IGCSE Physics · Motion, Forces and Energy — Teacher Copy
Work and Energy — Solutions
Formula shown first, then short numbered steps with the working on its own line, so you can glance down and pick up the thread instantly while teaching.
1Work Done
Question 1
Calculate the work done when a force of 250 N moves an object 8 m in the direction of the force.
Formula: W = Fd
1
Substitute the values directly
W = 250 × 8
✓W = 2000 J
Teaching tip: Point out this is the simplest possible work-done question — no unit conversions, no rearranging. Use it to nail the formula before adding any complexity in Question 2.
Question 2
A crane lifts a 500 kg load a height of 12 m. Calculate the work done against gravity. (g = 9.8 N/kg)
Formula: W = Fd, where the force needed is the load’s weight, F = mg
1
Find the weight of the load first
F = mg = 500 × 9.8 = 4900 N
2
Now use W = Fd, with d = height lifted
W = 4900 × 12
✓W = 58800 J
Teaching tip: This is really “find the weight, then use W=Fd” in disguise — the same two-step pattern reappears constantly whenever “lifting” is involved, so it’s worth naming explicitly.
Question 3
A force of 60 N pushes a box 3.5 m across a floor at constant speed. Calculate the work done by the applied force.
Formula: W = Fd
1
Substitute the values directly
W = 60 × 3.5
✓W = 210 J
Question 4
A weightlifter lifts a 120 kg barbell from the floor to a height of 2 m. Calculate the work done against gravity. (g = 9.8 N/kg)
Formula: W = Fd, where F = weight = mg
1
Find the weight of the barbell
F = mg = 120 × 9.8 = 1176 N
2
Use W = Fd, with d = 2 m
W = 1176 × 2
✓W = 2352 J
Question 5
A shopping trolley is pushed with a force of 45 N over a distance of 12 m. Calculate the work done, then give your answer in kJ as well as J.
Formula: W = Fd
1
Substitute the values directly
W = 45 × 12 = 540 J
2
Convert to kJ by dividing by 1000
540 J = 0.54 kJ
✓W = 540 J = 0.54 kJ
Question 6
A machine does 3600 J of work moving a load 9 m. Calculate the force involved.
Formula: W = Fd, rearranged to F = W/d
1
Rearrange and substitute
F = 3600 / 9
✓F = 400 N
Teaching tip: Every formula in this topic (W=Fd, Eₖ=½mv², ΔEₚ=mgΔh, P=W/t) can be rearranged for any variable — mix in “rearranged” questions like this one regularly so students don’t just memorise one direction.
2Kinetic and Gravitational Potential Energy
Question 7
Calculate the kinetic energy of a 0.6 kg ball moving at 15 m/s.
Formula: Eₖ = ½mv²
1
Square the velocity first
15² = 225
2
Substitute into the formula
Eₖ = ½ × 0.6 × 225
✓Eₖ = 67.5 J
Teaching tip: The single most common slip in this whole topic is forgetting to square v before multiplying — get students into the habit of calculating v² as its own separate step, exactly as shown here.
Question 8
A 1200 kg car increases its speed from 10 m/s to 25 m/s. Calculate the increase in its kinetic energy.
Formula: Eₖ = ½mv², calculated at both speeds, then subtracted
1
Find the kinetic energy at the initial speed (10 m/s)
Eₖ₁ = ½ × 1200 × 10² = ½ × 1200 × 100 = 60000 J
2
Find the kinetic energy at the final speed (25 m/s)
Eₖ₂ = ½ × 1200 × 25² = ½ × 1200 × 625 = 375000 J
3
Subtract to find the increase
ΔEₖ = 375000 − 60000
✓ΔEₖ = 315000 J (315 kJ)
Teaching tip: A great one to catch students who try to shortcut with ½m(v₂−v₁)² — that’s wrong, since kinetic energy doesn’t work that way. Emphasise: calculate each Eₖ fully and separately, THEN subtract.
Question 9
A crate of mass 8 kg is lifted 4 m onto a shelf. Calculate its gain in gravitational potential energy. (g = 9.8 N/kg)
Formula: ΔEₚ = mgΔh
1
Substitute the values directly
ΔEₚ = 8 × 9.8 × 4
✓ΔEₚ = 313.6 J
Question 10
A cyclist and bicycle have a combined mass of 75 kg and travel at 8 m/s. Calculate their kinetic energy.
Formula: Eₖ = ½mv²
1
Square the velocity first
8² = 64
2
Substitute into the formula
Eₖ = ½ × 75 × 64
✓Eₖ = 2400 J
Question 11
A 1000 kg car has a kinetic energy of 125000 J. Calculate its speed.
Formula: Eₖ = ½mv², rearranged to find v
1
Substitute known values into Eₖ = ½mv²
125000 = ½ × 1000 × v²
2
Simplify the right side
125000 = 500v²
3
Divide both sides by 500
v² = 250
4
Take the square root
v = √250
✓v ≈ 15.8 m/s
Teaching tip: The reverse of Question 7 — here students rearrange for v instead of calculating Eₖ directly. Watch for the common error of forgetting to square-root at the very end (leaving the answer as 250 instead of ≈15.8).
Question 12
A 60 kg hiker climbs a hill, gaining 850 m in height. Calculate the increase in her gravitational potential energy. (g = 9.8 N/kg)
Formula: ΔEₚ = mgΔh
1
Substitute the values directly
ΔEₚ = 60 × 9.8 × 850
✓ΔEₚ = 499800 J (≈ 500 kJ)
Question 13
A ball of mass 900 g is thrown at 20 m/s. Calculate its kinetic energy in joules.
Formula: Eₖ = ½mv² (mass must be in kg)
1
Convert the mass to kilograms first
900 g = 0.9 kg
2
Square the velocity
20² = 400
3
Substitute into the formula
Eₖ = ½ × 0.9 × 400
✓Eₖ = 180 J
Teaching tip: The unit conversion (g → kg) has to happen BEFORE substituting — plugging in 900 directly is the single most common error on a question like this.
Question 14
A 5 kg object gains 245 J of gravitational potential energy after being raised. Calculate the height it was raised through. (g = 9.8 N/kg)
Formula: ΔEₚ = mgΔh, rearranged to h = ΔEₚ / (mg)
1
Rearrange and substitute
h = 245 / (5 × 9.8) = 245 / 49
✓h = 5 m
3Conservation of Energy
Question 15
A ball of mass 0.2 kg is dropped from a height of 10 m. Ignoring air resistance, calculate its speed just before it hits the ground. (g = 9.8 N/kg)
Idea: GPE lost = KE gained, so mgΔh = ½mv²
1
Set GPE lost equal to KE gained
mgΔh = ½mv²
2
The mass m cancels from both sides — no need to even use 0.2 kg!
gΔh = ½v²
3
Substitute known values
9.8 × 10 = ½v² → 98 = ½v² → v² = 196
4
Take the square root
v = √196
✓v = 14 m/s
Teaching tip: Students often panic when the mass cancels out — reassure them this is expected and correct. Every object, regardless of mass, reaches the same speed falling from the same height (ignoring air resistance) — a nice physical fact to highlight here.
Question 16
A ball of mass 0.5 kg is thrown straight upward with a speed of 12 m/s. Ignoring air resistance, use conservation of energy to calculate the maximum height it reaches. (g = 9.8 N/kg)
Idea: KE at launch = GPE at the top, so ½mv² = mgh
1
Set KE at launch equal to GPE at maximum height
½mv² = mgh
2
The mass m cancels from both sides again
½v² = gh
3
Rearrange to make h the subject
h = v² / (2g)
4
Substitute the values
h = 12² / (2 × 9.8) = 144 / 19.6
✓h ≈ 7.35 m
Teaching tip: This is the “reverse” of Question 15 — here KE converts into GPE going up, instead of GPE converting into KE coming down. Pair the two questions together to show the same idea working in both directions.
Question 17
A skier starts from rest at the top of a frictionless slope 25 m high. Calculate her speed at the bottom. (g = 9.8 N/kg, ignore air resistance)
Idea: GPE lost = KE gained, so gh = ½v² (mass cancels)
1
Set up the energy equation with mass already cancelled
gh = ½v²
2
Substitute known values
9.8 × 25 = ½v² → 245 = ½v² → v² = 490
3
Take the square root
v = √490
✓v ≈ 22.1 m/s
Question 18
A pendulum bob of mass 0.3 kg swings from a point 0.2 m above its lowest point. Ignoring air resistance, calculate its speed at the lowest point. (g = 9.8 N/kg)
Idea: GPE lost = KE gained, so gh = ½v² (mass cancels)
1
Substitute into gh = ½v²
9.8 × 0.2 = ½v² → 1.96 = ½v² → v² = 3.92
2
Take the square root
v = √3.92
✓v ≈ 1.98 m/s
Teaching tip: Same method as Question 17, just smaller numbers and a different real-world context — a good pair to show students the physics doesn’t care whether it’s a skier or a pendulum, only the height matters.
Question 19
A 400 kg roller-coaster car starts from rest at the top of a 20 m high hill. At the bottom, its speed is measured as 18 m/s. Calculate the energy lost to friction and air resistance during the descent. (g = 9.8 N/kg)
Idea: energy lost = GPE at top − KE actually gained at the bottom
1
Calculate the GPE at the top (this is the total energy available)
GPE = mgh = 400 × 9.8 × 20 = 78400 J
2
Calculate the KE actually measured at the bottom
KE = ½mv² = ½ × 400 × 18² = ½ × 400 × 324 = 64800 J
3
The difference is the energy lost to friction/air resistance
energy lost = 78400 − 64800
✓Energy lost = 13600 J
Teaching tip: This is the first question in the set where energy ISN’T fully conserved between GPE and KE — a good bridge into real-world conservation of energy, where “lost” energy hasn’t vanished, just transferred to heat/sound instead of useful motion.
Question 20
A ball is thrown straight upward, reaches its highest point, and falls back down into the thrower’s hand. Ignoring air resistance, describe the energy transfers taking place throughout this motion.
Idea: conservation of energy — no calculation needed, just tracking the stores
1
On the way up
kinetic energy → gravitational potential energy
2
At the highest point
velocity = 0, so KE = 0; all the energy is now GPE
3
On the way down
gravitational potential energy → kinetic energy
4
Back in the hand
all GPE has converted back to KE — same speed as it was thrown
✓KE ⇌ GPE throughout, with total energy conserved (no air resistance)
Teaching tip: A purely descriptive question like this is worth including precisely BECAUSE it has no numbers — it tests whether students actually understand the physics behind Questions 15, 16, 17, and 18, rather than just plugging into a formula.
Question 21
A 0.4 kg ball is dropped from a height of 5 m onto a hard floor. It rebounds to a height of 3.2 m. Calculate the percentage of its gravitational potential energy that is “lost” during the bounce. (g = 9.8 N/kg)
Idea: compare GPE before the drop and GPE at the top of the rebound
1
Calculate the initial GPE (mass and g will cancel in the final ratio, but let’s keep them for clarity)
GPE₁ = mgh₁ = 0.4 × 9.8 × 5 = 19.6 J
2
Calculate the GPE at the top of the rebound
GPE₂ = mgh₂ = 0.4 × 9.8 × 3.2 = 12.544 J
3
Find the energy lost, then express as a percentage of the original
% lost = (19.6 − 12.544) / 19.6 × 100
✓≈ 36% of the energy is lost during the bounce
Teaching tip: Faster students can be shown that mass and g cancel entirely, so % lost = (h₁−h₂)/h₁ × 100 = (5−3.2)/5 × 100 = 36% — the same answer without ever calculating an energy value in joules. Worth revealing AFTER they’ve done it the long way once.
4Power and Efficiency
Question 22
A motor does 4500 J of work in 15 s. Calculate its power output.
Formula: P = W / t
1
Substitute the values directly
P = 4500 / 15
✓P = 300 W
Question 23
A pump has a power rating of 2 kW and runs for 3 minutes. Calculate the total energy transferred, in joules.
Formula: E = P × t (time must be in seconds)
1
Convert power to watts and time to seconds
P = 2 kW = 2000 W, t = 3 min = 180 s
2
Substitute into E = Pt
E = 2000 × 180
✓E = 360000 J (360 kJ)
Teaching tip: Both unit conversions (kW→W and min→s) happen before any calculation — make this a fixed first step students write down automatically, since forgetting either one is the most common way marks are lost here.
Question 24
A machine is supplied with 800 J of energy and produces 600 J of useful output. Calculate its efficiency.
Formula: % efficiency = (useful output / total input) × 100%
1
Substitute the values
efficiency = (600 / 800) × 100
✓Efficiency = 75%
Teaching tip: Ask “where did the missing 200 J go?” — the answer (wasted as heat/sound to the surroundings) reinforces conservation of energy alongside the efficiency calculation itself.
Question 25
A weightlifter raises a 100 kg barbell 2 m in 0.8 s. Calculate the useful power output. (g = 9.8 N/kg)
Method: find the work done first (W = mgh), then P = W/t
1
Find the work done lifting the barbell
W = mgh = 100 × 9.8 × 2 = 1960 J
2
Now find power using P = W/t
P = 1960 / 0.8
✓P = 2450 W
Teaching tip: This question quietly combines Question 2’s method (finding work done against gravity) with Question 22’s formula (power) — a good one to use as a final check that students can chain two ideas together, not just apply one formula at a time.
Question 26
An electric motor has a power input of 500 W and a useful power output of 350 W. Calculate its efficiency.
Formula: % efficiency = (useful power output / total power input) × 100%
1
Substitute the values
efficiency = (350 / 500) × 100
✓Efficiency = 70%
Teaching tip: Point out this uses the POWER version of the efficiency formula (Question 24 used the ENERGY version) — same structure, different quantities, so students see both forms on the formula sheet are really the same idea.
Question 27
A light bulb transfers 100 J of electrical energy, of which 20 J is usefully transferred as light and the rest as heat. Calculate its efficiency, and state where the “lost” energy goes.
Formula: % efficiency = (useful output / total input) × 100%
1
Substitute the values
efficiency = (20 / 100) × 100
2
The remaining 80 J is not destroyed — it’s transferred elsewhere
80 J → thermal energy (heat) in the surroundings
✓Efficiency = 20%; the other 80 J becomes heat
Question 28
A pump has a power rating of 750 W. Calculate how long it takes to transfer 45000 J of energy.
Formula: P = E/t, rearranged to t = E/P
1
Rearrange and substitute
t = 45000 / 750
✓t = 60 s
Question 29
A 3 kW heater runs for 2.5 hours. Calculate the energy transferred in kWh, and the total cost if electricity costs $0.18 per kWh.
Formula: energy (kWh) = power (kW) × time (h); cost = energy × price per kWh
1
Calculate the energy directly in kWh (no unit conversion needed here)
energy = 3 × 2.5 = 7.5 kWh
2
Multiply by the price per kWh
cost = 7.5 × $0.18
✓Energy = 7.5 kWh; Cost = $1.35
Teaching tip: Contrast this with Question 23 — there, power and time both had to be converted (to W and s) to get an answer in J; here, kW and hours are used directly to get an answer in kWh. Neither is “more correct,” but mixing units mid-calculation is the error to watch for.
Question 30
A 1500 kg car accelerates from rest to 20 m/s in 8 s. Assuming all the work done by the engine goes into kinetic energy (ignore friction and air resistance), calculate the average power developed by the engine.
Method: find the kinetic energy gained (= work done by the engine), then P = W/t
1
Calculate the kinetic energy gained (car starts from rest, so initial KE = 0)
Eₖ = ½mv² = ½ × 1500 × 20² = ½ × 1500 × 400 = 300000 J
2
This kinetic energy gained equals the work done by the engine — now find power
P = W/t = 300000 / 8
✓Average power = 37500 W (37.5 kW)
Teaching tip: This is the most “joined-up” question in the whole set — it needs Eₖ=½mv² AND P=W/t AND the idea that work done = energy transferred, all in one problem. Good as a final, capstone question to check everything has actually connected.
