The formula

Work-energy theorem
W_net = ΔKE = ½mv² - ½mv₀²
Work done by a constant force
W = Fd·cos θ
Kinetic energy
KE = ½mv²

What the symbols mean

SymbolMeaningUnit
W_netNet work from all forces actingJ
WWork done by one constant forceJ
ΔKEChange in kinetic energyJ
KEKinetic energy at a given speedJ
mMass of the objectkg
vFinal speedm/s
v₀Initial speedm/s
FMagnitude of the forceN
dMagnitude of the displacement while the force actsm
θAngle between the force and the displacementdegrees or radians

When it applies

  • You care about speeds and distances but not about how long the process took. Time never appears in the theorem.
  • Every force that does work is included, friction included, since friction does negative work.
  • A force perpendicular to the motion does no work, which is why the normal force and the tension in a circular path drop out.

Worked example

Problem. A 1200 kg car traveling at 20 m/s brakes to a stop in 45 m. What average braking force acted on it?

  1. Kinetic energy at the start: KE₀ = ½(1200 kg)(20 m/s)² = 240,000 J.
  2. The car ends at rest, so ΔKE = 0 - 240,000 = -2.4 × 10⁵ J. That is the net work done on it.
  3. The braking force opposes the displacement, so cos θ = -1 and W_net = -Fd. Set -F(45 m) = -2.4 × 10⁵ J.
  4. Solve: F = 240,000 / 45 = 5333 N, which is 5.3 × 10³ N to two significant figures.

Answer. About 5.3 × 10³ N of average braking force.

Common mistakes

  • Using only the applied force. The theorem is about the net work from every force acting on the object.
  • Dropping cos θ when the force is at an angle to the displacement.
  • Forgetting that kinetic energy goes with speed squared, so doubling the speed quadruples the stopping distance at the same braking force.
  • Giving work a direction. Work and energy are scalars, and the sign only says whether energy went in or out.

Related formulas

Sources