Physics becomes easier when you represent a situation carefully before using a formula. Draw a simple diagram, choose a positive direction, list known values with units and identify the quantity the question requests. This approach prevents many avoidable calculation errors.
These worked examples use idealised conditions such as negligible air resistance or constant acceleration. Real observations require stated assumptions and measurements. Compare your assigned PHY101 material for the full set of topics and any prescribed constants.
What you will be able to do
- Use consistent SI units and check dimensional meaning.
- Solve one-dimensional constant-acceleration problems.
- Apply Newton’s second law with a clearly stated net force.
- Distinguish work, power, temperature change and heat energy.
Lessons with worked examples
1. Measurements, units and physical meaning
A physical quantity includes a number and a unit. SI base units include the metre for length, kilogram for mass and second for time. Derived quantities combine them: speed is measured in metres per second, acceleration in metres per second squared and force in newtons, where 1 N = 1 kg·m/s².
Dimensional analysis helps catch mistakes. If a formula for distance produces units of seconds, it cannot be correct as written. Convert measurements before calculation, and round only at the end unless a question specifies significant figures. Distinguish scalar quantities, which have magnitude alone, from vectors, which also have direction.
- Convert kilometres per hour into metres per second before using SI equations.
- Write the unit after each calculated result.
- Represent directed quantities with signs or arrows.
Worked example — speed conversion
A cyclist travels at 72 km/h. Convert using 72 × 1000 m / 3600 s = 20 m/s. Dividing by 3.6 is a useful shortcut for converting km/h to m/s; multiplying by 3.6 reverses the conversion.
Takeaway: Carry units through the working, not only into the final line.
2. Constant acceleration and motion
In one-dimensional motion with constant acceleration, the main equations include v = u + at and s = ut + ½at². Here u is initial velocity, v is final velocity, a is signed acceleration, t is time and s is displacement. Choose a direction as positive and keep the convention consistent.
Distance and displacement are not always the same. Distance totals the length of a path, while displacement measures net change in position with direction. A car that moves forward and then returns can travel a non-zero distance yet have zero displacement. Equations for constant acceleration cannot be applied unchanged to an arbitrary varying acceleration.
- List initial and final velocity separately.
- Choose a positive direction.
- Check whether the given acceleration is constant.
Worked example — bus acceleration
A bus starts at 4 m/s and accelerates at 2 m/s² for 5 s. Final velocity v = 4 + 2×5 = 14 m/s. Displacement s = 4×5 + ½×2×5² = 20 + 25 = 45 m. This describes a constant-acceleration interval and assumes the chosen forward direction is positive.
Takeaway: The same data can provide both final velocity and displacement.
3. Forces and Newton’s laws
Newton’s second law relates net force to acceleration: ΣF = ma. “Net” means the vector sum of the forces acting on the chosen object. Weight, normal reaction, tension and friction may all be relevant, depending on the question. Isolate one body at a time and draw a free-body diagram before assigning equations.
Newton’s third law says interacting objects exert equal and opposite forces on each other, but those forces act on different bodies; they therefore do not cancel in the net-force equation for one body. Friction and air resistance can make measured acceleration lower than a simplified model predicts.
- Identify which object the force equation describes.
- Sum forces with signs for the selected direction.
- Do not confuse action–reaction pairs with balanced forces on one object.
Worked example — horizontal push
A 3 kg trolley is pushed horizontally with 18 N while friction opposes motion with 6 N. Net horizontal force = 18 − 6 = 12 N. Acceleration = 12/3 = 4 m/s². Ignoring friction would wrongly give 6 m/s².
Takeaway: Acceleration follows the net force, not any one applied force.
4. Work, energy, power and heating
For a constant force parallel to displacement, work equals force multiplied by displacement: W = Fs. When force and displacement meet at angle θ, use W = Fs cosθ. Power is the rate of doing work, P = W/t. Kinetic energy is ½mv², and a change in energy can come from several transfers.
Temperature and heat are not synonyms. Temperature describes the thermal state of a system; heat is energy transferred because of a temperature difference. For a substance with constant specific heat capacity and no phase change, Q = mcΔT. State the units and assumptions; a phase change requires a different treatment.
- Use joules for energy and watts for power.
- Distinguish heat transfer from temperature.
- Check whether the process includes a phase change.
Worked example — heating water
Ignoring heat losses, how much energy raises 0.5 kg of water by 10°C if c = 4,200 J/(kg·°C)? Q = 0.5 × 4,200 × 10 = 21,000 J = 21 kJ. A real kettle usually uses more electrical energy because some heat warms the container and surroundings.
Takeaway: Make losses and idealisations explicit in numerical answers.
Apply what you learned — PHY101 Practice Clinic
Original independent learning exercise. Try it before opening the worked explanation; this is not an official NOUN examination question.
Calculate final speed and distance travelled
Situation: A body moves in a straight line at an initial velocity of 5 m/s and accelerates uniformly at 2 m/s² for 6 seconds.
Your task: Find its final velocity and displacement, stating the assumptions that make the equations valid.
Reveal the step-by-step solution
- Identify u = 5 m/s, a = 2 m/s² and t = 6 s.
- Final velocity v = u + at = 5 + (2 × 6) = 17 m/s.
- Displacement s = ut + ½at² = (5 × 6) + (0.5 × 2 × 36) = 66 m.
- Check units: velocity is m/s and displacement is metres; acceleration is constant throughout the interval.
Answer: Final velocity is 17 m/s. Displacement is 66 m under constant straight-line acceleration.
Common mistake to avoid: Do not use s = vt here; speed changes throughout the six-second interval.
Test yourself — original practice questions
These are independently created learning questions, not official NOUN TMA or examination questions. Try each one before revealing the explanation.
Question 1. 72 km/h is equal to which speed?
- 2 m/s
- 20 m/s
- 72 m/s
- 259.2 m/s
Show answer and explanation
Answer B: 20 m/s
72 ÷ 3.6 = 20 m/s.
Question 2. A body starts at 2 m/s and accelerates at 3 m/s² for 4 s. Final velocity is:
- 5 m/s
- 8 m/s
- 14 m/s
- 24 m/s
Show answer and explanation
Answer C: 14 m/s
v = u + at = 2 + 3×4 = 14 m/s.
Question 3. A 2 kg mass has a horizontal net force of 10 N. Its acceleration is:
- 0.2 m/s²
- 5 m/s²
- 12 m/s²
- 20 m/s²
Show answer and explanation
Answer B: 5 m/s²
a = F/m = 10/2 = 5 m/s².
Question 4. Heating 2 kg of substance by 5°C with c = 100 J/(kg·°C) requires:
- 10 J
- 100 J
- 500 J
- 1000 J
Show answer and explanation
Answer D: 1000 J
Q = mcΔT = 2×100×5 = 1000 J.
Three things to practise this week
- Convert three speeds from km/h to m/s and verify the units.
- Draw a free-body diagram for a sliding box including friction, then state the net-force equation.
- Solve a heating problem with a stated specific heat and explain one reason measured electrical energy may be higher.
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