Work, energy & power — Physical Sciences

5 exam-style question sets on Work, energy & power (Paper 1), each with a hint and a fully worked answer. The app holds 25 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Work-energy theorem: force at an angleWork–energy theorem (qualitative)Frictionless push then down a rough rampPower & mechanical energy at constant velocityNet work & the work–energy theorem

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Question 1

Work-energy theorem: force at an angleComprehension12 marksPaper 1

A cleaner drags an 8 kg laundry basket across a rough horizontal floor from point P to point Q using a rope that makes an angle of 25° with the horizontal. The distance PQ is 2 m.

A constant frictional force of 12 N acts on the basket. The basket starts from rest at P and moves at 1,5 m·s⁻¹ as it passes Q. The tension F in the rope is constant.

Ignore air resistance.

aDefine the term work done by a force. (2)
bDraw a labelled free-body diagram showing ALL the forces acting on the basket while it moves from P to Q. (4)
cUsing ENERGY PRINCIPLES ONLY, calculate the magnitude of the tension F. (4)
dThe rope is now raised so that it makes a LARGER angle with the horizontal, while the tension F and the distance PQ stay the same. How will the work done by force F from P to Q be affected? Write down INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. (2)

This question has a diagram, shown in the app.

Hint

Apply the work-energy theorem over PQ: only the horizontal component of F and the friction do net work on the basket.

Worked answer

a) The work done on an object by a constant force is the product of the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between the force and the displacement ( $W = F Δ x cos θ$ ). ✓✓

b) Free-body diagram (arrows from a dot, labelled): weight $w$ (or $F_{g} = m g$ ) vertically downwards ✓; normal force $N$ vertically upwards ✓; tension $F$ at 25° above the horizontal, in the direction of motion ✓; friction $f$ horizontally opposite to the motion ✓.

c) Work-energy theorem: $W_{n e t} = Δ E_{k}$ ✓
$F cos 2 5^{\circ} (2) - 12 (2) = \frac{1}{2} (8) (1, 5)^{2} - 0$ ✓✓
$1, 81 F = 9 + 24 = 33$
$F = 18, 21$ N ✓

( $N$ and $w$ are perpendicular to the displacement and do no work.)

d) DECREASES ✓ — $W = F Δ x cos θ$ and $cos θ$ becomes smaller as the angle increases, so F does less work over the same displacement ✓.

Modelled on Nov 2024 Paper 1, Question 5 — same skills, new scenario.

Question 2

Work–energy theorem (qualitative)Application2 marksPaper 1

A worker pushes a crate UP a rough incline at CONSTANT VELOCITY with a force P directed parallel to the incline, as shown.

Choose the option that correctly completes the statement:

The work done by P is … the sum of the magnitudes of the work done by the gravitational force and the work done by the frictional force.

This question has a diagram, shown in the app.

Hint

Constant velocity means the net work over any section is zero.

Worked answer

A. Constant velocity means $Δ K = 0$ , so the net work is zero: $W_{P} + W_{g r a v i t y} + W_{f r i c t i o n} = 0$ . Both gravity and friction do negative work on the way up, so $W_{P}$ must equal the sum of their magnitudes exactly. B would give a net positive work and the crate would speed up. C would make the crate slow down. D has no basis — doubling would also accelerate the crate.

Modelled on Nov 2024 Paper 1, Question 1.4 — same skill, new scenario.

Question 3

Frictionless push then down a rough rampApplication13 marksPaper 1

A 4 kg parcel is at rest at point P on a frictionless horizontal roller track. A constant horizontal force of 15 N pushes the parcel a distance of 6 m from P to Q.

At Q the force is removed and the parcel moves down a rough ramp inclined at 25° to the horizontal. The length of the ramp from Q to R is 5 m. While on the ramp the parcel experiences a constant frictional force of 24 N.

aState the work-energy theorem in words. (2)
bUsing ENERGY PRINCIPLES ONLY, calculate the kinetic energy of the parcel at point Q. (3)
cDraw a labelled free-body diagram showing ALL the forces acting on the parcel as it moves down the ramp. (3)
dUsing ENERGY PRINCIPLES ONLY, determine whether the parcel will pass point R. (5)

This question has a diagram, shown in the app.

Hint

On the frictionless stretch the net work equals the work done by the 15 N force; on the ramp include the (positive) work done by gravity and the (negative) work done by friction.

Worked answer

a) The work-energy theorem: the net work done on an object is equal to the change in the object's kinetic energy. ✓✓

b) From P to Q (frictionless, horizontal — normal force and weight do no work):
$W_{n e t} = Δ E_{k}$ ✓
$F Δ x cos θ = E_{k Q} - 0$
$15 \times 6 \times cos 0°$ ✓ $\Rightarrow E_{k Q} = 90$ J ✓

c) Free-body diagram of the parcel on the ramp: weight $w$ (or $F_{g}$ ) vertically downwards ✓; normal force $N$ perpendicular to the ramp surface ✓; friction $f = 24$ N up the ramp (opposite to the motion) ✓.

d) From Q to R: $W_{n e t} = Δ E_{k}$ ✓
Work by gravity: $W_{g} = m g sin 25° \times d = 4 \times 9, 8 \times sin 25° \times 5 = 82, 83$ J ✓
Work by friction: $W_{f} = - 24 \times 5 = - 120$ J ✓
$E_{k R} = 90 + 82, 83 - 120$ ✓ $= 52, 83$ J
Since $E_{k R} > 0$ , the parcel is still moving at R, so it WILL pass point R. ✓

Modelled on Nov 2025 Paper 1, Question 5 — same skills, new scenario.

Question 4

Power & mechanical energy at constant velocityComprehension2 marksPaper 1

A tower crane lifts a pallet of bricks vertically upwards at CONSTANT VELOCITY.

Consider the following statements about the pallet while it moves upwards:

(i) The power delivered to the pallet by the lifting cable is constant.

(ii) The power delivered to the pallet by the lifting cable increases.

(iii) The kinetic energy of the pallet increases.

(iv) The gravitational potential energy of the pallet increases.

Which of the statements above are CORRECT?

Hint

Use $P = F v$ with both $F$ and $v$ constant, and think about $E_{k}$ and $E_{p}$ separately.

Worked answer

B. At constant velocity the cable force is constant (equal to the weight) and $P = F v$ is therefore constant — statement (i). The height increases, so $E_{p} = m g h$ increases — statement (iv). Statements (ii) and (iii) are false: neither the power nor the kinetic energy changes at constant speed, which eliminates A, C and D. Modelled on Nov 2025 Paper 1, Question 1.3 — same skills, new scenario.

Question 5

Net work & the work–energy theoremComprehension2 marksPaper 1

A trolley moves in a straight line on a rough horizontal floor. A constant horizontal force of 25 N acts on it in the direction of motion, while a constant frictional force of 10 N opposes the motion.

Which ONE of the following combinations, for the ACCELERATION and the KINETIC ENERGY of the trolley, is correct?

Hint

Find the net force first — is it constant, and is it zero?

Worked answer

A. The net force is $25 - 10 = 15$ N, constant and non-zero, so the acceleration is constant and the trolley keeps speeding up — its kinetic energy increases. B wrongly lets the acceleration grow even though the net force is fixed. C and D treat the forces as balanced, but 25 N > 10 N, so the net force is not zero. Modelled on Nov 2025 Paper 1, Question 1.5 — same skills, new scenario.

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Grade 12 Physical Sciences: Work, energy & power

Question 1

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