy=½×200×0.100²=1.00 J.
½×0.500×v²=1.00 gives v=2.00 m/s.
At 1.80 m/s, K=0.810 J; difference from prediction=0.190 J.
The 0.190 J difference is an account relative to the ideal model, not proof of one particular loss mechanism. Track rolling resistance, spring motion, sound, and measurement uncertainty before assigning a cause. The result assumes all modeled elastic energy becomes cart translational kinetic energy.
Scientific references and source-model foundations
Parent / teacher teaching notes
Require students to distinguish an energy account from a causal diagnosis. A plausible loss mechanism still needs evidence. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
The key idea
Kinetic energy depends on speed/frame; potential energy depends on interaction and chosen reference. Conserve a consistent energy account rather than mixing force with energy.
Kinetic Energy
KE = ½mv²
m is mass in kg; v is speed in m/s; KE is in joules (J). Speed is squared, so KE grows fast as v increases.
Gravitational Potential Energy
PE = mgh
m is mass in kg; g = 9.8 m/s² near Earth's surface; h is height above the reference level in metres; PE is in joules (J).
Elastic Potential Energy
PE = ½kx²
k is the spring constant in N/m; x is the stretch or compression from the natural length in metres; PE is in joules (J).
Worked examples
Example 1
- Use KE = ½mv².
- KE = ½ × 1,200 × 20².
- = ½ × 1,200 × 400 = 240,000.
Example 2 · Worked application
- KE=½mv²=½×2×25=25 J.
- PE=mgh=2×10×3=60 J.
- Potential energy depends on the chosen reference; changes are physically useful.
Do not add height to speed or confuse their squares.
Example 3
- Use PE = ½kx².
- PE = ½ × 400 × 0.10².
- = ½ × 400 × 0.01 = 2.0.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Do not add height to speed or confuse their squares.
- Its zero is chosen; differences between states are the meaningful quantity in the model.
Do not add height to speed or confuse their squares.
Example 5 · Reason from the visual model
- For an ideal linear spring, elastic potential energy relative to its unstrained state is Us=½kx². This is the area under the balancing force–extension curve, not the endpoint force itself. A spring stretched twice as far stores four times the energy within the same elastic range. Compression can also store energy because x² is positive, while the restoring force reverses direction according to −kx.
- Us=0.50 J; force magnitude=10 N. Their units and meanings differ.
Calculated linear-spring energy curve, valid only over the stated elastic range.
Avoid these mistakes
- Gravitational potential energy needs a chosen reference; negative values can be valid.
Check your understanding
A 2 kg mass at 3 m has what U usingg=9.8 andU=0 atground?
Discuss the answer
U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.
Return to the phenomenon
Return to the initial case. Where does a falling object’s kinetic energy come from? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Gravitational potential decreases while kinetic energy rises; resistance would require additional energy destinations.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Where does a falling object’s kinetic energy come from? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Gravitational potential decreases while kinetic energy rises; resistance would require additional energy destinations.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
A 1.5 kg cart has speed 4 m/s. It compresses a loss-free spring with k=600 N/m on a level track. Find initial K and maximum compression.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
K=½×1.5×16=12 J. Set 12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.
Words to know
- kinetic energy
- The energy an object has because of its motion: KE = ½mv², measured in joules (J).
- gravitational potential energy
- Stored energy an object has because of its height in a gravitational field: PE = mgh.
- elastic potential energy
- Energy stored in a stretched or compressed elastic object such as a spring: PE = ½kx².
- spring constant (k)
- A measure of a spring's stiffness, in newtons per metre (N/m); it appears in Hooke's law F = kx.
- reference level
- The chosen height where gravitational potential energy is defined as zero; only changes in PE have physical meaning.
The 'vis viva' controversy: was motion measured by mv or mv²?
In the early 1700 s, Europe's best thinkers fought over how to quantify a moving object's 'force.' Followers of Descartes backed quantity of motion, mv (what we now call momentum); Leibniz argued for vis viva, or 'living force,' proportional to mv². Experiments by Willem 's Gravesande — dropping brass balls into soft clay and measuring the depth of the dent — showed the effect grew with the square of the speed, and the mathematician Émilie du Châtelet helped make the case decisively. The resolution was that both quantities are real and useful: mv is conserved as momentum, and ½mv² is the kinetic energy. This whole unit is, in a sense, the happy ending to that 300-year-old argument.
Misconception: potential energy is not an absolute number
It is tempting to think an object 'has' a definite amount of gravitational potential energy, but PE = mgh depends entirely on where you set h = 0. A book on a second-floor desk has one PE value measured from the floor, a larger one measured from the ground outside, and a negative one measured from the ceiling. None is more 'correct' than the others. What is physically real — and always comes out the same — is the change in PE as the object rises or falls. Choosing a convenient reference level is a freedom that often makes a hard problem easy.
Where the ½ in ½kx² comes from: the area under Hooke's law
A spring obeys Hooke's law, F = kx: the more you stretch it, the harder it pulls back, in a straight-line relationship. To find the energy stored you cannot just multiply force by distance, because the force keeps changing as you stretch. Instead you take the area under the force-versus-stretch graph — and since that graph is a straight line from 0 up to kx, the area is a triangle: ½ × base × height = ½ × x × (kx) = ½kx². That same 'area under a changing force' reasoning is why kinetic energy also carries a ½, and it is your first taste of the calculus that underlies all of energy physics.
Storing energy at grid scale: pumped hydro and giant springs
The formulas in this lesson run some of the world's largest machines. Pumped-storage plants use surplus night-time electricity to pump enormous volumes of water uphill, banking it as gravitational PE = mgh; at peak demand they let it fall back through turbines, recovering about 80% of the energy. Elastic potential energy is stored at scale too — from the coiled springs in mechanical watches to experimental flywheel and spring systems that smooth out power grids. Whenever engineers need to save energy now and release it later, they reach for exactly the potential-energy equations you are practicing here.
In the real world
Pumped-storage power stations bank surplus electricity by pumping water uphill into a reservoir as gravitational potential energy, then release it through turbines at peak demand — a giant rechargeable battery built out of PE = mgh.
Exit check
A2 kg mass at 3 m has what U usingg=9.8 andU=0 atground?
Check your explanation after answering
U=mgh=58.8 J; changing the reference shifts absolute U but not energy differences.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Answer and explanation
100 J
KE = ½mv² = ½ × 2.0 × 10² = ½ × 2.0 × 100 = 100 J.
Problem 2
Answer and explanation
98 J
PE = mgh = 0.50 × 9.8 × 20 = 98 J.
Problem 3
Answer and explanation
10 J
PE = ½kx² = ½ × 500 × 0.20² = ½ × 500 × 0.04 = 10 J.
Problem 4
- The motorcycle (62,500 J)
- They are equal
- It cannot be determined
- The car (50,000 J)
Answer and explanation
The motorcycle (62,500 J)
Car KE = ½ × 1,000 × 10² = 50,000 J. Motorcycle KE = ½ × 200 × 25² = 62,500 J. The motorcycle's higher speed (squared) wins out over its smaller mass.
Problem 5
Answer and explanation
Because PE = mgh measures height h from a chosen zero level, the numerical value of PE changes if you pick a different reference. This is not a problem because only changes in PE (as an object rises or falls) affect motion, and those changes come out the same no matter where you place the reference level.
Problem 6
Hints
Do not add height to speed or confuse their squares.
Answer and explanation
½×4×9=18 J.
Problem 7
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
It increases ninefold because of the square dependence.
Problem 8
Answer and explanation
Us=0.50 J; force magnitude=10 N. Their units and meanings differ.
For an ideal linear spring, elastic potential energy relative to its unstrained state is Us=½kx². This is the area under the balancing force–extension curve, not the endpoint force itself. A spring stretched twice as far stores four times the energy within the same elastic range. Compression can also store energy because x² is positive, while the restoring force reverses direction according to −kx.
Us=0.50 J; force magnitude=10 N. Their units and meanings differ.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 9
Answer and explanation
K=½×1.5×16=12 J. Set 12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.
K=½×1.5×16=12 J.
Set 12=½×600x², giving x=0.20 m within the spring’s assumed elastic range.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.