A-Level — physicsMedium4 min read

Work, Energy, and Power

W = Fs cos θ; kinetic and potential energy; P = Fv; work-energy theorem

Tutor AI Team
Tutor AI Team

# Work, Energy, and Power — A-Level Physics

Work, energy, and power form a central pillar of mechanics. The work-energy theorem provides an alternative to Newton's Laws for solving motion problems, and conservation of energy is one of the most powerful principles in all of physics.

1. Work Done

W=Fscos⁡θ\boxed{W = Fs\cos\theta}

Where:

  • WW = work done (J)
  • FF = force (N)
  • ss = displacement (m)
  • θ\theta = angle between force and displacement

Key Points

  • Work is a scalar quantity
  • 1 joule = 1 N × 1 m = 1 kg m²/s²
  • If θ=0°\theta = 0°: W=FsW = Fs (force along displacement)
  • If θ=90°\theta = 90°: W=0W = 0 (force perpendicular — e.g., centripetal force, normal force on flat surface)
  • If θ>90°\theta > 90°: W<0W < 0 (force opposes motion — e.g., friction)

For a variable force, work = area under the force-displacement graph: W=∫F dsW = \int F \, ds

2. Work-Energy Theorem

Wnet=ΔKE=12mv2−12mu2\boxed{W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2}

The net work done on an object equals its change in kinetic energy.

3. Kinetic Energy

KE=12mv2KE = \frac{1}{2}mv^2

Derived from work-energy theorem: W=Fs=masW = Fs = mas. Using v2=u2+2asv^2 = u^2 + 2as with u=0u = 0: W=m⋅v22s⋅s=12mv2W = m \cdot \frac{v^2}{2s} \cdot s = \frac{1}{2}mv^2.

4. Potential Energy

Gravitational PE (near Earth's surface)

GPE=mghGPE = mgh

Elastic PE (spring)

EPE=12ke2EPE = \frac{1}{2}ke^2

Where kk = spring constant (N/m), ee = extension (m).

This comes from: W=∫0ekx dx=12ke2W = \int_0^e kx \, dx = \frac{1}{2}ke^2

5. Conservation of Energy

In a closed system with only conservative forces (gravity, elastic):

KE1+PE1=KE2+PE2KE_1 + PE_1 = KE_2 + PE_2

With non-conservative forces (friction): KE1+PE1=KE2+PE2+WfrictionKE_1 + PE_1 = KE_2 + PE_2 + W_{\text{friction}}

6. Power

P=Wt=Et\boxed{P = \frac{W}{t} = \frac{E}{t}}

Also, for constant force: P=Fv\boxed{P = Fv}

Where vv is the velocity.

This is particularly useful for:

  • A car moving at constant velocity against resistive forces: P=Fresistance×vP = F_{\text{resistance}} \times v
  • Maximum speed: when driving force = resistance → P=FresvmaxP = F_{\text{res}} v_{\text{max}}

7. Efficiency

Efficiency=useful output powertotal input power=useful output energytotal input energy\text{Efficiency} = \frac{\text{useful output power}}{\text{total input power}} = \frac{\text{useful output energy}}{\text{total input energy}}

Worked Example: Work Done at an Angle

Problem

A 100 N force pulls a sled 20 m at 35° to the horizontal. Find the work done.

W=Fscos⁡θ=100×20×cos⁡35°=2000×0.819=1638 JW = Fs\cos\theta = 100 \times 20 \times \cos35° = 2000 \times 0.819 = 1638 \text{ J}

Solution

Worked Example: Conservation of Energy

Problem

A 500 g ball is dropped from 15 m. Find the speed at 5 m above the ground (ignore air resistance).

mgh1=12mv2+mgh2mgh_1 = \frac{1}{2}mv^2 + mgh_2 g(h1−h2)=12v2g(h_1 - h_2) = \frac{1}{2}v^2 v=2g(h1−h2)=2×9.81×10=14.0 m/sv = \sqrt{2g(h_1 - h_2)} = \sqrt{2 \times 9.81 \times 10} = 14.0 \text{ m/s}

Solution

Worked Example: Car Power Problem

Problem

A car of mass 1200 kg moves at constant velocity of 30 m/s on a level road. The total resistive force is 800 N. Find the engine power.

P=Fv=800×30=24,000 W=24 kWP = Fv = 800 \times 30 = 24{,}000 \text{ W} = 24 \text{ kW}

Solution

Worked Example: Maximum Speed

Problem

The same car has a maximum engine power of 60 kW. Find its maximum speed against the same resistance.

vmax=PF=60,000800=75 m/sv_{\text{max}} = \frac{P}{F} = \frac{60{,}000}{800} = 75 \text{ m/s}

Solution

Worked Example: Friction and Energy

Problem

A 5 kg block slides 3 m down a 30° incline with μ=0.25\mu = 0.25. Find the speed at the bottom (starting from rest).

Height lost: h=3sin⁡30°=1.5h = 3\sin30° = 1.5 m N=mgcos⁡30°=5(9.81)(0.866)=42.5N = mg\cos30° = 5(9.81)(0.866) = 42.5 N Friction work: Wf=μN×s=0.25×42.5×3=31.9W_f = \mu N \times s = 0.25 \times 42.5 \times 3 = 31.9 J

mgh=12mv2+Wfmgh = \frac{1}{2}mv^2 + W_f 5(9.81)(1.5)=12(5)v2+31.95(9.81)(1.5) = \frac{1}{2}(5)v^2 + 31.9 73.6=2.5v2+31.973.6 = 2.5v^2 + 31.9 v2=16.68v^2 = 16.68 v=4.08v = 4.08 m/s

Solution

9. Practice Questions

    1. A 2 kg mass slides down a frictionless ramp from a height of 4 m. Calculate its speed at the bottom. (2 marks)
    1. A force of 50 N at 60° to the horizontal pulls a box 8 m. Calculate the work done by this force. (2 marks)
    1. A 1500 kg car accelerates from 10 m/s to 25 m/s. Calculate the work done by the engine. (3 marks)
    1. A car of mass 1000 kg travels at constant speed up a hill inclined at 5° to the horizontal at 20 m/s. The resistive force is 600 N. Calculate the power output. (4 marks)

    Answers

    1. v=2gh=2(9.81)(4)=8.86v = \sqrt{2gh} = \sqrt{2(9.81)(4)} = 8.86 m/s.

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Summary

  • Work: W=Fscos⁡θW = Fs\cos\theta (scalar; area under F-s graph for variable force)
  • Work-energy theorem: Wnet=ΔKEW_{\text{net}} = \Delta KE
  • Conservation: KE1+PE1=KE2+PE2+WfrictionKE_1 + PE_1 = KE_2 + PE_2 + W_{\text{friction}}
  • Power: P=W/t=FvP = W/t = Fv
  • At maximum speed: driving force = resistance, so P=FresvmaxP = F_{\text{res}}v_{\text{max}}
Tags:
#a-level#physics#mechanics#work#energy#power

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