Energy & Fields: stores, work, fields and power
Key idea: Track energy transfers consistently, connect work to kinetic and potential-energy change, then apply power and efficiency.
Before you start: Forces & DynamicsMotion & Forces
By the end, you can
- Describe energy stores and transfers and apply conservation of energy.
- Define work, derive kinetic energy from work and uniformly accelerated motion, and use Eₖ = ½mv².
- Represent gravitational and electric fields and relate field work to potential-energy change.
- Distinguish gravitational, electric and elastic potential energy and use force–extension graph area.
- Apply energy-transfer rate, mechanical power and efficiency, including practical energy losses.
Starting-point self-check
1. Check your starting point
Attempt all three groups without notes and mark the first equation or energy transfer you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Energy stores, work and kinetic energy 4(a)–(e)
Question 1
A 2.0 kg trolley starts from rest and a constant 6.0 N resultant force moves it 3.0 m. State the energy transfer, find the work done and hence find its speed.
Check the model response
Work is a mechanical transfer into the trolley’s kinetic-energy store. W = Fs = 18 J. With 18 = ½(2.0)v², v = 4.24 m s⁻¹.
Question 2
A falling ball loses 12 J of gravitational potential energy and air resistance transfers 3 J to thermal stores. Find its kinetic-energy gain and explain whether energy was lost.
Check the model response
Kinetic-energy gain = 12 − 3 = 9 J. Energy was not destroyed; 3 J was transferred to thermal stores in the ball, air and surroundings.
repair
2. Repair the six common breaks
Use only the repair matching an error, then repeat the corresponding diagnostic model.
Energy stores, work and kinetic energy 4(a)–(e)
Check this idea
Misconception: An inefficient device destroys energy.
Repair: Energy is conserved. Inefficiency means some input is transferred to less useful stores, commonly thermal stores in the device and surroundings.
Check this idea
Misconception: Every force does work equal to force times path length.
Repair: For a constant force, use displacement in the force direction: W = Fs cos θ. A perpendicular force does no work.
worked example
3. Follow three worked models
Track the system, transfer, direction and graph area before substituting values.
Energy stores, work and kinetic energy 4(a)–(e)
Model 1
Derive Eₖ = ½mv² for a body accelerated from rest by a constant resultant force.
Check the model response
W = Fs and F = ma, so W = mas. From v² = u² + 2as with u = 0, as = v²/2. Therefore W = m(v²/2) = ½mv². The work is the kinetic-energy increase.
guided practice
4. Guided practice
Use each hint only to select the governing relationship.
Energy stores, work and kinetic energy 4(a)–(e)
Question 1
A 20 N force acts 60° to a 5.0 m displacement. Find the work done and the kinetic-energy increase if it is the only transfer.
Hint: Use the force component along displacement.
Check the model response
W = Fs cos 60° = 20(5.0)(0.5) = 50 J, so kinetic energy increases by 50 J.
independent practice
5. Independent practice
Solve without repair notes and state every energy store, transfer and sign convention used.
Energy stores, work and kinetic energy 4(a)–(e)
Question 1
A 1200 kg car slows from 20 to 10 m s⁻¹. Find the kinetic-energy change and describe where that energy may be transferred during braking.
Check the model response
ΔEₖ = ½(1200)(10² − 20²) = −1.80 × 10⁵ J. Energy is transferred mainly to thermal stores in brakes, tyres, road and surrounding air.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Energy stores, work and kinetic energy 4(a)–(e)
Question 1
A constant 45 N force moves a 3.0 kg body 8.0 m from rest on a frictionless surface. Derive the kinetic-energy relationship needed, then calculate final speed.
Check the model response
W = Fs = 360 J. Combining W = mas with v² = 2as gives W = ½mv². Thus 360 = ½(3.0)v² and v = 15.5 m s⁻¹.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Energy stores, work and kinetic energy 4(a)–(e)
Question 1
A 4.0 kg object moving at 3.0 m s⁻¹ receives 82 J of net work. Find its final kinetic energy and speed.
Check the model response
Initial Eₖ = ½(4.0)(3.0²) = 18 J. Final Eₖ = 100 J, so 100 = ½(4.0)v² and v = 7.07 m s⁻¹.