Apply the work-energy theorem ($W_{\rm net} = \Delta KE$) to motion problems, distinguishing situations where energy methods are more efficient than kinematic methods

What this dot point is asking

VCAA wants you to apply the work-energy theorem to motion problems and to recognise when energy methods solve a problem more efficiently than kinematic methods.

Work-energy theorem

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

For a constant force at angle $\theta$ to the displacement:

Positive work ($\theta < 90°$) increases KE. Negative work ($\theta > 90°$, e.g. friction) decreases KE.

When to use energy vs kinematics

Energy methods are more efficient when:

• The path is complex (curved, multi-stage) but the start and end speeds are needed.
• The forces are known but not the time or detailed trajectory.
• A non-constant force does work via an area under a force-displacement graph.

Kinematics is more efficient when:

• The acceleration is constant.
• Time, displacement, velocity are all asked for explicitly.

For most problems either method works; choose whichever gives the shortest path to the answer.

Conservation of mechanical energy

With only conservative forces (gravity, ideal springs):

With friction or other non-conservative forces:

where $E_{\text{lost}}$ equals the work done against friction (for constant friction $f$ over distance $d$, $E_{\text{lost}} = f d$).

Worked example (roller coaster style)

A $0.50$ kg cart starts from rest at height $h = 2.0$ m on a frictionless track. Find its speed at the bottom.

Conservation: $mgh = \frac{1}{2} m v^2$. $v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 2.0} = 6.26$ m s$^{-1}$.

The mass cancels: any object falling the same height through gravity reaches the same speed in the absence of friction.

If friction does $4.0$ J of negative work over the descent:

$mgh - 4.0 = \frac{1}{2} m v^2$

$(0.50)(9.8)(2.0) - 4.0 = 0.25 v^2$

$9.8 - 4.0 = 0.25 v^2$

$v^2 = 23.2$, so $v = 4.82$ m s$^{-1}$.

Common traps

Forgetting $\cos\theta$ in $W = Fd\cos\theta$

A force at $90°$ to motion does no work.

Treating $W$ as scalar but signed

Work has a sign even though it is a scalar.

Adding KE and PE on the wrong side

Conservation balances total mechanical energy before and after; friction work goes on the "after" side as $E_{\text{lost}}$.

Confusing $\frac{1}{2}mv^2$ with $mv$

KE involves $v^2$ and a factor of $\frac{1}{2}$; momentum is linear in $v$.

In one sentence

The work-energy theorem $W_{\text{net}} = \Delta KE$ relates net work to change in kinetic energy and underpins conservation of mechanical energy when only conservative forces act, with friction or other non-conservative forces dissipating energy at a rate of $f d$ for constant friction.

Examples in context

Example 1. Loy Yang lignite conveyor friction work. A Loy Yang coal conveyor carries $10$ tonnes per minute (167 kg s$^{-1}$) at $4$ m s$^{-1}$ on a $5$ km belt. The belt experiences friction with rollers giving an average drag of $50$ N per kg of moving belt-plus-load. Treating the steady-state operation as zero net work on the coal (no KE change), the work-energy theorem gives $W_{\rm motor} = W_{\rm friction}$ per unit time. Friction power $P = Fv = F \times 4$ m s$^{-1}$; for total moving mass on the belt of about $40\,000$ kg, friction force is $2 \times 10^6$ N and power consumption is $8$ MW. This is why bulk-handling conveyors prefer continuous belts over reciprocating systems where kinetic energy is repeatedly accelerated and discarded.

Example 2. Bathurst hill-climb fuel-energy budget. A Bathurst Supercar (1300 kg) climbs Mountain Straight at constant $250$ km h$^{-1}$ ($69.4$ m s$^{-1}$), with a $174$ m vertical rise. KE is unchanged at constant speed, so by the work-energy theorem, $W_{\rm engine} - W_{\rm drag} - W_{\rm gravity} = 0$. Work against gravity: $W_g = mgh = 1300 \times 9.8 \times 174 = 2.22 \times 10^6$ J. Aerodynamic drag at this speed gives roughly $5000$ N force over the $1.9$ km straight: $W_{\rm drag} = 5000 \times 1900 = 9.5 \times 10^6$ J. Engine delivers $\approx 1.17 \times 10^7$ J. At $90$ s climb time, average power is $\approx 130$ kW, well within the $470$ kW engine output.

Try this

Q1. State the work-energy theorem in equation form. [2 marks]

• Cue. $W_{\rm net} = \Delta KE = \tfrac{1}{2} m v_f^2 - \tfrac{1}{2} m v_i^2$.

Q2. A $5.0$ kg block slides from rest down a frictionless $30^\circ$ incline of length $4.0$ m. Use the work-energy theorem to calculate the speed at the bottom. [3 marks]

• Cue. $W = mg \sin 30^\circ \times 4 = 5 \times 9.8 \times 0.5 \times 4 = 98$ J. $v = \sqrt{2W/m} = \sqrt{39.2} = 6.26$ m s$^{-1}$.

Q3. Refer to the Bathurst hill-climb example. (a) Calculate gravitational work for a $1300$ kg car rising $174$ m. (b) Estimate drag work over $1900$ m at $5000$ N force. (c) Explain why energy methods are more efficient than kinematics for this problem. [2+2+2 marks]

• Cue. (a) $2.22 \times 10^6$ J. (b) $9.5 \times 10^6$ J. (c) Avoids breaking motion into segments; computes total work directly without integrating acceleration.