Define work, kinetic energy and gravitational potential energy, and apply the principle of conservation of mechanical energy to one-dimensional problems including those with friction

What this dot point is asking

QCAA wants you to apply work-energy principles to one-dimensional problems. The dot point combines four ideas: work done by a force, kinetic energy, gravitational potential energy, and the conservation of mechanical energy (with friction as the energy-loss term when present).

Work

Work done by a constant force is:

where $F$ is the force, $s$ is the displacement, and $\theta$ is the angle between the force and the displacement. SI unit: joule (J = N m).

• IMATH_9 : maximum positive work ($W = Fs$).
• IMATH_11 : zero work (centripetal force does no work on an object in uniform circular motion).
• IMATH_12 : negative work ($W = -Fs$, e.g. friction opposing motion).

For a variable force, work is the area under the force-displacement graph.

Kinetic energy

Kinetic energy is the energy of motion:

It is a scalar with SI unit joule.

Work-energy theorem

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

This connects forces (which do work) to motion (which has kinetic energy).

Gravitational potential energy

Near the Earth's surface, gravitational potential energy is:

where $h$ is the height above an agreed reference level. Only differences in $PE_g$ matter; the reference level is your choice.

Conservation of mechanical energy

In the absence of friction and air resistance:

Mechanical energy converts between kinetic and potential but the total stays the same. A roller coaster at the top of a hill has maximum $PE$ and minimum $KE$; at the bottom, the reverse.

With friction (or any non-conservative force), the energy equation becomes:

where $E_{\text{lost}}$ is the energy dissipated as heat, sound or deformation. For a constant friction force $f$ acting over a distance $d$, $E_{\text{lost}} = f d$.

Worked example

A pendulum bob of mass $0.50$ kg is released from rest at height $0.20$ m above its lowest point. Find the speed at the lowest point (ignore air resistance).

Conservation of mechanical energy: $mgh = \frac{1}{2} m v^2$.

$v = \sqrt{2 g h} = \sqrt{2 (9.8)(0.20)} = \sqrt{3.92} = 1.98$ m s$^{-1}$.

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

Common traps

Forgetting the cosine in $W = Fs\cos\theta$. Force at an angle to motion does less work than $Fs$.

Treating $mgh$ as absolute. $PE_g$ is always relative to a chosen reference. State your reference at the start.

Adding $KE$ and $PE$ separately when energy is lost. With friction, total mechanical energy decreases. The decrease equals the work done against friction.

Using only $KE$ for a ball at the top of a swing. At the highest point of a pendulum or projectile, $v$ has its lowest value but is not necessarily zero. Identify the geometry carefully.

Try it: Try the SUVAT calculator for one-dimensional motion (use angle $= 90°$) to cross-check problems where energy conservation gives a final speed.

In one sentence

Work done by a constant force is $W = Fs\cos\theta$, kinetic energy is $\frac{1}{2}mv^2$, gravitational potential energy is $mgh$, and mechanical energy ($KE + PE$) is conserved unless friction or another non-conservative force removes energy as heat, in which case $KE_i + PE_i = KE_f + PE_f + fd$.