Work, Energy, Impulse, and Momentum

Key Takeaways

  • Work-energy theorem: the net work done on a body equals its change in kinetic energy.
  • Kinetic energy: KE = ½mv² (translational) + ½Iω² (rotational).
  • Potential energy: gravitational PE = mgh, elastic PE = ½kx².
  • Conservation of energy: KE₁ + PE₁ = KE₂ + PE₂ (if no non-conservative forces do work).
  • Linear impulse-momentum: ΣF·Δt = mΔv (impulse equals change in momentum).
  • Conservation of momentum: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' (if no external impulse).
Last updated: March 2026

Work, Energy, Impulse, and Momentum

Work and Energy

Work Done by a Force

W=Fd=FdcosθW = \vec{F} \cdot \vec{d} = Fd\cos\theta

where θ is the angle between the force and displacement vectors.

Force DirectionWork
Same as displacement (θ = 0°)W = Fd (positive)
Perpendicular (θ = 90°)W = 0
Opposite to displacement (θ = 180°)W = -Fd (negative)

Work done by a spring: W = -½k(x₂² - x₁²)

Kinetic Energy

KE=12mv2+12Iω2KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

Potential Energy

Gravitational: PE = mgh (h = height above reference datum)

Elastic (spring): PE = ½kx² (x = deformation from natural length)

Work-Energy Theorem

W=ΔKE=KE2KE1\sum W = \Delta KE = KE_2 - KE_1

Conservation of Energy

When only conservative forces (gravity, springs) do work:

KE1+PE1=KE2+PE2KE_1 + PE_1 = KE_2 + PE_2

12mv12+mgh1+12kx12=12mv22+mgh2+12kx22\frac{1}{2}mv_1^2 + mgh_1 + \frac{1}{2}kx_1^2 = \frac{1}{2}mv_2^2 + mgh_2 + \frac{1}{2}kx_2^2

Power

P=dWdt=Fv=FvcosθP = \frac{dW}{dt} = \vec{F} \cdot \vec{v} = Fv\cos\theta

For rotational motion: P = τω (torque × angular velocity)

UnitEquivalent
1 Watt (W)1 J/s = 1 N·m/s
1 horsepower (hp)745.7 W = 550 ft·lb/s

Impulse and Momentum

Linear Impulse-Momentum Theorem

t1t2Fdt=mv2mv1\int_{t_1}^{t_2} \vec{F} \, dt = m\vec{v}_2 - m\vec{v}_1

For constant force: F·Δt = m(v₂ - v₁)

Conservation of Linear Momentum

When no external impulse acts on the system: m1v1+m2v2=m1v1+m2v2m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

Collisions

Coefficient of Restitution: e=v2v1v1v2=relative velocity of separationrelative velocity of approache = \frac{v_2' - v_1'}{v_1 - v_2} = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}

e ValueCollision TypeEnergy
e = 1Perfectly elasticKE conserved
0 < e < 1InelasticKE lost
e = 0Perfectly plastic (bodies stick)Maximum KE loss

For perfectly plastic collisions: m1v1+m2v2=(m1+m2)vm_1 v_1 + m_2 v_2 = (m_1 + m_2) v'

Angular Impulse-Momentum

t1t2Mdt=Iω2Iω1\int_{t_1}^{t_2} M \, dt = I\omega_2 - I\omega_1

For constant moment: M·Δt = I(ω₂ - ω₁)

Vibrations (Natural Frequency)

For an undamped spring-mass system:

Natural frequency: ωn=km(rad/s)\omega_n = \sqrt{\frac{k}{m}} \quad (\text{rad/s})

Period: T=2πωn=2πmk(seconds)T = \frac{2\pi}{\omega_n} = 2\pi\sqrt{\frac{m}{k}} \quad (\text{seconds})

Frequency: f=1T=12πkm(Hz)f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \quad (\text{Hz})

For a simple pendulum: ωn=gLT=2πLg\omega_n = \sqrt{\frac{g}{L}} \quad T = 2\pi\sqrt{\frac{L}{g}}

Test Your Knowledge

A 2 kg ball moving at 10 m/s collides head-on with a 3 kg ball at rest. If e = 0 (perfectly plastic), what is the velocity after collision?

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Test Your Knowledge

A 1,000 kg car traveling at 20 m/s brakes to a stop. How much kinetic energy is dissipated?

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Test Your Knowledge

A spring with k = 500 N/m supports a 2 kg mass. What is the natural frequency of vibration?

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