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How do work, energy and power connect forces, motion and time?

Work done by a force, kinetic and potential energy, the work-energy principle, the conservation of mechanical energy, and power as the rate of doing work.

A focused answer to the Edexcel A-Level Further Mathematics Further Mechanics content on work, energy and power, covering work done by a force, kinetic and gravitational potential energy, the work-energy principle, the conservation of mechanical energy, and power as the rate of doing work.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Work and energy
  3. The work-energy principle
  4. Power
  5. Examples in context
  6. Try this

What this dot point is asking

Edexcel Further Mechanics wants you to compute work done by a force, use kinetic and gravitational potential energy, apply the work-energy principle, use conservation of mechanical energy where only conservative forces act, and relate power to force and velocity. Power problems for vehicles climbing slopes against resistance are a recurring exam favourite.

Work and energy

Work is the transfer of energy by a force as its point of application moves. Only the component of the force along the displacement does work, hence the cosθ\cos\theta factor; a force perpendicular to the motion (such as a normal reaction) does no work. Kinetic energy and gravitational potential energy are the two mechanical energy stores at this level.

The work-energy principle

The net work done by all forces on a body equals its change in kinetic energy. This is often the cleanest route when forces vary or when you do not need the time taken, only the speeds and distances. When the only forces doing work are conservative (gravity, ideal springs), mechanical energy is conserved and you can equate total energy at two points. When friction or air resistance acts, include the work done against those forces explicitly.

Power

Power is the rate at which work is done, measured in watts (W\text{W}). For a force driving motion in the direction of travel, power equals force times speed, P=FvP = Fv. This relationship explains why a vehicle at constant engine power accelerates less as it speeds up: the driving force F=PvF = \frac{P}{v} falls as vv rises, until at maximum speed the driving force just balances the total resistance.

Examples in context

Work, energy and power underpin much of mechanics. Conservation of energy is the standard tool for finding speeds in a vertical circle (the circular-motion dot point) and for projectile or slope problems where time is not needed. The kinetic energy lost in a collision, computed in the elastic-collisions dot point, is exactly the mechanical energy not conserved when e<1e < 1. The integral definition of work, W=FdxW = \int F\,dx for a variable force, connects to further calculus, as does the work done by a spring, 12kx2\frac{1}{2}kx^2. Power-speed problems for cars and trains on inclines are a classic application combining all three concepts.

Try this

Q1. A car engine works at 12000W12\,000\,\text{W}, driving the car at 20m s120\,\text{m s}^{-1}. Find the driving force. [2 marks]

  • Cue. F=Pv=1200020=600NF = \frac{P}{v} = \frac{12000}{20} = 600\,\text{N}.

Q2. Find the kinetic energy of a 3kg3\,\text{kg} mass moving at 4m s14\,\text{m s}^{-1}. [1 mark]

  • Cue. KE=12(3)(42)=24J\operatorname{KE} = \frac{1}{2}(3)(4^2) = 24\,\text{J}.

Q3. A 1kg1\,\text{kg} ball is dropped from 5m5\,\text{m}. Use energy conservation to find its speed on landing (ignore air resistance, g=9.8g = 9.8). [2 marks]

  • Cue. 12v2=gh\frac{1}{2}v^2 = gh, so v2=2×9.8×5=98v^2 = 2 \times 9.8 \times 5 = 98 and v9.9m s1v \approx 9.9\,\text{m s}^{-1}.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20187 marksA car of mass 1200kg1200\,\text{kg} travels up a slope inclined at arcsin120\arcsin\frac{1}{20} to the horizontal against a constant resistance of 400N400\,\text{N}. The engine works at a constant rate of 24kW24\,\text{kW}. Find the maximum speed of the car up the slope. Take g=9.8m s2g = 9.8\,\text{m s}^{-2}.
Show worked answer →

At maximum speed the acceleration is zero, so the driving force balances gravity and resistance.

Driving force at speed vv: F=Pv=24000vF = \frac{P}{v} = \frac{24000}{v} (M1).

The component of weight down the slope is mgsinθ=1200×9.8×120=588Nmg\sin\theta = 1200 \times 9.8 \times \frac{1}{20} = 588\,\text{N} (M1 A1).

At maximum speed, F=mgsinθ+resistance=588+400=988NF = mg\sin\theta + \text{resistance} = 588 + 400 = 988\,\text{N} (M1 A1).

So 24000v=988\frac{24000}{v} = 988, giving v=2400098824.3m s1v = \frac{24000}{988} \approx 24.3\,\text{m s}^{-1} (A1 A1).

Edexcel 20226 marksA particle of mass 0.5kg0.5\,\text{kg} slides from rest down a smooth slope, dropping a vertical height of 2m2\,\text{m}, then travels along a rough horizontal surface where the coefficient of friction is 0.40.4. Find the distance it travels on the rough surface before stopping. Take g=9.8m s2g = 9.8\,\text{m s}^{-2}.
Show worked answer →

Use conservation of energy on the smooth slope, then the work-energy principle against friction.

On the smooth slope, PE converts fully to KE: 12mv2=mgh\frac{1}{2}mv^2 = mgh, so v2=2gh=2×9.8×2=39.2v^2 = 2gh = 2 \times 9.8 \times 2 = 39.2 (M1 A1).

KE at the bottom =12(0.5)(39.2)=9.8J= \frac{1}{2}(0.5)(39.2) = 9.8\,\text{J} (A1).

Friction force =μmg=0.4×0.5×9.8=1.96N= \mu mg = 0.4 \times 0.5 \times 9.8 = 1.96\,\text{N} (M1). Work done against friction over distance dd equals the KE: 1.96d=9.81.96 d = 9.8 (M1), so d=5md = 5\,\text{m} (A1).

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