Energy is a scalar quantity measured in joules (J \mathrm{J} J
Form Description Kinetic Energy of motion Gravitational potential Energy due to position in a gravitational field Elastic potential Energy stored in a deformed object Thermal Energy associated with temperature Chemical Energy stored in chemical bonds Nuclear Energy stored in atomic nuclei Electrical Energy of moving charges Radiant (electromagnetic) Energy of electromagnetic waves
Kinetic Energy
Translational Kinetic Energy
For an object of mass m m m v v v
E k = 1 2 m v 2 E_k = \frac{1}{2}mv^2 E k = 2 1 m v 2 Key Points
Kinetic energy is always non-negative (E k ≥ 0 E_k \ge 0 E k ≥ 0
It depends on speed squared, so doubling speed quadruples kinetic energy.
It is a scalar (no direction).
Work-Energy Theorem
The net work done on an object equals the change in its kinetic energy:
W n e t = Δ E k = 1 2 m v f 2 − 1 2 m v i 2 W_{\mathrm{net}} = \Delta E_k = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 W net = Δ E k = 2 1 m v f 2 − 2 1 m v i 2 A 1500 k g 1500\mathrm{ kg} 1500 kg 20 m / s 20\mathrm{ m/s} 20 m/s
W = Δ E k = 0 − 1 2 ( 1500 ) ( 400 ) = − 300000 J = − 300 k J W = \Delta E_k = 0 - \frac{1}{2}(1500)(400) = -300000\mathrm{ J} = -300\mathrm{ kJ} W = Δ E k = 0 − 2 1 ( 1500 ) ( 400 ) = − 300000 J = − 300 kJ The negative sign indicates the brakes do negative work (remove energy from the car).
Gravitational Potential Energy
Near Earth's Surface
For small height changes (where g g g
E p = m g h E_p = mgh E p = m g h where h h h
Universal Gravitational Potential Energy
For any two masses separated by distance r r r
E_p = -\frac`\{GMm}`{r}
where G = 6.674 × 10 − 11 N ⋅ m 2 / k g 2 G = 6.674 \times 10^{-11}\mathrm{ N}\cdot\mathrm{m}^2/\mathrm{kg}^2 G = 6.674 × 1 0 − 11 N ⋅ m 2 / kg 2
Key Differences
Property Near Surface (m g h mgh m g h Universal (− G M m / r -GMm/r − GM m / r Reference level Arbitrary (usually ground) Zero at infinity Sign Positive above reference Negative (bound state) Validity h ≪ R E a r t h h \ll R_{\mathrm{Earth}} h ≪ R Earth Any distance Gradient − d E p d h = − m g -\dfrac{dE_p}{dh} = -mg − d h d E p = − m g − d E p d r = − G M m r 2 -\dfrac{dE_p}{dr} = -\dfrac{GMm}{r^2} − d r d E p = − r 2 GM m
Find the gravitational potential energy of a 70 k g 70\mathrm{ kg} 70 kg 50 m 50\mathrm{ m} 50 m m g h mgh m g h
E p = 70 × 9.81 × 50 = 34335 J ≈ 34.3 k J E_p = 70 \times 9.81 \times 50 = 34335\mathrm{ J} \approx 34.3\mathrm{ kJ} E p = 70 × 9.81 × 50 = 34335 J ≈ 34.3 kJ A satellite of mass 500 k g 500\mathrm{ kg} 500 kg 300 k m 300\mathrm{ km} 300 km M E = 5.97 × 10 24 k g M_E = 5.97 \times 10^{24}\mathrm{ kg} M E = 5.97 × 1 0 24 kg R E = 6.37 × 10 6 m R_E = 6.37 \times 10^6\mathrm{ m} R E = 6.37 × 1 0 6 m
r = 6.37 × 10 6 + 3.0 × 10 5 = 6.67 × 10 6 m r = 6.37 \times 10^6 + 3.0 \times 10^5 = 6.67 \times 10^6\mathrm{ m} r = 6.37 × 1 0 6 + 3.0 × 1 0 5 = 6.67 × 1 0 6 m E p = − 6.674 × 10 − 11 × 5.97 × 10 24 × 500 6.67 × 10 6 E_p = -\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{6.67 \times 10^6} E p = − 6.67 × 1 0 6 6.674 × 1 0 − 11 × 5.97 × 1 0 24 × 500 = − 1.992 × 10 17 6.67 × 10 6 = − 2.987 × 10 10 J = -\frac{1.992 \times 10^{17}}{6.67 \times 10^6} = -2.987 \times 10^{10}\mathrm{ J} = − 6.67 × 1 0 6 1.992 × 1 0 17 = − 2.987 × 1 0 10 J Escape Velocity
The minimum speed needed to escape a gravitational field:
v e s c = 2 G M r v_{\mathrm{esc}} = \sqrt{\frac{2GM}{r}} v esc = r 2 GM For Earth: v e s c ≈ 11200 m / s ≈ 11.2 k m / s v_{\mathrm{esc}} \approx 11200\mathrm{ m/s} \approx 11.2\mathrm{ km/s} v esc ≈ 11200 m/s ≈ 11.2 km/s
Elastic Potential Energy
Hooke's Law
For a spring obeying Hooke's law (within the elastic limit):
F = − k x F = -kx F = − k x where k k k x x x
Elastic Potential Energy
E e = 1 2 k x 2 E_e = \frac{1}{2}kx^2 E e = 2 1 k x 2 This is also the work done in compressing or extending the spring by x x x
W = ∫ 0 x k x ′ d x ′ = 1 2 k x 2 W = \int_0^x kx'\,dx' = \frac{1}{2}kx^2 W = ∫ 0 x k x ′ d x ′ = 2 1 k x 2 A spring with k = 200 N / m k = 200\mathrm{ N/m} k = 200 N/m 0.1 m 0.1\mathrm{ m} 0.1 m
E e = 1 2 ( 200 ) ( 0.01 ) = 1.0 J E_e = \frac{1}{2}(200)(0.01) = 1.0\mathrm{ J} E e = 2 1 ( 200 ) ( 0.01 ) = 1.0 J Force-Extension Graphs
For a spring obeying Hooke's law, the force-extension graph is a straight line through the origin.
The area under the graph equals the elastic potential energy.
Conservation of Mechanical Energy
Energy Skate Park: Basics
Observe how kinetic energy and gravitational potential energy interchange as a skater moves along a
track. Try changing the skater's mass and the track shape to see how energy is conserved.
Principle
In a system with only conservative forces (gravity, elastic forces), the total mechanical energy is
conserved:
E k + E p + E e = c o n s t a n t E_k + E_p + E_e = \mathrm{constant} E k + E p + E e = constant 1 2 m v 1 2 + m g h 1 + 1 2 k x 1 2 = 1 2 m v 2 2 + m g h 2 + 1 2 k x 2 2 \frac{1}{2}mv_1^2 + mgh_1 + \frac{1}{2}kx_1^2 = \frac{1}{2}mv_2^2 + mgh_2 + \frac{1}{2}kx_2^2 2 1 m v 1 2 + m g h 1 + 2 1 k x 1 2 = 2 1 m v 2 2 + m g h 2 + 2 1 k x 2 2 Conservative vs Non-Conservative Forces
Conservative Forces Non-Conservative Forces Gravity Friction Elastic (spring) Air resistance Electrostatic Applied pushes/pulls Work is path-independent Work is path-dependent
Applications
A 2 k g 2\mathrm{ kg} 2 kg 10 m 10\mathrm{ m} 10 m
E k ( t o p ) + E p ( t o p ) = E k ( b o t t o m ) + E p ( b o t t o m ) E_k(\mathrm{top}) + E_p(\mathrm{top}) = E_k(\mathrm{bottom}) + E_p(\mathrm{bottom}) E k ( top ) + E p ( top ) = E k ( bottom ) + E p ( bottom ) 0 + 2 ( 9.81 ) ( 10 ) = 1 2 ( 2 ) v 2 + 0 0 + 2(9.81)(10) = \frac{1}{2}(2)v^2 + 0 0 + 2 ( 9.81 ) ( 10 ) = 2 1 ( 2 ) v 2 + 0 196.2 = v 2 196.2 = v^2 196.2 = v 2 v = 14.0 m / s v = 14.0\mathrm{ m/s} v = 14.0 m/s A pendulum of length 1.5 m 1.5\mathrm{ m} 1.5 m
Taking the lowest point as reference (E p = 0 E_p = 0 E p = 0
E p ( t o p ) = m g L = m ( 9.81 ) ( 1.5 ) E_p(\mathrm{top}) = mgL = m(9.81)(1.5) E p ( top ) = m g L = m ( 9.81 ) ( 1.5 ) E k ( b o t t o m ) = 1 2 m v 2 E_k(\mathrm{bottom}) = \frac{1}{2}mv^2 E k ( bottom ) = 2 1 m v 2 By conservation: m g L = 1 2 m v 2 mgL = \dfrac{1}{2}mv^2 m g L = 2 1 m v 2
v = 2 g L = 2 ( 9.81 ) ( 1.5 ) = 29.43 = 5.42 m / s v = \sqrt{2gL} = \sqrt{2(9.81)(1.5)} = \sqrt{29.43} = 5.42\mathrm{ m/s} v = 2 g L = 2 ( 9.81 ) ( 1.5 ) = 29.43 = 5.42 m/s A block of mass 0.5 k g 0.5\mathrm{ kg} 0.5 kg 3 m 3\mathrm{ m} 3 m μ k = 0.4 \mu_k = 0.4 μ k = 0.4
At the bottom of the ramp, all E p E_p E p E k E_k E k
1 2 ( 0.5 ) v 2 = 0.5 ( 9.81 ) ( 3 ) ⟹ v 2 = 58.86 ⟹ v = 7.67 m / s \frac{1}{2}(0.5)v^2 = 0.5(9.81)(3) \implies v^2 = 58.86 \implies v = 7.67\mathrm{ m/s} 2 1 ( 0.5 ) v 2 = 0.5 ( 9.81 ) ( 3 ) ⟹ v 2 = 58.86 ⟹ v = 7.67 m/s Friction does work to stop the block:
μ k m g ⋅ d = 1 2 m v 2 \mu_k mg \cdot d = \frac{1}{2}mv^2 μ k m g ⋅ d = 2 1 m v 2 0.4 ( 0.5 ) ( 9.81 ) d = 1 2 ( 0.5 ) ( 58.86 ) 0.4(0.5)(9.81)d = \frac{1}{2}(0.5)(58.86) 0.4 ( 0.5 ) ( 9.81 ) d = 2 1 ( 0.5 ) ( 58.86 ) 1.962 d = 14.72 ⟹ d = 7.50 m 1.962d = 14.72 \implies d = 7.50\mathrm{ m} 1.962 d = 14.72 ⟹ d = 7.50 m Alternatively, using energy directly:
m g h = μ k m g d ⟹ d = h μ k = 3 0.4 = 7.5 m mgh = \mu_k mg d \implies d = \dfrac{h}{\mu_k} = \dfrac{3}{0.4} = 7.5\mathrm{ m} m g h = μ k m g d ⟹ d = μ k h = 0.4 3 = 7.5 m
Work Done by Non-Conservative Forces
When non-conservative forces (like friction) are present:
W n c = Δ E k + Δ E p W_{\mathrm{nc}} = \Delta E_k + \Delta E_p W nc = Δ E k + Δ E p Or equivalently:
E k , i + E p , i + W n c = E k , f + E p , f E_{k,i} + E_{p,i} + W_{\mathrm{nc}} = E_{k,f} + E_{p,f} E k , i + E p , i + W nc = E k , f + E p , f where W n c W_{\mathrm{nc}} W nc
A 3 k g 3\mathrm{ kg} 3 kg 5 m 5\mathrm{ m} 5 m 30 ° 30\degree 30° μ k = 0.2 \mu_k = 0.2 μ k = 0.2
m g h − μ k m g cos θ ⋅ d = 1 2 m v 2 mgh - \mu_k mg\cos\theta \cdot d = \frac{1}{2}mv^2 m g h − μ k m g cos θ ⋅ d = 2 1 m v 2 3 ( 9.81 ) ( 5 sin 30 ° ) − 0.2 ( 3 ) ( 9.81 ) cos 30 ° × 5 sin 30 ° = 1 2 ( 3 ) v 2 3(9.81)(5\sin 30\degree) - 0.2(3)(9.81)\cos 30\degree \times \frac{5}{\sin 30\degree} = \frac{1}{2}(3)v^2 3 ( 9.81 ) ( 5 sin 30° ) − 0.2 ( 3 ) ( 9.81 ) cos 30° × sin 30° 5 = 2 1 ( 3 ) v 2 Wait, d = h sin θ = 5 sin 30 ° sin 30 ° = 5 m d = \dfrac{h}{\sin\theta} = \dfrac{5\sin 30\degree}{\sin 30\degree} = 5\mathrm{ m} d = sin θ h = sin 30° 5 sin 30° = 5 m h = 5 sin 30 ° = 2.5 m h = 5\sin 30\degree = 2.5\mathrm{ m} h = 5 sin 30° = 2.5 m
3 ( 9.81 ) ( 2.5 ) − 0.2 ( 3 ) ( 9.81 ) ( 0.866 ) ( 10 ) = 1 2 ( 3 ) v 2 3(9.81)(2.5) - 0.2(3)(9.81)(0.866)(10) = \frac{1}{2}(3)v^2 3 ( 9.81 ) ( 2.5 ) − 0.2 ( 3 ) ( 9.81 ) ( 0.866 ) ( 10 ) = 2 1 ( 3 ) v 2 73.58 − 50.97 = 1.5 v 2 73.58 - 50.97 = 1.5v^2 73.58 − 50.97 = 1.5 v 2 22.61 = 1.5 v 2 ⟹ v = 3.88 m / s 22.61 = 1.5v^2 \implies v = 3.88\mathrm{ m/s} 22.61 = 1.5 v 2 ⟹ v = 3.88 m/s Power
Definition
Power is the rate at which work is done or energy is transferred:
P = W t = Δ E t P = \frac{W}{t} = \frac{\Delta E}{t} P = t W = t Δ E For a constant force:
P = F v P = Fv P = F v Units
SI unit: watt (W \mathrm{W} W 1 W = 1 J / s 1\mathrm{ W} = 1\mathrm{ J/s} 1 W = 1 J/s
Other units: kilowatt (1 k W = 1000 W 1\mathrm{ kW} = 1000\mathrm{ W} 1 kW = 1000 W 1 h p ≈ 746 W 1\mathrm{ hp} \approx 746\mathrm{ W} 1 hp ≈ 746 W
Power and Inclined Planes
For an object moving up an incline at constant speed v v v
P = F v = ( m g sin θ + f ) v P = Fv = (mg\sin\theta + f)v P = F v = ( m g sin θ + f ) v Power and Vehicles
For a car on a level road at maximum speed (driving force equals drag):
P = F v = f d r a g ⋅ v P = Fv = f_{\mathrm{drag}} \cdot v P = F v = f drag ⋅ v Since drag increases with speed, there is a maximum speed where
P = f d r a g ⋅ v max P = f_{\mathrm{drag}} \cdot v_{\max} P = f drag ⋅ v m a x
A car engine produces 80 k W 80\mathrm{ kW} 80 kW 800 N 800\mathrm{ N} 800 N
P = F v ⟹ 80000 = 800 v ⟹ v = 100 m / s P = Fv \implies 80000 = 800v \implies v = 100\mathrm{ m/s} P = F v ⟹ 80000 = 800 v ⟹ v = 100 m/s This is 360 k m / h 360\mathrm{ km/h} 360 km/h 80 k W 80\mathrm{ kW} 80 kW v 2 v^2 v 2
Efficiency of Energy Transfers
Definition
E f f i c i e n c y = u s e f u l e n e r g y o u t p u t t o t a l e n e r g y i n p u t × 100 % \mathrm{Efficiency} = \frac{\mathrm{useful energy output}}{\mathrm{total energy input}} \times 100\% Efficiency = totalenergyinput usefulenergyoutput × 100% Energy Degradation
In all real energy transfers, some energy is dissipated (usually as thermal energy due to friction).
This means:
Efficiency is always less than 100%.
Total energy is always conserved, but useful energy decreases.
The "lost" energy is not destroyed — it is transferred to the surroundings as heat.
Sankey Diagrams
Sankey diagrams visually represent energy flows:
The width of each arrow is proportional to the amount of energy.
The input energy splits into useful output and wasted energy.
A light bulb converts 100 J 100\mathrm{ J} 100 J 10 J 10\mathrm{ J} 10 J 90 J 90\mathrm{ J} 90 J
E f f i c i e n c y = 10 100 × 100 % = 10 % \mathrm{Efficiency} = \frac{10}{100} \times 100\% = 10\% Efficiency = 100 10 × 100% = 10% Power input = 100 W = 100\mathrm{ W} = 100 W = 10 W = 10\mathrm{ W} = 10 W
Common Efficiencies
Device Typical Efficiency Incandescent light bulb 5--10% LED light bulb 30--40% Electric motor 70--95% Car engine (petrol) 20--30% Diesel engine 30--40% Steam turbine 35--45% Solar cell 15--25% Human body 20--25%
Energy in Simple Harmonic Motion
Total Energy in SHM
In simple harmonic motion, energy continuously converts between kinetic and potential:
E t o t a l = E k + E p = 1 2 k A 2 = 1 2 m ω 2 A 2 E_{\mathrm{total}} = E_k + E_p = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2 E total = E k + E p = 2 1 k A 2 = 2 1 m ω 2 A 2 where A A A
Energy as a Function of Position
E k ( x ) = 1 2 k ( A 2 − x 2 ) E_k(x) = \frac{1}{2}k(A^2 - x^2) E k ( x ) = 2 1 k ( A 2 − x 2 ) E p ( x ) = 1 2 k x 2 E_p(x) = \frac{1}{2}kx^2 E p ( x ) = 2 1 k x 2 Energy as a Function of Time
E k ( t ) = 1 2 k A 2 cos 2 ( ω t ) E_k(t) = \frac{1}{2}kA^2\cos^2(\omega t) E k ( t ) = 2 1 k A 2 cos 2 ( ω t ) E p ( t ) = 1 2 k A 2 sin 2 ( ω t ) E_p(t) = \frac{1}{2}kA^2\sin^2(\omega t) E p ( t ) = 2 1 k A 2 sin 2 ( ω t ) The total energy remains constant at all times.
IB Exam-Style Questions
Question 1 (Paper 1 style)
A roller coaster car of mass 500 k g 500\mathrm{ kg} 500 kg A which is
30 m 30\mathrm{ m} 30 m B which is 10 m 10\mathrm{ m} 10 m B.
m g h A = m g h B + 1 2 m v B 2 mgh_A = mgh_B + \frac{1}{2}mv_B^2 m g h A = m g h B + 2 1 m v B 2 9.81 ( 30 ) = 9.81 ( 10 ) + 1 2 v B 2 9.81(30) = 9.81(10) + \frac{1}{2}v_B^2 9.81 ( 30 ) = 9.81 ( 10 ) + 2 1 v B 2 294.3 = 98.1 + 0.5 v B 2 294.3 = 98.1 + 0.5v_B^2 294.3 = 98.1 + 0.5 v B 2 v B 2 = 392.4 ⟹ v B = 19.8 m / s v_B^2 = 392.4 \implies v_B = 19.8\mathrm{ m/s} v B 2 = 392.4 ⟹ v B = 19.8 m/s Question 2 (Paper 2 style)
A spring-loaded launcher has spring constant k = 500 N / m k = 500\mathrm{ N/m} k = 500 N/m 0.15 m 0.15\mathrm{ m} 0.15 m 0.2 k g 0.2\mathrm{ kg} 0.2 kg
(a) Find the speed of the ball as it leaves the launcher.
1 2 k x 2 = 1 2 m v 2 \frac{1}{2}kx^2 = \frac{1}{2}mv^2 2 1 k x 2 = 2 1 m v 2 v = x k m = 0.15 500 0.2 = 0.15 2500 = 0.15 × 50 = 7.5 m / s v = x\sqrt{\frac{k}{m}} = 0.15\sqrt{\frac{500}{0.2}} = 0.15\sqrt{2500} = 0.15 \times 50 = 7.5\mathrm{ m/s} v = x m k = 0.15 0.2 500 = 0.15 2500 = 0.15 × 50 = 7.5 m/s (b) Find the maximum height reached (from the launch point).
1 2 m v 2 = m g h \frac{1}{2}mv^2 = mgh 2 1 m v 2 = m g h h = v 2 2 g = 56.25 19.62 = 2.87 m h = \frac{v^2}{2g} = \frac{56.25}{19.62} = 2.87\mathrm{ m} h = 2 g v 2 = 19.62 56.25 = 2.87 m Question 3 (Paper 2 style)
A pump lifts 500 k g 500\mathrm{ kg} 500 kg 15 m 15\mathrm{ m} 15 m
P_{\mathrm{useful}} = \frac`\{mgh}`{t} = \frac{500 \times 9.81 \times 15}{60} = 1226\mathrm{ W}
P i n p u t = P u s e f u l η = 1226 0.65 = 1887 W ≈ 1.89 k W P_{\mathrm{input}} = \frac{P_{\mathrm{useful}}}{\eta} = \frac{1226}{0.65} = 1887\mathrm{ W} \approx 1.89\mathrm{ kW} P input = η P useful = 0.65 1226 = 1887 W ≈ 1.89 kW Question 4 (Paper 1 style)
A satellite of mass 200 k g 200\mathrm{ kg} 200 kg 500 k m 500\mathrm{ km} 500 km M E = 5.97 × 10 24 k g M_E = 5.97 \times 10^{24}\mathrm{ kg} M E = 5.97 × 1 0 24 kg R E = 6.37 × 10 6 m R_E = 6.37 \times 10^6\mathrm{ m} R E = 6.37 × 1 0 6 m
(a) Find the orbital speed.
r = 6.87 × 10 6 m r = 6.87 \times 10^6\mathrm{ m} r = 6.87 × 1 0 6 m \frac`\{GMm}`{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac`\{GM}`{r}} = \sqrt{\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.87 \times 10^6}}
v = 5.80 × 10 7 = 7616 m / s v = \sqrt{5.80 \times 10^7} = 7616\mathrm{ m/s} v = 5.80 × 1 0 7 = 7616 m/s (b) Find the total mechanical energy.
E_{\mathrm{total}} = E_k + E_p = \frac{1}{2}mv^2 - \frac`\{GMm}`{r} = \frac{1}{2}m\frac`\{GM}`{r} - \frac`\{GMm}`{r} = -\frac`\{GMm}`{2r}
E = − 6.674 × 10 − 11 × 5.97 × 10 24 × 200 2 × 6.87 × 10 6 E = -\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 200}{2 \times 6.87 \times 10^6} E = − 2 × 6.87 × 1 0 6 6.674 × 1 0 − 11 × 5.97 × 1 0 24 × 200 = − 7.97 × 10 16 1.374 × 10 7 = − 5.80 × 10 9 J = -\frac{7.97 \times 10^{16}}{1.374 \times 10^7} = -5.80 \times 10^9\mathrm{ J} = − 1.374 × 1 0 7 7.97 × 1 0 16 = − 5.80 × 1 0 9 J Question 5 (Paper 2 style)
A 60 k g 60\mathrm{ kg} 60 kg 5 m 5\mathrm{ m} 5 m 0.5 m 0.5\mathrm{ m} 0.5 m
At the lowest point, all energy is elastic potential energy:
m g ( h + x ) = 1 2 k x 2 mg(h + x) = \frac{1}{2}kx^2 m g ( h + x ) = 2 1 k x 2 60 ( 9.81 ) ( 5.5 ) = 1 2 k ( 0.25 ) 60(9.81)(5.5) = \frac{1}{2}k(0.25) 60 ( 9.81 ) ( 5.5 ) = 2 1 k ( 0.25 ) 3237.3 = 0.125 k ⟹ k = 25898 N / m ≈ 26000 N / m 3237.3 = 0.125k \implies k = 25898\mathrm{ N/m} \approx 26000\mathrm{ N/m} 3237.3 = 0.125 k ⟹ k = 25898 N/m ≈ 26000 N/m Summary
Energy Type Formula Kinetic E k = 1 2 m v 2 E_k = \dfrac{1}{2}mv^2 E k = 2 1 m v 2 Gravitational PE (near surface) E p = m g h E_p = mgh E p = m g h Gravitational PE (universal) E p = − G M m r E_p = -\dfrac{GMm}{r} E p = − r GM m Elastic PE E e = 1 2 k x 2 E_e = \dfrac{1}{2}kx^2 E e = 2 1 k x 2 Conservation E k + E p + E e = c o n s t a n t E_k + E_p + E_e = \mathrm{constant} E k + E p + E e = constant Work-energy theorem W n e t = Δ E k W_{\mathrm{net}} = \Delta E_k W net = Δ E k Power P = W t = F v P = \dfrac{W}{t} = Fv P = t W = F v Efficiency η = E o u t E i n × 100 % \eta = \dfrac{E_{\mathrm{out}}}{E_{\mathrm{in}}} \times 100\% η = E in E out × 100%
For energy conservation problems, always define your reference level for gravitational PE. Show the
energy at two clearly chosen points (usually start and end). When friction is present, subtract the
work done by friction from the total mechanical energy.
Energy in Gravitational Fields (Extended)
Gravitational Potential Energy Curves
For two masses M M M m m m r r r
| Total Energy E E E E < 0 E \lt 0 E < 0 ∣ E ∣ < ∣ E p ∣ | E | \lt | E_p | ∣ E ∣ < ∣ E p ∣ E = 0 E = 0 E = 0 E > 0 E \gt 0 E > 0
For a circular orbit:
E_{\mathrm{total}} = E_k + E_p = \frac{1}{2}mv^2 - \frac`\{GMm}`{r} = \frac`\{GMm}`{2r} - \frac`\{GMm}`{r} = -\frac`\{GMm}`{2r}
Energy to Change Orbits
To move from one circular orbit to another, energy must be supplied. The minimum energy required is
the difference in total orbital energies.
Find the energy required to move a 1000 k g 1000\mathrm{ kg} 1000 kg 300 k m 300\mathrm{ km} 300 km 600 k m 600\mathrm{ km} 600 km
(M E = 5.97 × 10 24 k g M_E = 5.97 \times 10^{24}\mathrm{ kg} M E = 5.97 × 1 0 24 kg R E = 6.37 × 10 6 m R_E = 6.37 \times 10^6\mathrm{ m} R E = 6.37 × 1 0 6 m
r 1 = 6.67 × 10 6 m , r 2 = 6.97 × 10 6 m r_1 = 6.67 \times 10^6\mathrm{ m}, \quad r_2 = 6.97 \times 10^6\mathrm{ m} r 1 = 6.67 × 1 0 6 m , r 2 = 6.97 × 1 0 6 m E_1 = -\frac`\{GMm}`{2r_1} = -\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000}{2 \times 6.67 \times 10^6} = − 3.985 × 10 17 1.334 × 10 7 = − 2.988 × 10 10 J = -\frac{3.985 \times 10^{17}}{1.334 \times 10^7} = -2.988 \times 10^{10}\mathrm{ J} = − 1.334 × 1 0 7 3.985 × 1 0 17 = − 2.988 × 1 0 10 J E_2 = -\frac`\{GMm}`{2r_2} = -\frac{3.985 \times 10^{17}}{1.394 \times 10^7} = -2.858 \times 10^{10}\mathrm{ J} Δ E = E 2 − E 1 = − 2.858 × 10 10 − ( − 2.988 × 10 10 ) = 1.30 × 10 9 J \Delta E = E_2 - E_1 = -2.858 \times 10^{10} - (-2.988 \times 10^{10}) = 1.30 \times 10^9\mathrm{ J} Δ E = E 2 − E 1 = − 2.858 × 1 0 10 − ( − 2.988 × 1 0 10 ) = 1.30 × 1 0 9 J The energy required is 1.30 × 10 9 J 1.30 \times 10^9\mathrm{ J} 1.30 × 1 0 9 J 1.30 G J 1.30\mathrm{ GJ} 1.30 GJ
Work Done by a Variable Force
When the force varies with position, the work done is the area under the force-displacement graph:
W = ∫ x 1 x 2 F ( x ) d x W = \int_{x_1}^{x_2} F(x)\,dx W = ∫ x 1 x 2 F ( x ) d x A spring obeys Hooke's law: F = − k x F = -kx F = − k x x = 0 x = 0 x = 0 x = − d x = -d x = − d
W = ∫ 0 − d ( − k x ) d x = [ − k x 2 2 ] 0 − d = − k ( − d ) 2 2 = − k d 2 2 W = \int_0^{-d} (-kx)\,dx = \left[-\frac{kx^2}{2}\right]_0^{-d} = -\frac{k(-d)^2}{2} = -\frac{kd^2}{2} W = ∫ 0 − d ( − k x ) d x = [ − 2 k x 2 ] 0 − d = − 2 k ( − d ) 2 = − 2 k d 2 The negative sign indicates work is done on the spring (energy stored). The elastic potential energy
is 1 2 k d 2 \dfrac{1}{2}kd^2 2 1 k d 2
Force-Extension Graphs for Non-Hookean Materials
For materials that do not obey Hooke's law, the area under the force-extension graph still equals
the elastic potential energy, but it must be found by integration or by counting squares.
Power in Rotational Systems
For rotational systems:
P = τ ω P = \tau\omega P = τ ω where τ \tau τ ω \omega ω
A motor delivers a torque of 50 N ⋅ m 50\mathrm{ N}\cdot\mathrm{m} 50 N ⋅ m 3000 r p m 3000\mathrm{ rpm} 3000 rpm
ω = 3000 × 2 π 60 = 314.2 r a d / s \omega = 3000 \times \frac{2\pi}{60} = 314.2\mathrm{ rad/s} ω = 3000 × 60 2 π = 314.2 rad/s P = 50 × 314.2 = 15708 W ≈ 15.7 k W P = 50 \times 314.2 = 15708\mathrm{ W} \approx 15.7\mathrm{ kW} P = 50 × 314.2 = 15708 W ≈ 15.7 kW Energy Dissipation and Thermal Effects
Friction and Heat
When friction does work W f W_f W f
Q = W f = μ k N d Q = W_f = \mu_k Nd Q = W f = μ k N d Air Resistance
Air resistance converts kinetic energy to thermal energy:
Δ E k = W d r a g = ∫ F d r a g d x \Delta E_k = W_{\mathrm{drag}} = \int F_{\mathrm{drag}}\,dx Δ E k = W drag = ∫ F drag d x Additional IB Exam-Style Questions
Question 6 (Paper 2 style)
A 0.5 k g 0.5\mathrm{ kg} 0.5 kg 1.0 m 1.0\mathrm{ m} 1.0 m
(a) Find the tension in the string at the lowest point.
At the lowest point, all E p E_p E p E k E_k E k
1 2 ( 0.5 ) v 2 = 0.5 ( 9.81 ) ( 1.0 ) ⟹ v 2 = 19.62 ⟹ v = 4.43 m / s \frac{1}{2}(0.5)v^2 = 0.5(9.81)(1.0) \implies v^2 = 19.62 \implies v = 4.43\mathrm{ m/s} 2 1 ( 0.5 ) v 2 = 0.5 ( 9.81 ) ( 1.0 ) ⟹ v 2 = 19.62 ⟹ v = 4.43 m/s For circular motion at the lowest point:
T − m g = m v 2 r T - mg = \frac{mv^2}{r} T − m g = r m v 2 T = m g + m v 2 r = 0.5 ( 9.81 ) + 0.5 ( 19.62 ) 1.0 = 4.905 + 9.81 = 14.72 N T = mg + \frac{mv^2}{r} = 0.5(9.81) + \frac{0.5(19.62)}{1.0} = 4.905 + 9.81 = 14.72\mathrm{ N} T = m g + r m v 2 = 0.5 ( 9.81 ) + 1.0 0.5 ( 19.62 ) = 4.905 + 9.81 = 14.72 N (b) Find the speed when the string makes an angle of 45 ° 45\degree 45°
Height above lowest point: h = L − L cos 45 ° = 1.0 − 0.707 = 0.293 m h = L - L\cos 45\degree = 1.0 - 0.707 = 0.293\mathrm{ m} h = L − L cos 45° = 1.0 − 0.707 = 0.293 m
m g h = 1 2 m v 2 mgh = \frac{1}{2}mv^2 m g h = 2 1 m v 2 v = 2 g h = 2 ( 9.81 ) ( 0.293 ) = 5.75 = 2.40 m / s v = \sqrt{2gh} = \sqrt{2(9.81)(0.293)} = \sqrt{5.75} = 2.40\mathrm{ m/s} v = 2 g h = 2 ( 9.81 ) ( 0.293 ) = 5.75 = 2.40 m/s Question 7 (Paper 2 style)
A car of mass 1500 k g 1500\mathrm{ kg} 1500 kg 5 ° 5\degree 5° 20 m / s 20\mathrm{ m/s} 20 m/s 400 N 400\mathrm{ N} 400 N
(a) Calculate the driving force required.
F = m g sin θ + f = 1500 ( 9.81 ) sin 5 ° + 400 = 1283 + 400 = 1683 N F = mg\sin\theta + f = 1500(9.81)\sin 5\degree + 400 = 1283 + 400 = 1683\mathrm{ N} F = m g sin θ + f = 1500 ( 9.81 ) sin 5° + 400 = 1283 + 400 = 1683 N (b) Calculate the power output of the engine.
P = F v = 1683 × 20 = 33660 W ≈ 33.7 k W P = Fv = 1683 \times 20 = 33660\mathrm{ W} \approx 33.7\mathrm{ kW} P = F v = 1683 × 20 = 33660 W ≈ 33.7 kW (c) If the engine efficiency is 25 % 25\% 25%
P i n p u t = 33660 0.25 = 134640 W = 134.6 k W P_{\mathrm{input}} = \frac{33660}{0.25} = 134640\mathrm{ W} = 134.6\mathrm{ kW} P input = 0.25 33660 = 134640 W = 134.6 kW Question 8 (Paper 1 style)
A spring with k = 500 N / m k = 500\mathrm{ N/m} k = 500 N/m 0.1 k g 0.1\mathrm{ kg} 0.1 kg 0.08 m 0.08\mathrm{ m} 0.08 m
1 2 k x 2 = m g h \frac{1}{2}kx^2 = mgh 2 1 k x 2 = m g h h = k x 2 2 m g = 500 ( 0.0064 ) 2 ( 0.1 ) ( 9.81 ) = 3.2 1.962 = 1.63 m h = \frac{kx^2}{2mg} = \frac{500(0.0064)}{2(0.1)(9.81)} = \frac{3.2}{1.962} = 1.63\mathrm{ m} h = 2 m g k x 2 = 2 ( 0.1 ) ( 9.81 ) 500 ( 0.0064 ) = 1.962 3.2 = 1.63 m For the A-Level treatment of this topic, see Work, Energy and Power .
Diagnostic Test
Ready to test your understanding of Energy and Conservation ? The diagnostic test contains the hardest questions within the IB specification for this topic, each with a full worked solution.
Unit tests probe edge cases and common misconceptions. Integration tests combine Energy and Conservation with other physics topics to test synthesis under exam conditions.
See Diagnostic Guide for instructions on self-marking and building a personal test matrix.