4.5 - Work and energy

Work by a constant force

A rope pulls a box partly upwards as the box moves horizontally. Does only a horizontal pull do work? The rope does work through its horizontal component; the vertical component has no displacement in its direction.

For a $20\,\mathrm{N}$ pull at $60^\circ$ above a $5\,\mathrm{m}$ rightward displacement, the component along the displacement is $20\cos60^\circ=10\,\mathrm{N}$. Its work is $10\times5=50\,\mathrm{J}$.

Resolving the force along the displacement turns an angled work calculation into force times distance.

If the force has a component in the same direction as the displacement, work is positive. If it opposes the displacement, work is negative. A perpendicular force does zero work; so does any force whose point of application has no displacement. Magnitudes $F$ and $d$ are non-negative: the angle already handles direction.

[DIAGRAM: asset_name: Work follows the displacement direction; asset_slug: work-force-angle; caption: Reversing the displacement reverses the work of the same force; description: Two panels show the same 20 newton force at 60 degrees above the rightward horizontal, with a horizontal component of 10 newtons right. In the left panel the displacement is 5 metres right, the angle to the force is 60 degrees, and work is plus 50 joules. In the right panel the displacement is 5 metres left, the angle to the force is 120 degrees, and work is minus 50 joules. The displacement arrows are separate from the force arrows.]

Reverse only the displacement. The force arrow stays fixed, but the angle in the work formula changes.

A constant friction force of magnitude $r$ opposing a travelled distance $d$ does work $-rd$. The work done against resistance is the positive amount $rd$. These describe the same energy transfer with different sign conventions.

Identify the force–displacement angle in this reversed journey.

Kinetic and potential energy

A moving particle has energy associated with its speed. A raised particle also has gravitational potential energy relative to a chosen height level. Keeping these two quantities separate makes changes of speed and height easier to track.

For a $2\,\mathrm{kg}$ particle moving at $3\,\mathrm{m\,s^{-1}}$, its kinetic energy is $\tfrac12(2)(3^2)=9\,\mathrm{J}$. If it is $4\,\mathrm{m}$ above the chosen zero level, its potential energy is $2(10)(4)=80\,\mathrm{J}$. Here and below take $g=10\,\mathrm{m\,s^{-2}}$.

Kinetic energy is non-negative and is the same for opposite velocities of equal magnitude. Doubling speed multiplies kinetic energy by four. Gravitational potential energy depends on the zero level: a particle below that level has negative $U$, which is perfectly possible. Changing the zero level changes both initial and final energies by the same constant, so it leaves $\Delta U$ unchanged.

Why does a height increase store $mgh$? Lifting at constant speed against weight $mg$ through vertical height $h$ requires work $mgh$. Along a slope of length $s$ and angle $\alpha$, the height change is $s\sin\alpha$, not $s$ itself.