8867 / 2027
Energy and Fields overview

Lesson 2 of 6

Work and kinetic energy

  • I can explain work as mechanical energy transfer and use force times displacement in the force direction.Syllabus

    Syllabus 8867, 4(c). Explain work as mechanical energy transfer and use force times displacement in the force direction.

  • I can derive kinetic energy from work and the equations of straight-line uniform acceleration.Syllabus

    Syllabus 8867, 4(d). Derive kinetic energy from work and the equations of straight-line uniform acceleration.

  • I can recall and use E_k = (1/2)mv^2.Syllabus

    Syllabus 8867, 4(e). Recall and use E_k = (1/2)mv^2.

Work is a mechanical transfer of energy. Identify the force, the body it acts on and that body's displacement before calculating it.

For a constant force, work is the force multiplied by displacement in the force's direction. If the angle between force and displacement is θ:

W = Fs cos θ

Here F and s are the force and displacement magnitudes. Equivalently, multiply the force component along the displacement by s. The angle is between these two vectors, not automatically the angle to a vertical line or surface normal. Work is a scalar, in J = N m.

  • A force component along the displacement does positive work.
  • A force component opposite the displacement does negative work.
  • A force perpendicular to the displacement does zero work, even when the force is nonzero.

A force holding an object stationary does no mechanical work on that object because its displacement is zero. This does not imply that the person holding it uses no metabolic energy.

If the force component varies with displacement, divide the motion into small intervals. Add force component × displacement for those intervals: the result is the signed area under a force-component/displacement graph. Final force × total displacement is not a general replacement for that area.

A force model identifies all external forces on the chosen body. Work by one named force and net work by all the forces are different quantities.

Derive kinetic energy

Consider a constant-mass particle, or a body translating without relevant rotation or internal changes. It moves in a straight line under a constant resultant force. Choose a positive direction and use signed Fresultant, a and s consistently.

  1. Use work and Newton's second law:

    Wnet = Fresultants = mas

  2. Constant force and constant mass give constant acceleration. From the straight-line motion equations, v2 - u2 = 2as, so:

    Wnet = ½m(v2 - u2)

  3. This is the change in kinetic energy. Taking kinetic energy to be zero at rest gives:

    Ek = ½mv2
    Wnet = Ek,final - Ek,initial

The derivation uses constant acceleration at its middle step. It also explains why the resultant force matters. Substituting a pulling force while ignoring resistance will not give the kinetic-energy change.

Kinetic energy depends on speed squared and has no direction. Reversing velocity while keeping the same speed leaves kinetic energy unchanged, although momentum changes. Its units are kg m2 s-2 = J.

Worked force and energy calculation

An angled pull with resistance

A 2.0 kg crate initially moves right at 1.0 m/s. A constant 8.0 N pull acts at 60 degrees above the rightward horizontal. A supplied constant 2.0 N resistance acts left. The crate moves 3.0 m right along a fixed horizontal floor.

Pair each force with the same rightward displacement

An angled pull does positive work while resistance does negative workA point force model selects the two-kilogram crate translating on a horizontal floor. All purple force arrows use ten drawing units per newton. The eight-newton pull is sixty degrees above right, with components four newtons right and 6.928 newtons up. Resistance is two newtons left. Weight is 19.62 newtons down and the normal force 12.6918 newtons up, balancing the upward pull component and preserving contact. The blue arrow below gives a separate three-metre rightward displacement; its length is not a force scale. Pull work is positive twelve joules, resistance work negative six, and weight and support do zero work because they are perpendicular to the displacement.60°Pull: 8.0 NNormal: 12.69 NResistance2.0 N leftWeight: 19.62 N2.0 kg crateDisplacement: 3.0 m right

The angle is measured from the displacement direction. Purple force arrows share one force scale; the blue displacement arrow has different units. Weight and support are nonzero forces, yet each does zero work on this horizontal displacement.

The force arrows act on the crate. The separate movement arrow describes its 3.0 m displacement. Only horizontal force components contribute to work along this displacement.
Work on the crate over 3.0 m
ForceWork / J
8.0 N pull8.0 × 3.0 × cos 60° = +12
2.0 N resistance-2.0 × 3.0 = -6
Weight0: perpendicular to displacement
Normal support0: perpendicular to displacement
Net+12 - 6 = +6

Initial kinetic energy = ½(2.0)(1.0)2 = 1.0 J. Therefore final kinetic energy = 1.0 + 6 = 7.0 J.

v = √(2Ek/m) = √(2 × 7/2) = √7 m/s

This is 2.64575... m/s: 2.65 m/s as a working value, or about 2.6 m/s to two significant figures.

Check that the crate remains in contact. With g = 9.81 N/kg, vertical balance requires support = 19.62 - 8 sin 60° = 12.69 N upward. It is positive, so the assumed contact is consistent. This vertical balance contributes no extra horizontal work term.

Try it step by step

Work it out in steps: Work done by the pull F d cos theta in J, then Work done against friction in J, then Gain in kinetic energy in J.

  1. Problem 1. A sledge is pulled 10 m along level ground by a rope at 30 degrees above the horizontal with a tension of 50 N. Friction is 20 N. Find the gain in kinetic energy.
  2. Problem 2. A crate is pulled 5.0 m along level ground by a rope at 40 degrees above the horizontal with a tension of 80 N. Friction is 25 N. Find the gain in kinetic energy.
  3. Problem 3. A toy cart is pulled 8.0 m along level ground by a string at 60 degrees above the horizontal with a tension of 30 N. Friction is 8.0 N. Find the gain in kinetic energy.
Show the answer

Problem 1.

Work done by the pull F d cos theta in J: W = 50 x 10 x cos 30 = 433 J

Work done against friction in J: Against friction: 20 x 10 = 200 J

Gain in kinetic energy in J: Gain in KE = 433 - 200 = 233 J (3 sf)

Problem 2.

Work done by the pull F d cos theta in J: W = 80 x 5.0 x cos 40 = 306 J

Work done against friction in J: Against friction: 25 x 5.0 = 125 J

Gain in kinetic energy in J: Gain in KE = 306 - 125 = 181 J (3 sf)

Problem 3.

Work done by the pull F d cos theta in J: W = 30 x 8.0 x cos 60 = 120 J

Work done against friction in J: Against friction: 8.0 x 8.0 = 64 J

Gain in kinetic energy in J: Gain in KE = 120 - 64 = 56 J (2 sf)

If the system is enlarged to include the crate and floor, the 12 J pulling input gives a 6 J kinetic increase and a 6 J internal increase under the stated friction model. This is the same energy account with a different boundary. Do not subtract friction's 6 J work and then subtract the corresponding 6 J internal increase again.

Estimate the scale: a body of roughly 1 kg moving at about 1 m/s has Ek ≈ ½(1)(1)2 = 0.5 J, of order 1 J. This uses rough mass and speed assumptions; it is a magnitude check, not a precise measurement.

Check your understanding

A 2.0 kg crate moves 3.0 m right. An 8.0 N pull acts 60 degrees above the displacement and a constant 2.0 N resistance acts left. Weight and support are perpendicular to the displacement. What is the change in the crate's kinetic energy?

  1. +6 J
  2. +12 J
  3. +18 J
Show the answer

+6 J

Pulling work is +12 J and resistance work is -6 J. Weight and support do zero work. Net work, and therefore the kinetic-energy increase in this translating-crate model, is +6 J.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define work done by a force. A constant resultant force F acts on a body of mass m, initially at rest, and moves it a distance s in the direction of the force. Show that its kinetic energy is then (1/2)mv2, where v is its speed. [4]

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