Dynamics: Force, Energy and Momentum

Newton’s laws, work and energy, impulse and momentum, collisions and rotation.

On this page

  1. Newton’s laws
  2. Worked example: a block sliding down a slope
  3. Work and energy
  4. Worked example: stopping a car
  5. Impulse, momentum and collisions
  6. Rotational dynamics
  7. FAQ

Newton’s laws

ΣF = m a    ΣMG = IG α    weight W = m g

Worked example: a block sliding down a slope

Worked example: μ = 0.2 on a 30° slope

  • Along the slope: m a = m g sin θ − μ m g cos θ, so a = g (sin θ − μ cos θ) = 9.81 × (0.5 − 0.1732) = 3.21 m/s². The mass cancels.
  • After sliding 5 m from rest: v = √(2 a s) = 5.66 m/s, reached in 1.77 s.
  • The same answer comes from energy: m g s sin θ − μ m g cos θ s = ½ m v² (work of gravity minus friction equals the kinetic energy gained).

Work and energy

W = ∫ F · ds    KE = ½ m v²    PE = m g h    spring: ½ k x²    Wnet = ΔKE    power P = F v

Worked example: stopping a car

Worked example: energy of a moving car and of a spring

  • A 1 000 kg car at 100 km/h (27.78 m/s) carries ½ m v² = 385.8 kJ. Stopping it in 48.2 m needs an average braking force F = KE ÷ d = 8.00 kN (0.82 g), and the brakes turn all that energy into heat. At 200 km/h the energy is four times as much.
  • A spring of rate 800 N/m compressed 50 mm stores ½ k x² = 1.00 J; released against a 2 kg mass it gives it v = √(2E ÷ m) = 1.00 m/s (see Springs).
  • Power: pushing a load at 2 m/s against a 500 N resistance needs P = F v = 1 000 W.

Impulse, momentum and collisions

impulse J = ∫ F dt = Δ(m v)    momentum is conserved in a collision: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂    e = (v₂ − v₁) ÷ (u₁ − u₂)

Worked example: an impact and a collision

  • A 150 g ball at 40 m/s stopped by a glove in 5 ms: average force = m Δv ÷ Δt = 1 200 N (815 times its weight). Cushioning that stretches the time to 50 ms cuts the force ten times.
  • A 2 kg block at 3 m/s hits a 1 kg block at rest with e = 0.5: speeds after impact 1.50 m/s and 3.00 m/s. Momentum before 6.00, after 6.00 kg·m/s; kinetic energy falls from 9.00 J to 6.75 J, a loss of 2.25 J.

Rotational dynamics

T = I α    KE = ½ I ω²    I = Σ m r²    solid disc: ½ m r²    ring: m r²    angular momentum L = I ω

Worked example: spinning up a disc

  • A 10 kg steel disc of radius 0.2 m has I = ½ m r² = 0.20 kg·m². A constant 5 N·m torque gives α = T ÷ I = 25 rad/s² and reaches 1 500 rpm (157.1 rad/s) in 6.28 s. The energy stored then is ½ I ω² = 2.47 kJ. This is the calculation behind motor sizing and flywheel sizing (Flywheels, Sizing a Motion Axis).

Related

FAQ

When do I use energy instead of F = m a?

When you know the start and end states and the distance but not the time, or when forces vary along the path. Use F = m a when you need accelerations or times.

Is momentum conserved when there is friction?

During a short collision, yes (the impact forces dwarf friction). Over a long slide, no.

Why do I use kilograms and not weight?

Mass is the property that resists acceleration; weight is a force that depends on gravity. Mixing lb and lbm is the classic source of factor-of-32 errors.

References

  • Hibbeler, R. C., Engineering Mechanics: Dynamics, Pearson.
  • Meriam, J. L., Kraige, L. G., Engineering Mechanics: Dynamics, Wiley.
  • Beer, F. P., Johnston, E. R., Vector Mechanics for Engineers: Dynamics, McGraw-Hill.

Last reviewed: October 2026. Examples use constant coefficients; real friction and drag vary.