Only the parallel part works. W = F·d = Fd cosθ — a force at right angles to the motion does no work at all (normal force, centripetal force, magnetic force).
Work can be negative. When a force opposes the displacement (friction, gravity on the way up) it drains kinetic energy: W < 0.
Net work = ΔKE. Add the work of every force; the total equals the change in ½mv² — nothing else.
Potential energy needs a conservative force. Only gravity, springs and the like store energy you can get back; friction turns it irreversibly into heat.
Drag a crate a distance d with a force at angle θ to the floor. Only the component along the motion, F cosθ, does work: W = Fd cosθ. Swing the angle past 90° and the work turns negative — the force now takes energy away.
Bench · a force at an angle
force F{{ wkFval }} N
angle θ{{ wkAngVal }}°
distance d{{ wkDval }} m
work W = Fd cosθ
{{ wkWork }} J
{{ wkState }} {{ wkNote }}
§02
Variable force — work is an integral
When the force changes along the path, the tidy product Fd fails and you must sum the work slice by slice: W = ∫F dx — the signed area under the force–displacement graph. Where the curve sits above the axis the force pumps energy in; below it, the force draws energy back out.
{{ vfBenchLabel }}
{{ vfKlabel }}{{ vfKval }}
displacement x{{ vfXval }} m
work W = area
{{ vfExpr }}
F at x
{{ vfFnow }} N
W = ∫F dx
{{ vfWork }} J
{{ vfState }} {{ vfNote }}
§03
The work–energy theorem — net work is ΔKE
Push a block over a rough floor. The applied work feeds kinetic energy while friction siphons it off; whatever is left over — the net work — equals the change in ½mv². The area of the force–distance graph is the work.
Bench · work as area (m = 2 kg · g = 10)
applied W
{{ thWapp }} J
friction W
{{ thWfric }} J
net = ΔKE
{{ thWnet }} J
initial speed u{{ thUval }} m/s
applied force F{{ thFval }} N
distance d{{ thDval }} m
friction μ{{ thMuVal }}
KE start → end
{{ thKEi }} → {{ thKEf }}
final speed
{{ thVf }}
{{ thState }} {{ thNote }}
§04
The chain over the edge — stepping the equation
A chain of mass M and length L lies on a table with a length a hanging over. Released, the overhang grows and so does the pull — so the force is never constant. Energy settles the smooth case in one line. Make the table rough, or start the chain in a heap, and the tidy line fails: then you step the equation in x and read the speed off the end.
An engine delivering a fixed P does not give a fixed force: F = P/v, so the harder it is already going, the less it can push. Off the line the force is enormous and the acceleration fierce; a few seconds later the same engine is barely gaining. Speed grows as √t and distance as t3/2 — which is why a car's 0–60 is quick and its 60–120 is not.
{{ pBenchLabel }}
power P{{ pPval }} W
mass m{{ pMval }} kg
speed now
{{ pSpeed }}
driving force P/v
{{ pForce }}
distance
{{ pDist }}
kinetic energy
{{ pKE }}
{{ pState }} {{ pNote }}
§06
Reading a potential well — the whole motion at a glance
Draw U(x) and a horizontal energy line and you have the motion without solving anything: the gap between them is the kinetic energy, the crossings are the turning points, the valleys are stable equilibria and the hilltop is an unstable one. Raise the line above the central barrier and a ball trapped in one valley is suddenly free to visit both.