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.