Grade 12 Physical Sciences · Mechanics

Use the work-energy theorem

Use the fact that net work equals the change in kinetic energy (W_net = ΔEk) to find a speed, force or distance.

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Question 5.2 · November 2025 · Paper 1 · National (DBE) · 3 marks

A crate of mass

3

kg is at rest at point A on a frictionless horizontal surface. A constant horizontal force F of

12

N acts on the crate and moves it from point A to point B. The crate then moves down a rough plane, inclined at

20°

to the horizontal. The distance from point A to point B is

7

m and from point B to point C is

6, 8

m, as shown in the diagram below. Diagram for this question

5.2 Using ENERGY PRINCIPLES ONLY, calculate the kinetic energy of the crate at point B.(3)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use the work-energy theorem

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3. Question 1.4 · November 2024 · Paper 1 · National (DBE) · 2 marks
1.4 A force F acts on a box as the box moves from rest down a rough incline at a constant acceleration. The force is parallel to the incline, as shown in the diagram below.

Choose the option that CORRECTLY completes the following statement.

The work done by the gravitational force is ... the work done by the frictional force and the work done by F.(2)
1. A equal to the sum of
2. B less than the sum of
3. C greater than the sum of
4. D equal to the difference between
Show the memo (1 tick)Memo
1. ✓
  Correct optionThe box speeds up from rest, so $Δ E_{k} > 0$ and $W_{n e t} > 0$ .
  
  $W_{n e t} = W_{g} + W_{f} + W_{F}$ . Gravity does positive work (the box moves down), while friction and F both act up the incline, against the motion, so they do negative work.
  
  For $W_{n e t}$ to be positive, the work done by gravity must be greater than the (magnitudes of the) work done by friction and F together.
  
  A would only be true at constant velocity ( $Δ E_{k} = 0$ ).
  
  Answer: C
  2 marks · Use the work-energy theorem
4. Question 5.2 · May/June 2023 · Paper 1 · National (DBE) · 5 marks
An electric motor pulls a $20$ kg crate from rest at point A up an inclined plane by means of a light inextensible rope. The inclined plane makes an angle of $18°$ with the horizontal. B, C and D are points on the inclined plane and the distance between points A and C is $15, 6$ m, as shown in the diagram below.

The motor exerts a constant force of $96, 8$ N parallel to the inclined plane on the rope.

A constant frictional force of $13, 5$ N acts on the crate as it moves on the inclined plane.
5.2 Use ENERGY PRINCIPLES to calculate the speed of the crate when it reaches point C.(5)
Show the memo (5 ticks)Memo
1. ✓
  METHOD 1: Formula $W_{n e t} = Δ K$
  $W_{w} + W_{f} + W_{F} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ (any one)
  1 mark · Use the work-energy theorem
2. ✓
  METHOD 1: Work done by gravity $(20) (9, 8) (sin 18°) (15, 6) cos 180° + \dots$
  1 mark · Use the work-energy theorem
3. ✓
  METHOD 1: Work done by friction and the motor $\dots + (13, 5) (15, 6) cos 180° + (96, 8) (15, 6) cos 0°$
  1 mark · Use the work-energy theorem
4. ✓
  METHOD 1: Substitution of $Δ K$ $= \frac{1}{2} (20) (v_{f}^{2} - 0^{2})$
  1 mark · Use the work-energy theorem
5. ✓
  METHOD 1: Answer with unit $v_{f} = 5, 96$ m·s $^{- 1}$
  1 mark · Use the work-energy theorem
5. Question 5.3 · May/June 2022 · Paper 1 · National (DBE) · 5 marks
Arrestor beds are used to help moving trucks to come to a stop when their brakes fail.

The driver of a $30000$ kg truck driving down a steep road drives onto an ASCENDING arrestor bed inclined at $28°$ to the horizontal, as shown in the diagram below.
5.3 Use ENERGY PRINCIPLES to calculate the minimum length of the arrestor bed needed to bring the truck to a stop.(5)
Show the memo (5 ticks)Memo
1. ✓
  METHOD 1: Formula $W_{n e t} = Δ K$
  
  (Any one: $W_{n e t} = Δ K$ , $W_{w} + W_{f} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ or $m g sin θ Δ x cos θ + W_{f} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ )
  1 mark · Use the work-energy theorem
2. ✓
  METHOD 1: Work done by the weight $(30000) (9, 8) sin 28°Δ x cos 180°$
  1 mark · Use the work-energy theorem
3. ✓
  METHOD 1: Work done by friction $+ (31000) Δ x cos 180°$
  1 mark · Use the work-energy theorem
4. ✓
  METHOD 1: Change in kinetic energy $= \frac{1}{2} (30000) (0^{2} - 3 3^{2})$
  1 mark · Use the work-energy theorem
5. ✓
  METHOD 1: Answer with unit $Δ x = 96, 64$ m
  1 mark · Use the work-energy theorem
6. Question 5.4 · November 2023 · Paper 1 · National (DBE) · 6 marks
Learners conduct an experiment to determine how the initial kinetic energy given to a trolley affects the distance the trolley moves on a rough horizontal surface. A learner pushes a trolley of unknown mass until it reaches point $A$ with kinetic energy $E_{k A}$ . The horizontal distance ( $Δ x$ ) travelled by the trolley before it comes to rest is then measured. See the diagram below.

The experiment is repeated with the trolley moving on the same rough horizontal surface but with different initial kinetic energies at point $A$ .

The results obtained are shown in the graph below.

Ignore the rotational effects of the wheels of the trolley.
5.4 Calculate the mass of the trolley if the coefficient of kinetic friction between the wheels of the trolley and the rough horizontal surface is $0, 18$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Relating frictional force to gradientOption 1:
  gradient $= \frac{Δ y}{Δ x} = \frac{Δ x}{E _{ki}} = \frac{1}{f}$
  1 mark · Use the work-energy theorem
2. ✓
  Substitution of values from the graph $\frac{1 , 5}{6}$ OR $\frac{3}{12}$ OR $\frac{4 , 5}{18}$
  1 mark · Read a graph of results
3. ✓
  Gradient value $= \frac{1}{4}$
  $f = 4$ N
  1 mark · Read a graph of results
4. ✓
  Formula for the mass $f_{k} = μ_{k} F_{N}$
  1 mark · Calculate and explain frictional force
5. ✓
  Substitution of μ and 9,8 $4 = (0, 18) (m) (9, 8)$
  1 mark · Calculate and explain frictional force
6. ✓
  Answer with unit $m = 2, 27$ kg
  1 mark · Calculate and explain frictional force

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