Work, energy and power HL Paper 1&2

A.3 Work, Energy and Power

IB-style Paper 1 and Paper 2 question bank • SL and HL

This original practice bank contains 15 multiple-choice questions, 15 extended-response questions and a lab-style investigation. Each question has an expandable worked answer. These are practice questions, not official IB examination questions.

Instructions and assumptions

Choose one answer for each multiple-choice question. For extended responses, show working, units and reasoning.
Give numerical answers to 2–3 significant figures and retain unrounded values during calculations.
Use $g=9.81\,\mathrm{m\,s^{-2}}$ . Treat bodies as particles unless stated otherwise; neglect rotational kinetic energy.
Use near-surface gravitational potential energy. Springs are light and obey Hooke’s law within their elastic limit.
A.3 has the same syllabus content at SL and HL. More demanding problems are labelled “Challenge”. No calculus is required.

Formula reference

$W=Fs\cos\theta$

$E_k=\frac12mv^2=\frac{p^2}{2m}$

$\Delta E_p=mg\Delta h$

$E_H=\frac12k(\Delta x)^2$

$W_{\mathrm{net}}=\Delta E_k$

$P_{\mathrm{avg}}=\frac{\Delta W}{\Delta t}$

$P=Fv\quad\text{(force parallel to velocity)}$

$\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}=\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}$

$\text{Energy density}=\frac{\text{energy available}}{\text{mass of fuel}}$

Here $s$ is displacement along the chosen direction, $p$ is momentum, $k$ is spring constant and $\Delta x$ is spring extension or compression. For an angled force, instantaneous power is $P=Fv\cos\theta$ . Efficiency is a fraction; multiply by 100 to express it as a percentage.

Common misconceptions

“A force always does work.” Only its component along the displacement does work.
“Constant speed means the motor does no work.” Net work is zero, but a driving or lifting force can do positive work while another force does negative work.
“Friction destroys energy.” Mechanical energy decreases while internal energy increases; total energy is conserved.
“Energy and power are the same.” Energy is measured in joules; power is the rate of energy transfer, measured in watts.

Part 1 — Multiple-choice questions

Multiple choice 1 — Work at an angle [1 mark]

A constant 40 N force acts at 60° to a 6.0 m displacement. What is the work done by this force?

240 J
0 J
120 J
480 J

Show worked answer

Correct option: C.

$W=(40)(6.0)\cos60^\circ=120\,\mathrm{J}$

Multiple choice 2 — Negative work [1 mark]

A sliding block moves to the right while kinetic friction acts to the left. Which statement is correct?

Friction does negative work on the block.
Friction does positive work on the block.
Friction does no work because it is a contact force.
The block’s total energy is destroyed.

Show worked answer

Correct option: A.

The angle between friction and displacement is 180°, so $W_f=-fs$ . Mechanical energy is transferred into internal energy of the contacting bodies.

Multiple choice 3 — Force–displacement graphs [1 mark]

On a graph of force component along displacement against displacement, which quantity represents the work done?

The gradient
The maximum force
The time taken
The signed area between the graph and the displacement axis

Show worked answer

Correct option: D.

An area has units N m = J. Regions above the axis contribute positive work; regions below contribute negative work.

Multiple choice 4 — Kinetic-energy scaling [1 mark]

A body’s mass is tripled and its speed is halved. Its new kinetic energy is what fraction of its original kinetic energy?

1.5
0.75
3.0
0.25

Show worked answer

Correct option: B.

$\frac{E_{k,2}}{E_{k,1}}=3\left(\frac12\right)^2=\frac34$

Multiple choice 5 — Energy from momentum [1 mark]

A particle of mass 3.0 kg has momentum magnitude 6.0 kg m/s. What is its kinetic energy?

6.0 J
12 J
18 J
54 J

Show worked answer

Correct option: A.

$E_k=\frac{p^2}{2m}=\frac{6.0^2}{2(3.0)}=6.0\,\mathrm{J}$

Multiple choice 6 — Gravitational energy [1 mark]

A 2.0 kg object is raised vertically by 3.0 m near Earth’s surface. What is the increase in gravitational potential energy of the object–Earth system?

6.0 J
19.6 J
58.9 J
117.7 J

Show worked answer

Correct option: C.

$\Delta E_p=(2.0)(9.81)(3.0)=58.9\,\mathrm{J}$

Multiple choice 7 — Conservation with friction [1 mark]

A block slides down a rough slope. Which statement must be correct for the block, slope and Earth considered together as an isolated system?

Mechanical energy remains constant.
Total energy remains constant.
Kinetic energy remains constant.
Gravitational potential energy remains constant.

Show worked answer

Correct option: B.

Friction transfers mechanical energy into internal energy. Total energy of the isolated system is conserved, although mechanical energy need not be.

Multiple choice 8 — Elastic potential energy [1 mark]

A spring has spring constant 200 N/m and is compressed by 0.10 m. How much elastic potential energy does it store?

20 J
2.0 J
0.10 J
1.0 J

Show worked answer

Correct option: D.

$E_H=\frac12(200)(0.10)^2=1.0\,\mathrm{J}$

Multiple choice 9 — Instantaneous power [1 mark]

A 100 N force acts at 60° to a particle’s velocity. The particle’s instantaneous speed is 3.0 m/s. What power is transferred by the force?

300 W
600 W
150 W
0 W

Show worked answer

Correct option: C.

$P=Fv\cos\theta=(100)(3.0)\cos60^\circ=150\,\mathrm{W}$

Multiple choice 10 — Lifting at constant speed [1 mark]

A motor lifts a load vertically at constant speed. Air resistance is negligible. Which statement is correct?

The motor does positive work; net work on the load is zero.
The motor does no work; gravity does positive work.
The motor and gravity both do positive work.
The load’s kinetic energy increases.

Show worked answer

Correct option: A.

The lifting force equals the weight. The motor’s positive work equals the magnitude of gravity’s negative work, so the load’s kinetic energy is unchanged.

Multiple choice 11 — Efficiency [1 mark]

A device receives 500 J of energy and transfers 350 J usefully. What is its efficiency?

30%
143%
35%
70%

Show worked answer

Correct option: D.

$\eta=\frac{350}{500}=0.70=70\%$

Multiple choice 12 — Fuel energy density [1 mark]

A fuel has energy density 46 MJ/kg. What energy is available from complete combustion of 0.20 kg, before conversion losses?

230 MJ
9.2 MJ
46 MJ
0.0043 MJ

Show worked answer

Correct option: B.

$E=(46)(0.20)=9.2\,\mathrm{MJ}$

Multiple choice 13 — Successive efficiencies [1 mark]

A generator is 80% efficient and supplies a motor that is 75% efficient. What is the overall efficiency from generator input to useful motor output?

155%
77.5%
60%
5%

Show worked answer

Correct option: C.

$\eta_{\mathrm{overall}}=(0.80)(0.75)=0.60=60\%$

Multiple choice 14 — Doubling spring compression [1 mark]

Within a spring’s elastic limit, its compression is doubled. What happens to its stored elastic potential energy?

It is unchanged.
It doubles.
It increases by a factor of eight.
It increases by a factor of four.

Show worked answer

Correct option: D.

$E_H\propto(\Delta x)^2\quad\Rightarrow\quad\frac{E_{H,2}}{E_{H,1}}=2^2=4$

Multiple choice 15 — Falling with negligible resistance [1 mark]

An object falls through a height h near Earth’s surface. Air resistance is negligible. Which relationship is correct?

The decrease in gravitational energy is greater than the kinetic-energy increase.
The kinetic-energy increase equals the magnitude of the gravitational-energy decrease.
The object’s kinetic energy remains constant.
The gravitational-energy increase equals mgh.

Show worked answer

Correct option: B.

$\Delta E_p=-mgh,\qquad\Delta E_k=+mgh$

The object–Earth system’s mechanical energy remains constant.

Part 2 — Extended-response questions

Extended response 1 — Energy stores and transfer pathways [6 marks]

A battery-powered motor raises a load at constant speed. The motor becomes warmer during operation.

Identify the initial energy store and two stores that increase. [3]
Describe the transfer pathways and explain how total energy is conserved. [3]

Show worked answer

(a) [3] The battery’s chemical energy store decreases. The gravitational potential energy of the load–Earth system increases. Internal energy of the motor and surroundings increases.

(b) [3] Energy is transferred electrically from the battery to the motor and mechanically from the motor to the load. Heating transfers energy to the surroundings. The decrease in the battery’s store equals the combined increases in the other stores; at constant speed the load’s kinetic energy is unchanged.

Answer: Chemical energy supplies the gravitational-energy increase and the internal-energy increase. Electrical and mechanical processes are transfer pathways.

Extended response 2 — Pulling a suitcase [6 marks]

A suitcase moves 12 m horizontally under a constant pulling force of 50 N directed 30° above the horizontal.

Calculate the work done by the pulling force. [2]
Explain why its vertical component does no work. [2]
The suitcase moves at constant speed. Determine the work done by resistance over the 12 m. [2]

Show worked answer

(a) [2] Use the component along the displacement.

$W=(50)(12)\cos30^\circ=519.6\,\mathrm{J}$

(b) [2] Vertical displacement is zero. The vertical force component is perpendicular to the horizontal displacement, so its work is zero.

(c) [2] Constant speed gives zero kinetic-energy change and zero net work. Weight and the normal force also do no work on this horizontal path.

$W_{\mathrm{res}}=-W_{\mathrm{pull}}=-519.6\,\mathrm{J}$

Answer: Pulling work = +520 J; resistance work = −520 J.

Extended response 3 — A force–displacement graph [6 marks]

A horizontal force acts in the direction of motion. It increases linearly from 0 N at displacement 0 m to 12 N at 3.0 m, then remains 12 N until 7.0 m. A 2.0 kg particle starts from rest on a frictionless horizontal surface.

Calculate the work done from 0 to 3.0 m. [2]
Calculate the total work done from 0 to 7.0 m. [2]
Determine the particle’s final speed. [2]

Show worked answer

(a) [2] The first region is a triangle.

$W_1=\frac12(3.0)(12)=18\,\mathrm{J}$

(b) [2] The second region is a rectangle of width 4.0 m.

$W_2=(7.0-3.0)(12)=48\,\mathrm{J}$

$W_{\mathrm{total}}=18+48=66\,\mathrm{J}$

(c) [2] The applied force does all the net work.

$66=\frac12(2.0)v^2\quad\Rightarrow\quad v=\sqrt{66}=8.12\,\mathrm{m\,s^{-1}}$

Answer: 18 J; 66 J; final speed 8.12 m/s.

Extended response 4 — Work and a car’s kinetic energy [6 marks]

A car of mass 1200 kg accelerates from 8.0 m/s to 22 m/s on a horizontal road.

Calculate the increase in kinetic energy. [2]
State the net work done on the car. [1]
During this motion, resistance does −48 kJ of work. Calculate the work done by the driving force. [2]
Explain why driving work differs from net work. [1]

Show worked answer

(a) [2]

$\Delta E_k=\frac12(1200)(22^2-8.0^2)=2.52\times10^5\,\mathrm{J}$

(b) [1] The work–energy theorem gives $W_{\mathrm{net}}=+252\,\mathrm{kJ}$ .

(c) [2]

$W_{\mathrm{drive}}+W_{\mathrm{res}}=\Delta E_k$

$W_{\mathrm{drive}}=252-(-48)=300\,\mathrm{kJ}$

(d) [1] Some driving work compensates for mechanical energy transferred to the surroundings by resistance.

Answer: Kinetic-energy increase and net work = 252 kJ; driving work = 300 kJ.

Extended response 5 — Downhill motion with energy dissipation [7 marks]

A 2.0 kg trolley starts from rest and descends through a vertical height of 1.5 m. Resistance transfers 4.0 J into internal energy. Neglect the wheels’ rotational kinetic energy.

Calculate the change in gravitational potential energy. [2]
Determine the final kinetic energy. [2]
Calculate the final speed. [2]
Explain whether total energy is conserved. [1]

Show worked answer

(a) [2] Height decreases, so the signed potential-energy change is negative.

$\Delta E_p=(2.0)(9.81)(-1.5)=-29.43\,\mathrm{J}$

(b) [2] The decrease in potential energy supplies kinetic energy and internal energy.

$E_{k,f}=29.43-4.0=25.43\,\mathrm{J}$

(c) [2]

$v=\sqrt{\frac{2(25.43)}{2.0}}=5.04\,\mathrm{m\,s^{-1}}$

(d) [1] Total energy of the trolley, track and Earth is conserved. Mechanical energy decreases by 4.0 J.

Answer: Potential-energy change = −29.4 J; final kinetic energy = 25.4 J; speed = 5.04 m/s.

Extended response 6 — A spring launches a cart [6 marks]

A light spring of spring constant 180 N/m is compressed by 0.120 m against a stationary 0.500 kg cart. The horizontal surface is frictionless and wheel rotational energy is negligible. The cart loses contact with the spring at its natural length.

Calculate the initial elastic potential energy. [2]
Determine the cart’s speed as it leaves the spring. [2]
If the initial compression is doubled within the elastic limit, determine the new speed and explain the change. [2]

Show worked answer

(a) [2]

$E_H=\frac12(180)(0.120)^2=1.296\,\mathrm{J}$

(b) [2] Conservation of mechanical energy gives $E_H=E_k$ .

$v=\sqrt{\frac{2(1.296)}{0.500}}=2.28\,\mathrm{m\,s^{-1}}$

(c) [2] Doubling compression quadruples spring energy. Since kinetic energy is proportional to speed squared, the speed doubles.

$v_2=2v_1=4.55\,\mathrm{m\,s^{-1}}$

Answer: Initial spring energy = 1.30 J; speeds = 2.28 m/s and 4.55 m/s.

Extended response 7 — A motor lifting a load [7 marks]

A motor lifts a 35 kg load vertically at a constant speed of 0.80 m/s for 12 s. Neglect air resistance.

Calculate the lifting force. [1]
Calculate the useful mechanical power. [2]
Calculate the energy transferred usefully in 12 s. [2]
State the net work on the load and explain why this does not contradict the positive useful power. [2]

Show worked answer

(a) [1] With zero acceleration, the lifting force equals the weight.

$F=mg=(35)(9.81)=343.35\,\mathrm{N}$

(b) [2] The force and velocity are parallel.

$P=Fv=(343.35)(0.80)=274.68\,\mathrm{W}$

(c) [2]

$E_{\mathrm{useful}}=Pt=(274.68)(12)=3296.16\,\mathrm{J}$

(d) [2] Net work on the load is zero because kinetic energy is unchanged. The motor does positive work while gravity does equal negative work. The useful energy increases the gravitational store of the load–Earth system.

Answer: Lifting force ≈ 343 N; useful power ≈ 275 W; useful energy ≈ 3.30 kJ; net work = 0 J.

Extended response 8 — Efficiency and a Sankey diagram [8 marks]

An electric heater receives 120 kJ in 300 s. The water being heated gains 90 kJ. Treat the remaining transferred energy as heating the surroundings.

Calculate the efficiency. [2]
Calculate the energy transferred to the surroundings. [1]
Calculate input power, useful power and the power transferred to the surroundings. [3]
Describe the widths of the input, useful and wasted branches of a Sankey diagram. [2]

Show worked answer

(a) [2]

$\eta=\frac{90}{120}=0.75=75\%$

(b) [1] The surroundings receive $120-90=30\,\mathrm{kJ}$ .

(c) [3]

$P_{\mathrm{input}}=\frac{120000}{300}=400\,\mathrm{W}$

$P_{\mathrm{useful}}=\frac{90000}{300}=300\,\mathrm{W}$

$P_{\mathrm{surroundings}}=\frac{30000}{300}=100\,\mathrm{W}$

(d) [2] Arrow widths represent energy transfers. Draw an input width of four units splitting into useful and wasted widths of three and one units. The output widths add to the input width.

Answer: Efficiency = 75%; surroundings gain 30 kJ; powers = 400 W, 300 W and 100 W; width ratio = 4:3:1.

Extended response 9 — Braking work and power [8 marks]

An 800 kg car travelling at 20 m/s stops over a horizontal distance of 40 m under a constant braking force. Neglect other resistance.

Calculate the work done by the braking force. [2]
Determine the magnitude of the braking force. [2]
Calculate the stopping time. [2]
Compare the signed average braking power with the signed instantaneous braking power at the start. [2]

Show worked answer

(a) [2]

$W_b=\Delta E_k=0-\frac12(800)(20)^2=-1.60\times10^5\,\mathrm{J}$

(b) [2] Braking force opposes displacement.

$F_b=\frac{|W_b|}{s}=\frac{160000}{40}=4000\,\mathrm{N}$

(c) [2] Acceleration is $a=-4000/800=-5.0\,\mathrm{m\,s^{-2}}$ .

$t=\frac{0-20}{-5.0}=4.0\,\mathrm{s}$

(d) [2]

$P_{\mathrm{avg}}=\frac{-160000}{4.0}=-40000\,\mathrm{W}$

$P_{\mathrm{start}}=-F_bv=-(4000)(20)=-80000\,\mathrm{W}$

The instantaneous magnitude decreases to zero as the car stops. For this constant-force case, the initial magnitude is twice the average magnitude.

Answer: Braking work = −160 kJ; force magnitude = 4.00 kN; stopping time = 4.0 s; powers = −40 kW average and −80 kW initially.

Extended response 10 — Challenge: Can a block reach the top? [6 marks]

A 3.0 kg block starts moving up a straight 30° slope at 5.0 m/s. A constant 5.0 N friction force opposes its motion. The top is 4.0 m along the slope from the starting point.

Calculate the initial kinetic energy. [1]
Write an energy equation for the distance s travelled before the block first stops. [2]
Calculate this distance. [2]
Determine whether the block reaches the top. [1]

Show worked answer

(a) [1]

$E_{k,i}=\frac12(3.0)(5.0)^2=37.5\,\mathrm{J}$

(b) [2] The height gained is $s\sin30^\circ$ , and friction dissipates $5.0s$ joules.

$37.5=mg s\sin30^\circ+5.0s$

(c) [2]

$s=\frac{37.5}{(3.0)(9.81)\sin30^\circ+5.0}=1.90\,\mathrm{m}$

(d) [1] It first stops after 1.90 m, short of the 4.0 m top. Reaching the top would require 78.9 J, greater than the available 37.5 J.

Answer: Initial kinetic energy = 37.5 J; stopping distance = 1.90 m; the block does not reach the top.

Extended response 11 — Challenge: Spring compression with friction [7 marks]

A 1.0 kg cart moving at 4.0 m/s first contacts an uncompressed light spring on a horizontal track. The spring constant is 200 N/m. During compression a constant 4.0 N resistive force acts. Neglect rotational kinetic energy.

Calculate the kinetic energy at first contact. [1]
Derive an equation for the maximum compression x, allowing for resistance. [3]
Calculate the maximum compression. [2]
Explain why setting all initial kinetic energy equal to spring energy gives an overestimate. [1]

Show worked answer

(a) [1]

$E_{k,i}=\frac12(1.0)(4.0)^2=8.0\,\mathrm{J}$

(b) [3] At maximum compression the cart is momentarily at rest. Initial kinetic energy supplies spring energy and internal energy.

$8.0=\frac12(200)x^2+4.0x$

$100x^2+4.0x-8.0=0$

(c) [2] Select the positive root because compression is a positive distance.

$x=\frac{-4.0+\sqrt{4.0^2+4(100)(8.0)}}{2(100)}=0.264\,\mathrm{m}$

(d) [1] Some energy is dissipated by resistance, so less is stored elastically. Ignoring resistance predicts $x=\sqrt{8.0/100}=0.283\,\mathrm{m}$ .

Answer: Maximum compression = 0.264 m.

Extended response 12 — Challenge: Falling onto a vertical spring [8 marks]

A 1.0 kg block is released from rest 0.50 m above the top of an uncompressed vertical spring. The spring constant is 300 N/m. The block remains in contact with the spring during compression; neglect resistance.

Write an energy equation for the maximum compression x, taking account of the complete fall distance. [3]
Calculate the maximum compression. [3]
At maximum compression, state whether acceleration is zero and justify your answer. [2]

Show worked answer

(a) [3] The block falls 0.50 m before contact and a further x during compression. At maximum compression its speed is zero.

$mg(0.50+x)=\frac12kx^2$

$150x^2-9.81x-4.905=0$

(b) [3] Solve the quadratic and reject the negative root.

$x=\frac{9.81+\sqrt{9.81^2+4(150)(4.905)}}{300}=0.216\,\mathrm{m}$

(c) [2] Acceleration is not zero. The upward spring force exceeds the downward weight; zero instantaneous velocity does not imply zero acceleration.

$F_{\mathrm{net,up}}=kx-mg=(300)(0.216464)-9.81=55.1\,\mathrm{N}$

Answer: Maximum compression = 0.216 m; acceleration is upward at that instant.

Extended response 13 — Challenge: Constant engine power [8 marks]

A 1000 kg car receives a constant mechanical driving power of 30.0 kW on a level road. The resistive force is a constant 600 N. Consider non-zero speeds.

Determine the driving force and acceleration when the speed is 15 m/s. [3]
Determine the acceleration when the speed is 25 m/s. [2]
Calculate the limiting speed in this model. [2]
Explain why acceleration decreases as speed increases. [1]

Show worked answer

(a) [3] At each instant, the driving force is $F=P/v$ .

$F_{\mathrm{drive}}=\frac{30000}{15}=2000\,\mathrm{N}$

$a=\frac{2000-600}{1000}=1.40\,\mathrm{m\,s^{-2}}$

(b) [2]

$F_{\mathrm{drive}}=\frac{30000}{25}=1200\,\mathrm{N}$

$a=\frac{1200-600}{1000}=0.600\,\mathrm{m\,s^{-2}}$

(c) [2] At the limiting speed, driving force equals resistance.

$v_{\mathrm{lim}}=\frac{P}{F_{\mathrm{res}}}=\frac{30000}{600}=50.0\,\mathrm{m\,s^{-1}}$

(d) [1] At constant power, driving force decreases as speed rises. Constant resistance therefore leaves a smaller resultant force.

Answer: Accelerations = 1.40 m/s² and 0.600 m/s²; limiting speed = 50.0 m/s. This simplified model is not applicable at zero speed.

Extended response 14 — Fuel consumption and successive efficiencies [8 marks]

A fuel-powered generator has efficiency 30%, and the motor it supplies has efficiency 80%. The motor lifts a 500 kg load vertically through 12 m in 30 s at constant speed. The fuel has energy density 46 MJ/kg. Neglect other losses beyond those included in the stated efficiencies.

Calculate the useful energy transferred to the load–Earth system. [2]
Calculate the overall efficiency. [1]
Determine the fuel energy required and the fuel mass consumed. [3]
Calculate useful average lifting power and average fuel-input power. [2]

Show worked answer

(a) [2]

$E_{\mathrm{useful}}=mgh=(500)(9.81)(12)=58860\,\mathrm{J}$

(b) [1]

$\eta_{\mathrm{overall}}=(0.30)(0.80)=0.24$

(c) [3]

$E_{\mathrm{fuel}}=\frac{58860}{0.24}=245250\,\mathrm{J}$

$m_{\mathrm{fuel}}=\frac{245250}{46\times10^6}=5.33\times10^{-3}\,\mathrm{kg}$

(d) [2]

$P_{\mathrm{useful}}=\frac{58860}{30}=1962\,\mathrm{W}$

$P_{\mathrm{fuel}}=\frac{245250}{30}=8175\,\mathrm{W}$

Answer: Useful energy = 58.9 kJ; overall efficiency = 24%; fuel energy = 245 kJ; fuel mass = 5.33 g; useful and input powers = 1.96 kW and 8.18 kW.

Extended response 15 — Challenge: Energy and a vertical loop [8 marks]

A small cart is released from rest on a frictionless track at height h above the lowest point of a vertical circular loop. The loop radius is 5.0 m. The cart moves on the inside of the loop and is not attached to the track. Neglect rotational energy. At the top, the contact force N and weight both act toward the centre.

For h = 20 m, calculate the speed at the top of the loop. [2]
Explain why the contact force does no work while the cart remains in contact. [1]
Use Newton’s second law at the top to find the minimum speed needed to maintain contact. [2]
Determine the minimum release height for completing the loop in this model. [3]

Show worked answer

At the top:

$N+mg=\frac{mv_{\mathrm{top}}^2}{r}$

(a) [2] The top is at height 2r = 10 m. Use conservation of mechanical energy.

$mg(20)=mg(10)+\frac12mv_{\mathrm{top}}^2$

$v_{\mathrm{top}}=\sqrt{2(9.81)(10)}=14.0\,\mathrm{m\,s^{-1}}$

(b) [1] The contact force is perpendicular to the instantaneous displacement, so it does no work.

(c) [2] Contact requires $N\geq0$ . The limiting case has N = 0.

$v_{\mathrm{top,min}}=\sqrt{gr}=\sqrt{(9.81)(5.0)}=7.00\,\mathrm{m\,s^{-1}}$

(d) [3] At the limiting case, the initial gravitational energy supplies the gravitational energy and minimum kinetic energy at the top.

$mgh_{\mathrm{min}}=mg(2r)+\frac12m(gr)$

$h_{\mathrm{min}}=\frac52r=12.5\,\mathrm{m}$

Answer: Speed at the top from 20 m = 14.0 m/s; minimum top speed = 7.00 m/s; minimum release height = 12.5 m.

Part 3 — Lab-style investigation

Investigating spring energy transfer [20 marks]

A student launches a 0.500 kg glider on a level air track using a light spring of spring constant 100 N/m. The glider starts from rest. Its speed is measured with a light gate at the position where the spring returns to its natural length and loses contact. The compression is varied. Assume spring behaviour remains elastic and air-track resistance is negligible.

The student tests a model in which a constant fraction $\eta$ of the initial spring energy becomes the glider’s kinetic energy. The data below are simulated practice data; they are not measurements from an actual experiment.

Measurement uncertainties: mass ±0.005 kg; spring constant ±2 N/m; compression ±0.001 m; mean speed ±0.010 m/s. Use conservative uncertainty propagation: add fractional uncertainties for products and powers.

Compression x / m	Mean speed v / m s⁻¹	x² / m²	v² / m² s⁻²
0.040	0.537	Calculate	Calculate
0.060	0.805	Calculate	Calculate
0.080	1.073	Calculate	Calculate
0.100	1.342	Calculate	Calculate
0.120	1.610	Calculate	Calculate

Identify the independent and dependent variables, and state two control variables. [2]
Derive a linear relationship between v² and x², and identify the expected gradient if all spring energy becomes kinetic energy. [3]
Complete the two calculated columns, showing the calculation for one row. [3]
Describe the graph to plot and estimate its gradient. State the gradient’s unit. [3]
Use the gradient to estimate the spring-to-glider energy transfer efficiency. [2]
For the 0.120 m compression, calculate the efficiency and its absolute uncertainty. Decide whether this result supports complete energy conversion. [4]
Suggest two improvements to the method and one plausible explanation for the apparent energy loss. [3]

Show worked answer

(a) [2] Independent variable: spring compression x. Dependent variable: measured glider speed v, analysed as v². Keep the glider mass and spring constant fixed; also maintain track level, gate position and release procedure.

(b) [3] If fraction $\eta$ is transferred into kinetic energy:

$\frac12mv^2=\eta\frac12kx^2$

$v^2=\frac{\eta k}{m}x^2$

The graph should be a straight line through the origin in this model. The gradient is $b=\eta k/m$ . Complete conversion would give:

$b_{\mathrm{ideal}}=\frac{100}{0.500}=200\,\mathrm{s^{-2}}$

(c) [3] Example for the first row: $x^2=(0.040)^2=0.001600\,\mathrm{m^2}$ and $v^2=(0.537)^2=0.288369\,\mathrm{m^2\,s^{-2}}$ .

x / m	v / m s⁻¹	x² / m²	v² / m² s⁻²
0.040	0.537	0.001600	0.2884
0.060	0.805	0.003600	0.6480
0.080	1.073	0.006400	1.1513
0.100	1.342	0.010000	1.800964
0.120	1.610	0.014400	2.5921

(d) [3] Plot v² vertically against x² horizontally with labelled axes and units. Draw a best-fit straight line rather than joining adjacent points. A large gradient triangle gives approximately:

$b\approx\frac{2.5921-0.288369}{0.014400-0.001600}=180\,\mathrm{s^{-2}}$

The points are consistent with an approximately linear relationship and a small intercept. In a real investigation, plot error bars before deciding whether the intercept is consistent with zero.

(e) [2]

$\eta=\frac{bm}{k}=\frac{(180)(0.500)}{100}=0.900=90.0\%$

(f) [4] At the largest compression:

$E_H=\frac12(100)(0.120)^2=0.720\,\mathrm{J}$

$E_k=\frac12(0.500)(1.610)^2=0.648025\,\mathrm{J}$

$\eta=\frac{mv^2}{kx^2}=0.9000$

Using the stated conservative propagation rule:

$\frac{\delta\eta}{\eta}=\frac{\delta m}{m}+2\frac{\delta v}{v}+\frac{\delta k}{k}+2\frac{\delta x}{x}$

$\frac{\delta\eta}{\eta}=\frac{0.005}{0.500}+\frac{2(0.010)}{1.610}+\frac{2}{100}+\frac{2(0.001)}{0.120}=0.0591$

$\delta\eta=(0.9000)(0.0591)=0.0532$

Thus $\eta=0.90\pm0.05$ , or approximately $(90\pm5)\%$ . The interval is below 100%, so this measurement does not support complete conversion within the stated uncertainties. Unrecognised systematic errors could still affect the conclusion.

(g) [3] Repeat releases at every compression and use the scatter to assess uncertainty. Calibrate the light gate’s interrupting length and check spring compression from its true natural length; independently calibrate k with a force–extension graph. A plausible loss is internal damping in the spring or launcher vibration/sound. These transfers can occur even when track resistance is negligible. Any two specific improvements and one physically justified explanation earn the marks.

Answer: Gradient ≈ 180 s⁻²; efficiency ≈ 90%; largest-compression efficiency = 0.90 ± 0.05. Total energy remains conserved.

Exam readiness checklist

Distinguish a force’s work from net work.
Use signed force–displacement areas.
Apply energy conservation with clearly defined stores and systems.
Include resistance and spring energy in multi-step balances.
Use power at a stated speed and combine successive efficiencies.
Analyse experimental gradients, units, uncertainties and model limitations.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Work, energy and power HL Paper 1&2

Instructions and assumptions

Formula reference

Common misconceptions

Part 1 — Multiple-choice questions

Multiple choice 1 — Work at an angle [1 mark]

Multiple choice 2 — Negative work [1 mark]

Multiple choice 3 — Force–displacement graphs [1 mark]

Multiple choice 4 — Kinetic-energy scaling [1 mark]

Multiple choice 5 — Energy from momentum [1 mark]

Multiple choice 6 — Gravitational energy [1 mark]

Multiple choice 7 — Conservation with friction [1 mark]

Multiple choice 8 — Elastic potential energy [1 mark]

Multiple choice 9 — Instantaneous power [1 mark]

Multiple choice 10 — Lifting at constant speed [1 mark]

Multiple choice 11 — Efficiency [1 mark]

Multiple choice 12 — Fuel energy density [1 mark]

Multiple choice 13 — Successive efficiencies [1 mark]

Multiple choice 14 — Doubling spring compression [1 mark]

Multiple choice 15 — Falling with negligible resistance [1 mark]

Part 2 — Extended-response questions

Extended response 1 — Energy stores and transfer pathways [6 marks]

Extended response 2 — Pulling a suitcase [6 marks]

Extended response 3 — A force–displacement graph [6 marks]

Extended response 4 — Work and a car’s kinetic energy [6 marks]

Extended response 5 — Downhill motion with energy dissipation [7 marks]

Extended response 6 — A spring launches a cart [6 marks]

Extended response 7 — A motor lifting a load [7 marks]

Extended response 8 — Efficiency and a Sankey diagram [8 marks]

Extended response 9 — Braking work and power [8 marks]

Extended response 10 — Challenge: Can a block reach the top? [6 marks]

Extended response 11 — Challenge: Spring compression with friction [7 marks]

Extended response 12 — Challenge: Falling onto a vertical spring [8 marks]

Extended response 13 — Challenge: Constant engine power [8 marks]

Extended response 14 — Fuel consumption and successive efficiencies [8 marks]

Extended response 15 — Challenge: Energy and a vertical loop [8 marks]

Part 3 — Lab-style investigation

Investigating spring energy transfer [20 marks]

Exam readiness checklist

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