Further Mechanics: Work, Energy and Power
As part of A-level Further Mechanics, the work done by a constant force is the product of the force and the distance moved in the direction of the force (W = Fd, or W = Fd cos(theta) if the force acts at angle theta to the direction of motion), measured in joules. The work-energy principle states that the total work done by all the forces acting on a body equals its change in kinetic energy: sum of work done = (1/2)mv^2 - (1/2)mu^2. Power is the rate at which work is done; for a force moving at speed v in its own direction, instantaneous power P = Fv. These ideas let problems with a varying number of forces (driving force, resistance, weight component on a slope) be solved without finding the acceleration directly at every stage.
Before you start
Make sure you're comfortable with these topics first:
Method
- List every force acting on the body and decide which do work: a force does no work if it acts perpendicular to the motion (for example the normal reaction on a flat or sloped surface), so only forces with a component along the direction of travel contribute.
- Find the work done by each contributing force over the distance travelled: W=Fd for a force along the motion; for weight on a slope, use W=mgh, where h is the vertical height gained or lost over that distance.
- Give work done against a force (resistance, friction, or gravity when climbing) a negative sign, and work done by a driving force in the direction of travel a positive sign.
- Apply the work-energy principle: total work done = final KE - initial KE = (1/2)mv^2 - (1/2)mu^2, and solve for the unknown quantity.
- For a power question, use P=Fv for the instantaneous power provided by a driving force moving at speed v.
- At maximum speed, the acceleration is zero, so the driving force exactly balances the total resistance; use this to link maximum power, maximum speed and resistance: P_max = F_max x v_max = (resistance at that speed) x v_max.
- If power is constant but speed is not, find the driving force at a given instant from F=P/v, then apply Newton's second law (driving force minus resistance minus any weight component) = ma to find the acceleration at that instant.
- Keep every quantity in consistent SI units (kg, m, s, so energy is in joules and power in watts) before substituting into an equation.
Worked example
A box of mass 8 kg is pulled from rest across a rough horizontal floor by a horizontal force of 70 N. The resistance to motion is a constant friction force of 30 N. Find the speed of the box after it has travelled 10 m, using the work-energy principle.
- Identify the forces doing work along the direction of motion: the 70 N pulling force (positive work) and the 30 N friction force (negative work, since it opposes the motion); the weight and normal reaction do no work, as they act perpendicular to the motion.
- Find the work done by the pulling force: W1 = 70 x 10 = 700 J.
- Find the work done against friction: W2 = -30 x 10 = -300 J.
- Find the total work done: W1 + W2 = 700 - 300 = 400 J.
- Apply the work-energy principle with the box starting from rest (u=0): 400 = (1/2)(8)v^2 - 0.
- Solve: 400 = 4v^2, so v^2 = 100, giving v = 10 m/s.
Practice questions
Try each question, then tap to reveal the answer.
Q1Define the work done by a constant force acting in the direction of motion.Show answer
Answer: Work done = force x distance moved in the direction of the force (W=Fd), measured in joules.
Q2State the work-energy principle.Show answer
Answer: The total work done by all the forces acting on a body equals the change in its kinetic energy: total work = (1/2)mv^2 - (1/2)mu^2.
Q3A crate of mass 5 kg is lifted vertically at constant speed through 3 m. Find the work done against gravity (take g = 9.8 m/s^2).Show answer
Answer: W = mgh = 5 x 9.8 x 3 = 147 J.
Q4A cyclist and bike have combined mass 75 kg and travel at a constant 6 m/s along a flat road against a total resistance of 18 N. Find the power the cyclist produces.Show answer
Answer: At constant speed, driving force = resistance = 18 N, so P = Fv = 18 x 6 = 108 W.
Q5A car of mass 1200 kg has a maximum speed of 30 m/s on a flat road against a constant resistance of 500 N. Find the maximum power of the engine.Show answer
Answer: At maximum speed, driving force = resistance = 500 N, so P = Fv = 500 x 30 = 15,000 W (15 kW).
Q6A ball of mass 0.3 kg is thrown vertically upward and rises 4 m before momentarily coming to rest. Use the work-energy principle to find its initial kinetic energy (ignore air resistance, take g = 9.8 m/s^2).Show answer
Answer: Work done against gravity over the rise = mgh = 0.3 x 9.8 x 4 = 11.76 J; since final KE = 0, this equals the initial KE, so initial KE = 11.76 J.
Q7A van travels at constant power against a constant resistance. Explain why its acceleration is greatest when its speed is lowest.Show answer
Answer: At constant power P, the driving force is F=P/v, which is largest when v is smallest. Since acceleration = (driving force - resistance)/mass, a larger driving force with the same resistance and mass gives a larger net force and so a larger acceleration, so acceleration is greatest at the lowest speed.
Q8A skier of mass 60 kg, starting from rest, skis down a slope, descending 20 m vertically. Ignoring resistance, find her speed at the bottom, using energy conservation (take g = 9.8 m/s^2).Show answer
Answer: mgh = (1/2)mv^2, so v^2 = 2gh = 2 x 9.8 x 20 = 392, giving v = sqrt(392) = 14sqrt(2) m/s exactly, approximately 19.8 m/s (3 s.f.).
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
A cyclist and her bicycle have a combined mass of 50 kg. She rides up a straight slope, travelling 50 m measured along the slope while rising 5 m vertically. She starts at 5 m/s and pedals with a constant driving force of 81 N; the total resistance to motion is a constant 20 N. Using the work-energy principle, find her speed after travelling the 50 m. (Take g = 9.8 m/s^2.)
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A van of mass 1500 kg experiences a constant resistance to motion of 600 N and has a maximum speed of 25 m/s on a flat road. (a) Find the power of the van's engine at maximum speed. (b) The van's engine works at this same constant power while travelling at 15 m/s on the same flat road. Find the van's acceleration at this instant.
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