The work–energy principle | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Total work by the real forces equals the change in kinetic energy

\sum W=\Delta\mathrm{KE}

Symbols

$\sum W$ = add the signed work done by all real forces; Σ means add (J)
$\Delta\mathrm{KE}$ = change in kinetic energy; Δ means final value minus initial value (J)

If $u$ is the initial speed and $v$ is the final speed, then:

\Delta\mathrm{KE}=\tfrac12mv^2-\tfrac12mu^2

Swipe left or right to see the whole diagram →

Do not add a separate resultant-work term. Add the work done by the real forces.

Choose the start and finish, then record $u$ and $v$ .
List the real forces acting throughout that interval.
Write one signed work term for each force.
Set their total equal to final KE minus initial KE.

Common mistake

“Resultant force” is not another force to add to the account. Add the work of the actual forces once each.

Distinguish work supplied from the energy that resistance removes

Worked example

A cyclist speeds up against resistance

A cyclist and bicycle have total mass 72 kg. Their speed rises from 8 m s⁻¹ to 16 m s⁻¹ over 100 m against a constant 28 N resistance. Find the work done by the cyclist.

\Delta\mathrm{KE}=\tfrac12(72)(16^2-8^2)

\Delta\mathrm{KE}=6912\text{ J}

The energy transferred by working against resistance is:

28(100)=2800\text{ J}

W_{\mathrm{cyclist}}=9712\text{ J}

The cyclist supplies energy both to increase kinetic energy and to replace the energy dissipated by resistance.

(9709/42/M/J/24 Q1)

Examiner note

The mark scheme awarded the method for a dimensionally correct four-term work–energy equation. Listing the energy destinations first makes that structure visible before any arithmetic. (9709/42/M/J/24 Q1 mark scheme)

On a slope, keep path distance and vertical height separate

For a body moving distance $s$ down a slope of angle $\theta$ , gravity does positive work $mgs\sin\theta$ . A constant resistance $R$ does work $-Rs$ .

mgs\sin\theta-Rs=\tfrac12mv^2-\tfrac12mu^2

Key idea

Use the distance along the path for resistance. Use the vertical height change for GPE or work by gravity. These are different lengths.

Your turnIndependent check [5 marks]

A 10 kg block moves 4 m down a plane inclined at $30^\circ$ . Its initial speed is 3 m s⁻¹ and a constant 20 N resistance opposes the motion. Find its final speed.

Show worked answer

200-80=5(v^2-9)

v^2=33

v=\sqrt{33}=5.74\text{ m s}^{-1}

Answer

v=5.74\text{ m s}^{-1}