Grade 12 Physical Sciences · Mechanics · Paper 1 (Physics)

Work, Energy and Power Grade 12

Real Work, Energy and Power past paper questions and answers from 2023 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

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Question 1.5 · November 2025 · Paper 1 · National (DBE) · 2 marks

Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E.

1.5 A block moves along a rough horizontal surface while a horizontal force F of magnitude

18

N and a constant kinetic frictional force of magnitude

6

N act on it, as shown in the diagram.

Which ONE of the following combinations of ACCELERATION and NET WORK DONE ON THE BLOCK is CORRECT?(2) Diagram for question 1.5

A Acceleration: Constant; Net work done on the block: Increases
B Acceleration: Increases; Net work done on the block: Constant
C Acceleration: Increases; Net work done on the block: Increases
D Acceleration: Constant; Net work done on the block: Constant

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More Work, Energy and Power Grade 12 questions and answers

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4. Question 5.3.1 · May/June 2025 · Paper 1 · National (DBE) · 3 marks
Water is drawn from a water source $12, 5$ m below the ground by an electric pump. The water is lifted vertically upwards at a constant speed, as shown in the simplified diagram below.

The pump lifts the water upwards at a rate of $2, 5$ kg·s $^{- 1}$ . Ignore ALL frictional and capillarity effects.
5.3 The pump lifts $200$ kg of water from the source to the ground at a constant speed. Calculate the:
5.3.1 Work done by the pump(3)
Show the memo (3 ticks)Memo
1. ✓
  METHOD 1: Formula $W_{n c} = Δ E_{p} + Δ E_{k}$ (or $W_{p u m p} = m g Δ h$ / mgh)
  1 mark · Solve energy problems with friction
2. ✓
  METHOD 1: Substitution $W_{p u m p} = 200 (9, 8) (12, 5)$
  1 mark · Solve energy problems with friction
3. ✓
  METHOD 1: Answer $W_{p u m p} = 24500$ J
  1 mark · Solve energy problems with friction
5. Question 7.2.2 · May/June 2025 · Paper 1 · National (DBE) · 3 marks
7.2 Sphere S is attached to the ceiling at point P by a light inextensible string. When sphere R, on an insulated stand, is brought close to sphere S they repel and sphere S comes to rest with the horizontal distance between the centres of the spheres equal to $0, 03$ m.

Sphere R is directly below point P. The string makes an angle θ with the vertical, as shown in the diagram below.

Answer the questions that follow.
7.2.2 Draw a labelled free-body diagram showing ALL the forces acting on sphere S when it is at rest.(3)
Show the memo (3 ticks)Memo
1. ✓
  WeightA dot with an arrow vertically downwards, labelled w.
  
  Accepted labels: $F_{w}$ / $F_{g}$ / mg / $1, 47 \times 1 0^{- 3}$ N / gravitational force / weight.
  1 mark · Draw a free-body or force diagram
2. ✓
  Electrostatic forceAn arrow from the dot horizontally to the right (away from R), labelled $F_{E}$ .
  
  Accepted labels: F / Electrostatic force / Coulomb force / $F_{R o n S}$ .
  1 mark · Draw a free-body or force diagram
3. ✓
  TensionAn arrow from the dot along the string towards P (up and to the left), labelled T.
  
  Accepted labels: $F_{T}$ / $F_{s t r in g}$ / tension.
  1 mark · Draw a free-body or force diagram
6. Question 5.1 · November 2024 · Paper 1 · National (DBE) · 2 marks
A constant force F is applied at an angle of $30°$ to the horizontal on a crate of mass $6$ kg that is initially at rest, as shown in the diagram below.

A constant frictional force of $10$ N acts on the crate as it moves from rest at point A along a horizontal surface to point B. The distance between point A and point B is $1, 5$ m. The speed of the crate at point B is $2$ m·s $^{- 1}$ .
5.1 Define the term work done by a force.(2)
Show the memo (1 tick)Memo
1. ✓
  Definition of workThe work done on an object by a constant force F is $F Δ x cos θ$ , where F is the magnitude of the force, $Δ x$ the magnitude of the displacement and $θ$ the angle between the force and the displacement.
  
  OR
  
  The work done on an object is the product of the force and the displacement of the object in the direction of the displacement.
  2 marks · State energy laws and definitions
7. Question 5.5 · May/June 2024 · Paper 1 · National (DBE) · 3 marks
A crate of mass $18$ kg, initially at rest, slides down a frictionless slope from point A to point B. The crate then moves along a rough horizontal surface from point B towards point C. Point A is $3$ m above the horizontal surface. See the diagram below.
The height of the track is now lowered so that point A is at a vertical height less than $3$ m. The same crate is again released from point A.
5.5 How will the distance now, travelled by the crate along the horizontal surface before it comes to rest, compare to the distance calculated in QUESTION 5.4? Write only GREATER THAN, SMALLER THAN or EQUAL TO. Explain the answer.(3)
Show the memo (3 ticks)Memo
1. ✓
  ComparisonSmaller than
  1 mark · Use the work-energy theorem
2. ✓
  Less mechanical energy at ATotal mechanical/Gravitational potential energy (at A) is less.
  1 mark · Use the work-energy theorem
3. ✓
  Less kinetic energy at BVelocity (speed) at B is less/Kinetic energy at B/ $Δ E_{k}$ will be less from B to C/rest.
  1 mark · Use the work-energy theorem
8. Question 1.5 · May/June 2024 · Paper 1 · National (DBE) · 2 marks
Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10).
1.5 A stone of mass m is dropped from a height h above the ground. Ignore the effects of air friction.

Which ONE of the following combinations CORRECTLY represents the kinetic energy and the total mechanical energy of the stone at the instant the stone has fallen through a distance of $\frac{1}{4}$ h?(2)
1. A Kinetic energy: $\frac{3}{4}$ mgh; Total mechanical energy: $\frac{1}{4}$ mgh
2. B Kinetic energy: $\frac{1}{4}$ mgh; Total mechanical energy: $\frac{3}{4}$ mgh
3. C Kinetic energy: $\frac{3}{4}$ mgh; Total mechanical energy: mgh
4. D Kinetic energy: $\frac{1}{4}$ mgh; Total mechanical energy: mgh
Show the memo (1 tick)Memo
1. ✓
  Choose the correct optionWith no air friction, mechanical energy is conserved: it stays equal to its starting value, $E_{m ec h} = m g h + 0 = m g h$ .
  
  After falling $\frac{1}{4}$ h the stone is at height $\frac{3}{4}$ h, so $E_{P} = \frac{3}{4} m g h$ and $E_{K} = m g h - \frac{3}{4} m g h = \frac{1}{4} m g h$ .
  
  Answer: D
  2 marks · Use conservation of mechanical energy
9. Question 8.2 · May/June 2024 · Paper 2 · National (DBE) · 2 marks
The relationship between the concentration of the electrolyte and the cell potential is investigated using the following electrochemical cell represented by the cell notation:

Aℓ(s) | $A ℓ^{3 +}$ (aq) || $M^{2 +}$ (aq) | M(s)

The concentration of $M^{2 +}$ is changed and the corresponding emf is measured. The concentration of $A ℓ^{3 +}$ (aq) and the temperature are at standard conditions. The graph below shows the results of this investigation.
8.2 Determine the concentration of $M^{2 +}$ (aq) that will produce an emf of $1, 87$ V.(2)
Show the memo (1 tick)Memo
1. ✓
  Value read from the graph $0, 325$ (mol·dm $^{- 3}$ )
  2 marks · Read a graph of results
10. Question 10.3.2 · May/June 2023 · Paper 1 · National (DBE) · 4 marks
In a photoelectric investigation, light of different frequencies was radiated on each of two metals, A and B. The graph of maximum kinetic energy of the ejected electrons from metal A and the frequency of the incident photons is shown below.

Point X on the graph represents an unknown maximum kinetic energy.
10.3 Calculate the:
10.3.2 Value of X shown on the graph(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Formula $E = W_{0} + E_{k (ma x)}$
  $h f = h f_{0} + \frac{1}{2} m v_{ma x}^{2}$ (any one)
  1 mark · Use the photoelectric equation
2. ✓
  METHOD 1: Substitution of hf $(6, 63 \times 1 0^{- 34}) (12, 54 \times 1 0^{14}) = \dots$
  1 mark · Use the photoelectric equation
3. ✓
  METHOD 1: Substitution of $W_{0}$ $\dots = 3, 32 \times 1 0^{- 19} + E_{k (ma x)}$
  1 mark · Use the photoelectric equation
4. ✓
  METHOD 1: Answer with unit $E_{k (ma x)} = 5 \times 1 0^{- 19}$ J
  $X = 5 \times 1 0^{- 19}$ (J)
  1 mark · Use the photoelectric equation

Work, Energy and Power Grade 12 videos

Short lessons from South African teachers first. Videos load only when you tap play, to save your data.

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