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This lesson covers AQA Spec 3.4.1.7–3.4.1.8 (Work, energy and power) in full depth, including work done at angles, energy transformations, efficiency, and the relationship P = Fv.
Spec mapping (AQA 7408 §3.4.1.7–§3.4.1.8): This lesson develops the definition of work done as the dot product W = Fs cos θ, the work-energy theorem (net work equals change in kinetic energy), gravitational and elastic potential energy, the principle of conservation of energy across mechanical systems, power as the rate of doing work (with the operating identity P = Fv at constant speed), and efficiency as the ratio of useful output to total input. The treatment includes work done by variable forces using force-displacement graph areas, which leads naturally into elastic potential energy as ½kx². (Refer to the official AQA specification document for exact wording.)
Synoptic links: Energy conservation is the most general tool in physics. (i) Kinematics (Lesson 1, §3.4.1.2) delivers v from u, a, s; energy conservation delivers v from u, h directly without going via SUVAT, and the two routes cross-check each other. (ii) Springs and stress-strain (Lessons 7-8, §3.4.2) uses the same ½kx² result for elastic potential energy, and the area under a force-extension graph is the energy stored — a direct generalisation of work as area under F-s graph. (iii) Electric circuits (Year 13, §3.7) treats P = IV (power in a resistor) as a special case of P = rate of energy transfer; the dimensional consistency is the same. (iv) Thermal physics (Year 13, §3.6.2) extends conservation of energy to include internal energy and heat flow, with the first law of thermodynamics ΔU = Q + W formalising the idea that work and heat are both ways of transferring energy.
Key Definition: Work done by a force is the product of the force and the displacement in the direction of the force.
W = Fs cos θ
where θ is the angle between the force and the displacement.
SI unit of work: joule (J). 1 J = 1 N m.
A person pulls a suitcase 50 m along a horizontal floor using a force of 40 N at 30° above the horizontal. Find the work done.
Solution:
W = Fs cos θ = 40 × 50 × cos 30° = 40 × 50 × 0.866 = 1732 J ≈ 1.73 kJ
Exam Tip: Only the component of force in the direction of motion does work. A force perpendicular to the motion (like the normal contact force on a horizontal surface) does zero work.
When an object of mass m is raised through a vertical height h:
W = mgh
This equals the gain in gravitational potential energy.
A crane lifts a 500 kg girder through a vertical height of 20 m in 15 s. Find (a) the work done against gravity and (b) the power of the crane.
Solution:
(a) W = mgh = 500 × 9.81 × 20 = 98 100 J ≈ 98.1 kJ
(b) P = W/t = 98 100/15 = 6540 W ≈ 6.54 kW
KE = ½mv²
This is the energy an object has due to its motion. Derived from the work-energy theorem: the net work done on an object equals its change in kinetic energy.
Derivation: From F = ma and v² = u² + 2as:
W = Fs = mas = m × (v² − u²)/(2s) × s = ½mv² − ½mu²
Therefore: W_net = ΔKE = ½mv² − ½mu² (the work-energy theorem)
GPE = mgh
This is the energy stored due to an object's position in a gravitational field, where h is the height above a chosen reference level.
A 600 kg roller coaster car starts from rest at the top of a 35 m hill. Assuming no friction, find its speed at the bottom.
Solution: By conservation of energy:
Loss in GPE = Gain in KE
mgh = ½mv²
v² = 2gh = 2 × 9.81 × 35 = 686.7
v = √686.7 = 26.2 m s⁻¹
Key Point: Mass cancels — the speed at the bottom is independent of mass (when friction is negligible).
The same roller coaster car loses 50 000 J to friction. Find the speed at the bottom.
Solution:
GPE = KE + Energy lost to friction
mgh = ½mv² + 50 000
600 × 9.81 × 35 = ½ × 600 × v² + 50 000
206 010 = 300v² + 50 000
300v² = 156 010
v² = 520.0
v = 22.8 m s⁻¹
Key Definition: Power is the rate of doing work (or the rate of energy transfer).
P = W/t = E/t
SI unit: watt (W). 1 W = 1 J s⁻¹.
If a constant force F moves an object at constant velocity v:
P = W/t = Fs/t = F × (s/t) = Fv
This is extremely useful for problems involving vehicles at constant velocity.
A car of mass 1200 kg travels at a constant speed of 30 m s⁻¹ on a horizontal road. The total resistive force is 800 N. Find (a) the driving force and (b) the power output of the engine.
Solution:
(a) At constant speed, the resultant force is zero (Newton's First Law):
Driving force = resistive force = 800 N
(b) P = Fv = 800 × 30 = 24 000 W = 24 kW
A car has a maximum engine power of 60 kW. The total resistive force is 1500 N at high speed. Find the maximum speed.
Solution: At maximum speed, all power goes to overcoming resistance (acceleration = 0):
P = Fv → v = P/F = 60 000 / 1500 = 40 m s⁻¹ (144 km h⁻¹)
Efficiency = (useful output energy / total input energy) × 100%
Or equivalently:
Efficiency = (useful output power / total input power) × 100%
Efficiency is always less than 100% (some energy is always dissipated as heat, sound, etc.).
An electric motor has an input power of 5.0 kW. It lifts a 200 kg load through 12 m in 8.0 s. Find the efficiency.
Solution:
Useful output power = mgh/t = 200 × 9.81 × 12 / 8.0 = 23 544 / 8.0 = 2943 W
Efficiency = (2943 / 5000) × 100% = 58.9%
When the force varies with displacement, the work done is the area under the force–displacement graph.
For a spring obeying Hooke's law (F = kx):
W = area of triangle = ½ × kx × x = ½kx²
This equals the elastic potential energy stored in the spring:
EPE = ½kx² = ½Fx (where x is the extension and F = kx)
A spring with spring constant k = 40 N m⁻¹ is stretched by 0.25 m. Find (a) the force applied and (b) the elastic potential energy stored.
Solution:
(a) F = kx = 40 × 0.25 = 10 N
(b) EPE = ½kx² = ½ × 40 × 0.25² = ½ × 40 × 0.0625 = 1.25 J
Check: EPE = ½Fx = ½ × 10 × 0.25 = 1.25 J ✓
The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another.
In a closed system, total energy is constant. In practice, energy is dissipated (spread out) into the surroundings, mainly as thermal energy due to friction.
A 0.50 kg pendulum bob is pulled to a height of 0.10 m above its rest position and released. Find (a) its speed at the lowest point (assuming no air resistance) and (b) its speed if 10% of its energy is dissipated by the time it reaches the lowest point.
Solution:
(a) GPE = KE: mgh = ½mv² → v = √(2gh) = √(2 × 9.81 × 0.10) = √1.962 = 1.40 m s⁻¹
(b) 90% of GPE converts to KE: 0.90 × mgh = ½mv²
v = √(2 × 0.90 × 9.81 × 0.10) = √1.766 = 1.33 m s⁻¹
For a vehicle accelerating with a driving force F at instantaneous velocity v, the instantaneous power delivered is P = Fv. As the vehicle speeds up, the available driving force decreases (since maximum engine power is fixed and P = Fv ≤ P_max means F ≤ P_max/v). This explains why acceleration falls off as a car approaches its top speed.
A car of mass 1200 kg has an engine producing a maximum power output of 80 kW. At a given moment, the car is moving at 20 m s⁻¹ on a level road against a total resistive force of 600 N. Find the instantaneous acceleration if the engine delivers maximum power.
Solution: Driving force at this speed:
F_drive = P_max / v = 80 000 / 20 = 4000 N.
Net forward force:
F_net = F_drive − F_resist = 4000 − 600 = 3400 N.
Acceleration:
a = F_net / m = 3400 / 1200 = 2.83 m s⁻².
For the same car, find the top speed on a level road if the resistive force depends on speed as F_resist = 0.50 v² (a typical aerodynamic-drag model).
Solution: At top speed, all power goes to overcoming resistance (a = 0):
P_max = F_drive × v_max = F_resist × v_max = 0.50 v_max² × v_max = 0.50 v_max³.
v_max³ = P_max / 0.50 = 80 000 / 0.50 = 160 000.
v_max = ∛160 000 = 54.3 m s⁻¹ (about 196 km h⁻¹).
This cubic dependence on power is why doubling the power output of a car only increases top speed by a factor of ∛2 ≈ 1.26. Aerodynamic drag at the third power of speed makes high top speeds expensive in engine power.
A 1500 kg car accelerates from rest. The driving force is constant at 4500 N for the first 100 m of the journey; the resistive force grows linearly with distance from 0 N at the start to 800 N at the 100 m mark. Find the car's speed at 100 m.
Solution: Work done by driving force = F × s = 4500 × 100 = 450 000 J.
Work done against resistance is the area under the resistance-versus-distance graph (a triangle from 0 to 800 N over 100 m):
W_resist = ½ × 100 × 800 = 40 000 J.
Net work = 450 000 − 40 000 = 410 000 J.
By the work-energy theorem: W_net = ½mv² − 0.
½ × 1500 × v² = 410 000 → v² = 820 000 / 1500 = 546.7 → v = 23.4 m s⁻¹ (about 84 km h⁻¹).
This example shows the power of the work-energy theorem for problems where the force varies and SUVAT alone would fail. The area under any force-distance graph gives the work done by that force; the net work equals the change in kinetic energy.
AQA Specification Reference: Section 3.4.1.7 (Work, energy and power), 3.4.1.8 (Conservation of energy).
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