Fundamental physics: motion, energy, pressure, and heat
Physics connects numbers and mathematics to observable behavior. Every equation is treated as a model with a domain of validity, not as an incantation.
The goal is to identify dominant quantities and know what an elementary calculation proves—and what it does not.
1. Position, velocity, acceleration
Position locates an object in a frame. Velocity is change of position per time; acceleration is change of velocity per time. Constant high velocity can have zero acceleration, while uniform circular motion is accelerated because direction changes.
Exercise 1
A rover goes from 0 to 6 m/s in 12 s. Average acceleration?
Solution : (6−0)/12 = 0.5 m/s².
An average acceleration of 0.5 m/s² means velocity increases by half a metre per second each second over the interval. It does not describe traction peaks, wheel slip, or the time history of acceleration.
Why this matters
Position, velocity, and acceleration answer different questions: where am I, how fast is position changing, and how fast is velocity changing. Mixing them creates braking, rendezvous, and control errors.
A rover going from 0 to 6 m/s in 12 s has an average acceleration of 0.5 m/s². At a constant 6 m/s for 50 s it travels 300 m. The calculation below is a consistency test, not a complete truth.
Average acceleration does not describe a traction spike or impact; the relevant time scale depends on the system.
2. Forces and Newton’s second law
Net force follows F = ma in the Newtonian model. The same force accelerates a smaller mass more strongly. When forces balance, net acceleration is zero even if individual forces are large.
A Mars rover combines traction, slope, rolling resistance, and inertia. The elementary equation frames the problem but does not replace wheel-soil interaction models.
Case study — accelerate a loaded cart
A 600 kg cart experiences a net horizontal force of 900 N. From F = ma, ideal acceleration is 1.5 m/s². After 8 s from rest at constant acceleration it would reach 12 m/s and travel 48 m. The calculation is deliberately ideal: a real vehicle is limited by traction, motor torque, losses, stability, and the need to stop safely.
The word “net” is essential. If the drive produces 1,200 N while resistive forces total 300 N, the force used in F = ma is 900 N. Ignoring opposing forces overestimates acceleration; treating every force as a loss can hide balancing contributions. A force diagram therefore comes before arithmetic when the system is more complicated than one object on one axis.
Physical reading
Newton’s second law links net force to acceleration of mass. Net force is a vector sum: traction, resistance, slope, and contact can act simultaneously.
For 2,000 kg and net acceleration of 0.25 m/s², net force is 500 N. If 300 N of resistance opposes motion, the actuator must provide about 800 N in this model.
The point-mass model ignores traction, load transfer, soil deformation, and torque limits; it first provides a force balance.
3. Mass and weight are different
Mass measures inertia and remains 80 kg from Earth to Mars. Weight is W = mg. Using 9.81 m/s² on Earth and 3.71 m/s² as a Mars calculation value, 80 kg corresponds to about 785 N and 297 N respectively.
Lower weight changes vertical load and traction, not inertia. A 500 kg machine still resists rapid lateral acceleration.
Move into calculation
Mass measures inertia and is expressed in kilograms; weight is a gravitational force in newtons. On Mars an object keeps the same mass while its weight is roughly 0.38 of its terrestrial value.
An 80 kg person weighs about 80×3.71=297 N on Mars versus about 785 N at g=9.81 m/s² on Earth. The teaching value is seeing units turn into a decision.
Lower gravity reduces weight but not the kinetic energy required to accelerate the same mass sideways to the same speed.
4. Work and kinetic energy
Work by a constant aligned force is W = Fd. Kinetic energy is Eₖ = ½mv². Because speed is squared, doubling speed quadruples kinetic energy.
Exercise 4
Kinetic energy of a 1,000 kg vehicle at 10 m/s?
Solution : 0.5 × 1,000 × 100 = 50,000 J = 50 kJ. At 20 m/s: 200 kJ.
Control point
Work transfers energy when force acts through a displacement. Kinetic energy scales with speed squared, making higher speeds rapidly more demanding for braking and safety.
A 1,000 kg vehicle at 5 m/s has 12.5 kJ of kinetic energy; at 10 m/s it has 50 kJ, four times more for twice the speed. An engineer keeps numerical result, margin, and validity domain separate.
This balance excludes rolling resistance, slope, and motor efficiency; battery energy will therefore exceed the kinetic-energy change.
Physics check for 4. Work and kinetic energy. Repeat the calculation with one boundary condition changed, then explain which conservation law or constitutive relation makes the answer move. Check whether the new value is physically possible before trusting the arithmetic, and identify the measurement that would most efficiently discriminate between the two cases.
5. Power, duration, efficiency
Power is an energy rate: P = E/t. A 5 kW device ideally transfers 5 kJ each second. Efficiency compares useful output with input and makes losses explicit.
Mechanical and electrical losses often become heat. Power budgeting and thermal control are therefore coupled.
Interpretation
Power is a rate of energy transfer. It sizes converters, cables, and actuators, while integrated energy sizes batteries, fuel, or thermal storage.
A 5 kW heater operating for 10 h provides 50 kWh useful. At 90% overall efficiency, about 55.6 kWh electrical input is required. The formula is useful because it makes a system dependency visible.
A system can have enough total energy yet fail if the battery or converter cannot deliver peak power.
Physics check for 5. Power, duration, efficiency.
6. Pressure: force distributed over area
Pressure p = F/A is measured in pascals, 1 Pa = 1 N/m². A 70 kPa pressure difference over 1 m² corresponds to 70,000 N ideal resultant force; over 2 m² it is 140 kN.
Shell sizing still needs geometry, stress, openings, fatigue, materials, and safety factors. The simple calculation reveals the load scale.
Case study — pressure load on a wall
A 2 m² wall separates a habitat at 70 kPa from an exterior close to vacuum. Ideal resultant force from pressure difference is ΔP×A = 70,000×2 = 140,000 N, or 140 kN. Pressure can look modest when written in kilopascals, yet over a large area it becomes a major structural load.
That force alone does not size the wall. Geometry, supports, stress concentrations, pressure cycles, temperature, defects, and safety factors still matter. The pressure calculation is the first step in a structural chain, not a certification. It does provide an immediate scale check when a hatch, airlock, or window changes area.
Field reasoning
Pressure creates large structural forces because it acts over entire surfaces. A habitat at moderate pressure therefore carries loads far larger than the word “air” suggests.
At ΔP=70 kPa, a 2 m² panel sees an ideal resultant of 140 kN. That is roughly the Earth weight of 14 tonnes, without the panel itself having that mass.
The real structure distributes load through a shell, frames, and attachments; F=PA does not directly give local stress.
7. Fluids, density, and flow
Volumetric flow Q has units m³/s; mass flow ṁ has kg/s. Density connects them: ṁ = ρQ. For water near 1,000 kg/m³, 1 L/s is approximately 1 kg/s.
Real loops add pressure drop, viscosity, pipes, valves, and pumps. Mass balance remains accumulation = inflow − outflow.
Why this matters
Density, mass flow, and volumetric flow track a fluid without confusing amount of matter with occupied volume. Pumps and pipes also add pressure loss, cavitation, viscosity, and thermal limits.
For water at ρ≈1,000 kg/m³, 0.002 m³/s corresponds to about 2 kg/s. Ten minutes at that flow moves about 1,200 kg. The calculation below is a consistency test, not a complete truth.
Density depends on temperature and composition; for compressible gases a fixed density across a large pressure change is poor modelling.
Physics check for 7. Fluids, density, and flow.
8. Ideal gases and absolute temperature
The model pV = nRT connects pressure, volume, amount, and temperature. At fixed n and V, absolute pressure rises with absolute temperature. Use kelvins: T(K) = θ(°C) + 273.15.
The ideal model has limits but is a powerful consistency check for stored gases and closed volumes.
Physical reading
The ideal-gas model links pressure, volume, amount of substance, and absolute temperature. It is a useful consistency check for atmospheres, tanks, and processes even though real gases can deviate.
At fixed volume and amount, warming from 293 K to 303 K ideally raises pressure by 303/293≈1.034, about 3.4%.
Temperature must be in kelvins; using Celsius directly in a gas-law ratio gives a meaningless result.
Physics check for 8. Ideal gases and absolute temperature.
9. Heat and thermal transfer
Without phase change, sensible heating is approximated by Q = mcΔT. Heat moves through conduction, convection, and radiation.
Mars’s thin external atmosphere changes external convection while a pressurized habitat creates internal convection. Radiators and heat exchangers are therefore functional parts of survival and industrial systems.
Case study — connect thermal power and stored energy
Suppose a habitat loses an average 5 kW of heat during a ten-hour cold period. Maintaining temperature requires at least 50 kWh of useful heat in this simplified model. If an electrical heating path delivers 90% of input electricity as useful heat at the point of interest, electrical energy required is about 55.6 kWh. Efficiency belongs on the correct side of the equation: divide useful demand by 0.90.
A real balance includes internal gains, sunlight, machinery, temperature gradients, and thermal storage. The simple model already separates power from energy. Five kilowatts is an instantaneous rate; fifty kilowatt-hours is an amount integrated across ten hours. Storage is sized in energy while converters must also tolerate peak power.
Move into calculation
Heat is energy transferred because of a temperature difference. Mars systems must distinguish thermal storage, conduction through structures, convection inside a habitat, and radiation to the environment.
Heating 100 kg of water by 20 K with c≈4.18 kJ/(kg·K) ideally needs 8.36 MJ, about 2.32 kWh, before losses. The teaching value is seeing units turn into a decision.
Sensible heat excludes phase changes and heat lost to the environment; those terms can dominate a real process.
10. Rotation and centripetal acceleration
For angular speed ω and radius r, centripetal acceleration is a = ω²r. Rotating artificial-gravity concepts use this trade: larger radius permits lower angular speed for the same acceleration.
The equation does not solve motion sickness, body gradients, or structural mass. It exposes only the first geometric trade.
Case study — rotation and artificial gravity
For a centrifuge radius of 10 m targeting 3.71 m/s² centripetal acceleration, comparable to Mars surface gravity, a = ω²r gives ω = √(3.71/10) ≈ 0.609 rad/s. Frequency is ω/(2π) ≈ 0.0969 Hz, about 5.8 revolutions per minute. This calculation does not establish physiological acceptability; it only converts a target acceleration into rotation rate for a chosen radius.
Increasing radius to 40 m for the same acceleration reduces rotation rate to about 2.9 rpm. The trade becomes visible: larger radius lowers rotation rate but increases structural size and mass. Vestibular effects, head-to-foot gradients, and mechanical costs require additional models. Fundamental physics gives the relationship needed to pose the problem without pretending it is solved.
Control point
Rotation provides centripetal acceleration a=ω²r. It supports artificial-gravity studies, but physiology also depends on rotation rate and gradients along the body.
For a=3.71 m/s² and r=10 m, ω≈0.609 rad/s, about 5.8 rpm. At r=40 m the same acceleration needs about 2.9 rpm. An engineer keeps numerical result, margin, and validity domain separate.
The calculation does not define a medically acceptable threshold; it converts an acceleration target into geometry and rotation rate.
11. Electricity: voltage, current, resistance, power
For a simplified ohmic component, V = RI. Electrical power is P = VI. A 120 V bus at 10 A transfers 1,200 W or 1.2 kW.
Real systems add conversion, protection, wiring, transients, grounding, and redundancy. The basic equations remain useful branch-level checks.
Interpretation
Voltage, current, resistance, and power describe a simple electrical network before converters, batteries, protection, and transients are added. Resistive losses scale as I²R.
A 2.4 kW load at 120 V draws an ideal 20 A. A total line resistance of 0.05 Ω then dissipates I²R=20 W. The formula is useful because it makes a system dependency visible.
The same transmitted power at higher voltage uses less current, but insulation, converters, and safety create other trade-offs.
Physics check for 11. Electricity: voltage, current, resistance, power.
12. Integrated case: move and pressurize
A 500 kg unit accelerated at 0.4 m/s² needs 200 N net force. At 4 m/s its kinetic energy is 4,000 J. A motor drawing 2 kW for 10 s receives 20 kJ, so input energy clearly does not become kinetic energy with 100% efficiency.
A 2 m² panel under 70 kPa pressure difference carries 140 kN ideal resultant force, far larger than the small equipment acceleration force. Physics helps identify dominant effects.
Guided project — a pressurized rover as a physical system
Consider a 4,000 kg pressurized vehicle on Mars. With g≈3.71 m/s², its weight is about 14.8 kN while its mass remains 4,000 kg. A net horizontal acceleration of 0.20 m/s² requires F=ma=800 N, independent of vertical weight in this simplified model. Keeping these quantities separate prevents substituting weight for mass in acceleration equations.
The vehicle has a 1.2 m² door and an internal pressure 70 kPa above the exterior. Ideal normal load on the door is about 84 kN. That is far larger than the horizontal traction force calculated above. It does not mean the door is 'harder' than propulsion; the loads act in different mechanical architectures. The calculation exposes scale and prevents comparison of newtons without structural context.
For a 30 km traverse at 10 km/h average, driving time is three hours. If average electrical power for traction and auxiliaries is 18 kW, energy is 54 kWh. Add 4 kW for thermal and life-support functions during the same three hours: another 12 kWh, total 66 kWh. A battery with 100 kWh genuinely usable cannot cover an identical return trip requiring another 66 kWh without recharge, lower loads, more storage, or changed assumptions. Range in kilometres cannot be separated from the power profile.
Suppose regenerative braking returns 20% of kinetic energy during selected stops. At 10 m/s, vehicle kinetic energy is 0.5×4,000×10² = 200 kJ, only 0.0556 kWh. Recovering 20% returns about 0.011 kWh per braking event from that speed. The order of magnitude shows that hours of cruise energy do not disappear because a vehicle has regenerative braking.
A thermal balance and peak-power limit must still be added before design conclusions. The project demonstrates how elementary laws combine: gravity for weight, Newton for acceleration, pressure for structure, integrated power for energy, kinetic energy for transients. None is difficult alone; engineering skill lies in putting them in the right order and refusing to make an equation answer a question it does not describe.
13. End-of-module check
- I distinguish velocity, acceleration, mass, and weight.
- I connect force, work, energy, and power.
- I calculate pressure force.
- I connect mass and volume flow.
- I use absolute temperature for gases.
- I connect voltage, current, and power.
Corrected drill set — independent check
1. Mass 80 kg: weight on Mars with g=3.71 m/s²?
W=mg=80×3.71≈296.8 N. Mass remains 80 kg while weight changes with gravitational field.
2. A net 600 N acts on 300 kg. Acceleration?
a=F/m=2 m/s². “Net” means relevant forces have already been combined vectorially.
3. A 0.5 m² wall sees 60 kPa pressure difference. Force?
F=PA=60,000×0.5 = 30,000 N = 30 kN. Moderate pressure over area creates a large load.
4. 10 kW for 3 h, then 2 kW for 5 h: energy?
10×3 + 2×5 = 40 kWh. Adding powers alone is wrong; each interval must be weighted by its duration.
5. Why does 80% efficiency increase required input energy?
If a function requires useful energy E, input must be E/0.8. Losses mean more energy enters than reaches the useful function.
Deep practice workshop
For every physics problem, draw the system boundary first: what enters, what leaves, and which forces or forms of energy cross that boundary? Then write the physical law with units before inserting values. Each solution checks the arithmetic, sign, order of magnitude, and physical meaning of the result.
1. Acceleration
0→8 m/s in 16 s: average acceleration?
Reasoned solution : 0.5 m/s².
2. Net force
1,500 kg at 0.3 m/s² with 250 N resistance: motor force?
Reasoned solution : ma=450 N; motor≈700 N including resistance.
The 450 N force produces the requested acceleration only before resistance is included. Adding 250 N exposes the model boundary: motor and structure need roughly 700 N before any additional margin.
3. Mars weight
250 kg on Mars at g=3.71?
Reasoned solution : W≈927.5 N. Mass remains 250 kg.
Weight changes with local gravity whereas mass does not. The 927.5 N value is the Martian gravitational force on 250 kg; the inertia that must be accelerated or stopped still corresponds to 250 kg.
4. Kinetic energy
800 kg at 12 m/s?
Reasoned solution : E=0.5×800×144=57,600 J=57.6 kJ.
The v² dependence is the key result: doubling speed quadruples kinetic energy. A modest vehicle can therefore create rapidly growing braking and safety requirements as speed rises.
5. Pressure
65 kPa over 0.8 m²?
Reasoned solution : F=65,000×0.8=52,000 N=52 kN.
Converting 65 kPa to 65,000 Pa makes the units compatible with square metres. The 52 kN is total force on the selected area; real structure must distribute it through geometry and attachments.
6. Flow
0.0015 m³/s water for 20 min?
Reasoned solution : V=1.8 m³; mass≈1,800 kg.
7. Gas
At fixed volume, 280 K→300 K: pressure ratio?
Reasoned solution : p₂/p₁=300/280≈1.071, about +7.1% ideally.
At fixed volume and gas amount, absolute pressure follows absolute temperature. The 7.1% rise is valid only under those assumptions and while ideal-gas behavior remains adequate.
8. Energy capstone
A rover uses 14 kW traction +3 kW habitat for 5 h. Usable battery nameplate is 100 kWh and 20% reserve is required. Feasible without recharge?
Reasoned solution : Need=17×5=85 kWh. With 20% reserve, only 80 kWh is available. No: the mission is short by 5 kWh even before losses and variability. Reduce duration/power, increase capacity, or recharge.
The 5 kWh shortfall exists before efficiency, cold conditions, ageing, or power peaks, so the scenario already fails its stated reserve constraint. Reserve must survive the arithmetic all the way to the decision.
Additional advanced problems
1. Braking
1,200 kg from 10 to 0 m/s: kinetic energy to dissipate?
Reasoned solution : 0.5×1,200×100=60 kJ before slope and wheel rotation.
2. Pump power
A pump provides 2 kW hydraulic at 70% efficiency. Ideal electrical input?
Reasoned solution : 2/0.70≈2.86 kW.
Dividing useful hydraulic power by 0.70 works backward to electrical input power. Efficiency must remain between zero and one; the difference becomes losses and heat that the system must reject.
3. Heat water
50 kg water, +30 K, c=4.18 kJ/kg/K.
Reasoned solution : Q≈6.27 MJ≈1.74 kWh before losses.
The heat calculation uses Q=m·c·ΔT with c treated as constant. The 6.27 MJ is ideal thermal energy; insulation, plumbing losses, and heater efficiency would increase electrical demand.
4. Electrical mini-project
3 kW load on 120 V bus for 4 h: ideal current and energy?
Reasoned solution : I=P/V=25 A; E=12 kWh. Cable sizing follows current/transients while battery sizing follows energy and power.
Current and energy constrain different hardware. The 25 A drives cable and protection sizing; the 12 kWh drives stored-energy capacity. A battery system has to satisfy both at the same time.
Sources and references
The sources anchor physical definitions and Mars context. Cart, wall, heater, and centrifuge cases are explicitly simplified teaching models.
- NIST — Definitions of SI Base Units — SI base units and exact defining constants.
- NASA Glenn — Newton and forces — Newtonian force and motion fundamentals.
- NASA Glenn — Pressure — Pressure as force distributed over area.
- NASA Glenn — Equation of State — Ideal-gas relation and state variables.
- NASA Glenn — Mars — Primary or institutional source used for definitions, technology status, or module data.