Q: What does IP Combined Science Notes (Lower Sec, Year 1-2): 08) Work, Energy, Power & Pressure cover? A: Apply work-energy relationships, efficiency, and pressure calculations in solids and fluids.
Energy never disappears - it changes form. Track these transformations with equations, careful unit usage, and thought-out assumptions.
These notes align with MOE's Lower Secondary Science syllabus themes commonly taught in IP Sec 1–2, and act as a bridge into upper-secondary Physics, Chemistry, and Biology.
Status: MOE Lower Secondary Science syllabus (current release) checked 2025-11-30 - scope unchanged; remains the reference for these combined science notes.
The core idea is simple: Energy changes form; pressure spreads force over area.
Use it as a working check: Track input energy, useful output energy, and losses. For pressure, check whether area, depth, or fluid density is the key variable.
Then go one layer deeper: Example: a sharp knife cuts better because the same force acts over a smaller area, giving higher pressure at the edge.
Learning targets
Calculate work done, kinetic energy, and gravitational potential energy in SI units.
Relate power to rate of energy conversion and discuss efficiency improvements.
Apply pressure formulas in solids (force over area) and fluids (depth dependence).
Interpret hydraulic system diagrams and solve transmission-of-force problems.
1 Work & energy
Choose the quantity before choosing the formula:
Question asks about...
Use this idea first
What to identify
Common trap
Energy transferred by a force over a distance
Work done
Force in the direction of motion and distance moved
Using the whole force when it is not along the motion.
Mass, gravitational field strength, and height change
Forgetting that only the change in height matters.
Energy of a moving object
Kinetic energy
Mass and speed
Using velocity in km⋅h−1 without converting if SI units are needed.
How quickly energy is transferred
Power
Energy or work done, and time taken
Treating power as another type of energy.
Force spread over a surface
Pressure in solids
Force and contact area
Smaller area gives larger pressure for the same force.
Pressure inside a liquid
Pressure in fluids
Density, gravitational field strength, and depth
Using container shape instead of depth.
This first-choice table prevents formula hunting. Name the physical situation, then write the equation.
Quantity
Formula
Unit
Work done
W=Fscosθ
J
Kinetic energy
Ek=21mv2
J
Gravitational potential energy
Ep=mgh
J
Assume θ=0∘ for force parallel to displacement unless stated otherwise.
Work done direction checkpoint
Before using W=Fs, check whether the force is acting along the displacement. Work done by a force depends on the component of that force in the direction of motion.
Force situation
First move
Work done by that force
Common trap
Force is in the same direction as motion
Use the full force.
W=Fs
Recalculating an angle when none is needed.
Force is opposite to motion
Decide whether the question wants signed work or work done against resistance.
Work by the opposing force is negative; work done against it is positive.
Dropping the sign without explaining the energy transfer.
Force is perpendicular to motion
Use θ=90∘.
W=0 for that force.
Assuming every force on the object does work.
Force is at an angle to motion
Resolve the force along the displacement.
W=Fscosθ
Using the whole force instead of the parallel component.
Worked check: a trolley is pulled with 50N at 30∘ above the horizontal for 4.0m. The useful force along the motion is 50cos30∘, so the work done along the floor is 50cos30∘×4.0=173J, not 200J.
Misconception check: distance moved is not enough. The force component must be along the displacement for that force to transfer energy by work done.
Energy accounting checkpoint
Use this map before writing an efficiency or power equation. It separates where energy starts, where the useful part goes, and where the wasted part goes.
input energy
-> useful output energy
-> wasted energy stores such as heat or sound
-> total output still equals input if all stores are counted
Question clue
First quantity to name
Next step
Common trap
"Useful energy output"
Useful output energy
Compare it with total input energy for efficiency.
Treating useful output as the same as input.
"Energy wasted"
Wasted energy
Subtract useful output from input if both are in joules.
Calling wasted energy "lost" instead of transferred to other stores.
"Power rating"
Energy transfer per second
Multiply by time to get input energy, or divide energy by time to get power.
Treating watts as joules.
"Efficiency"
Useful output divided by input
Convert to a percentage only after the ratio is formed.
Dividing by wasted energy.
Worked check: if a lamp takes in 200J and gives 40J of useful light, then 160J is transferred to other stores. Efficiency is 40/200×100%=20%, not 40/160×100%.
Misconception check: wasted energy is not destroyed. It is energy transferred to less useful stores, usually thermal energy in the surroundings.
Worked example - Roller coaster drop
A 450kg car descends 25m.
Loss in Ep: 450×9.81×25=1.10×105,J.
Neglecting losses, this becomes kinetic energy. Solve for v:
21×450×v2=1.10×105⇒v=22.2,m⋅s−1.
Discuss real-world losses (air resistance, friction) reducing final speed.
2 Power & efficiency
Before substituting, decide whether the question is asking for total energy, rate of transfer, or useful fraction. These are related, but they answer different questions.
Question wording
Start with
What your answer means
"How much energy is transferred?"
E=Pt or work done
Total amount converted during the whole time.
"How fast is energy transferred?"
P=tE
Rate of energy transfer, in watts.
"What percentage is useful?"
η=inputuseful output×100%
Fraction of input energy that becomes the intended output.
Misconception check: a high-power device transfers energy quickly, but it is not automatically efficient. Efficiency compares useful output with input; power compares energy with time.
P=tW=Fv.
Power-time unit checkpoint
Before using E=Pt, match the power unit with the time unit. This prevents answers that are numerically neat but in the wrong energy unit.
Given power
Given time
Energy route
Final unit
Common trap
W
s
Multiply directly: E=Pt
J
Using minutes or hours without converting.
kW
s
Convert kW to W, then multiply.
J
kW
h
Multiply directly for billing-style energy.
kWh
Calling a kWh answer joules.
Worked check: a 60W lamp runs for 5.0min. Convert time first:
t=5.0min=300s,E=Pt=60×300=18000J.
Misconception check: watts are joules per second. If the time is not in seconds, the answer is not automatically in joules.
Efficiency:
η=Input energyUseful output energy×100%.
Appliance example
An electric kettle rated 2.4kW heats 1.2kg of water from 22∘C to 100∘C in 4.0min.
Energy absorbed by water:
Q=mcΔT=1.2×4.18×(100−22)=392,kJ.
Electrical energy input:
E=Pt=2.4×103×(4.0×60)=576,kJ.
Efficiency =576392×100=68.1%.
3 Pressure in solids
P=AF.
Pressure area checkpoint
For pressure questions, check the contact area before substituting. The same force can give very different pressure if the area unit or contact surface changes.
Question clue
First check
What to do
Common trap
Area is given in cm2
SI pressure uses m2
Convert using 1cm2=1⋅10−4m2
Treating 50cm2 as 50m2.
Diameter or radius is given
Area is not the same as length
Find area first, then use P=F/A
Doubling diameter and saying area only doubles.
Sharp edge or snowshoe comparison
Force may be similar, area changes
Smaller area gives larger pressure; larger area gives smaller pressure
Saying a sharp object has a larger force instead of a smaller contact area.
Hydraulic piston areas are compared
Pressure is transmitted through the fluid
Use the same pressure on both pistons, then find force from F=PA
Multiplying force by the diameter ratio instead of the area ratio.
Worked check: a 120N force on 50cm2 acts over 5.0⋅10−3m2, so the pressure is 2.4⋅104Pa. If the output piston area is ten times larger, the output force is ten times larger only because the area is ten times larger, not because the pressure increases.
Example - Snowshoes
A person of mass 68kg stands on snowshoes with total area 0.32m2.
F=mg=68×9.81=667,N.
P=0.32667=2.08×103,Pa.
Compare with regular shoes area 0.08m2 to highlight fourfold increase in pressure.
4 Pressure in fluids
Hydrostatic pressure at depth h:
P=ρgh.
Fluid pressure checkpoint
Before calculating liquid pressure, ask what changes the depth term and what does not.
Situation
What changes pressure?
What stays irrelevant if depth is the same?
Trap to avoid
Same liquid, deeper point
Larger h gives larger pressure.
Container width and total water volume.
Saying the wider tank has higher pressure at the same depth.
Different liquids, same depth
Larger density ρ gives larger pressure.
Colour or shape of the container.
Comparing only depth and ignoring density.
Same liquid and same depth
Pressure is the same at that level.
Whether the point is near the side or middle.
Treating pressure as larger near the wall.
Worked check: two holes at the same depth in connected water containers have the same water pressure, even if one container is narrow and the other is wide. A lower hole has higher pressure because h is larger.
Misconception check: liquid pressure depends on depth below the surface, not on the total amount of liquid above the base in a differently shaped container.
Example: Water density 1000kg⋅m−3, depth 1.5m:
P=1000×9.81×1.5=1.47×104,Pa.
Hydraulic lift
If master piston area 50cm2 and slave piston area 500cm2:
Mechanical advantage of 10 achieved assuming negligible losses.
Piston diameter checkpoint
When a hydraulic question gives diameter instead of area, convert the diameter change into an area change before comparing forces.
Given comparison
Area comparison
Force comparison if pressure is the same
Common trap
Large piston diameter is twice small piston diameter
Large area is four times small area
Output force is four times input force
Saying force only doubles.
Large piston radius is three times small piston radius
Large area is nine times small area
Output force is nine times input force
Forgetting that area depends on radius squared.
Areas are given directly
Use the area ratio directly
Output force follows the area ratio
Taking the square root of an area ratio.
Worked check: if the large piston diameter is twice the small piston diameter, then its radius is also twice as large, so its area is 22=4 times as large. An input force of 80N would give an ideal output force of 320N, not 160N.
Misconception check: hydraulic systems transmit the same pressure through the fluid. The force changes because the piston area changes.
Try it yourself
A 65kg hiker climbs 420m in 35min. Determine change in gravitational potential energy and average useful power output.
Two people push a crate with force 210N over 12m, with 5% of the work lost to heat. Calculate useful work delivered to the crate.
A hydraulic press has a large piston diameter twice that of the small piston. If 80N is applied on the small piston, find the output force (ignore losses) and describe two practical limitations of this model.
