Work, Energy, Impulse, and Momentum

Work and Kinetic Energy

The work done by a force over a displacement is the dot product:

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

where θ is the angle between force and displacement. A force perpendicular to motion (θ = 90°) does no work — this is why the normal force and centripetal force never change kinetic energy.

The work-energy principle states that net work equals the change in kinetic energy:

$\sum W = \Delta KE = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2$

(add ½Iω² terms for rotating bodies). This is the method of choice when a problem connects speed to distance without asking about time, since it sidesteps acceleration entirely.

Worked Example — Work-Energy

A 1000 kg car decelerates from 20 m/s to rest. The kinetic energy that the brakes must dissipate is:

$\Delta KE = \tfrac{1}{2}(1000)(20)^2 = \tfrac{1}{2}(1000)(400) = 200{,}000\text{ J} = 200\text{ kJ}$

If the braking distance is 40 m, the average braking force is F = ΔKE/d = 200{,}000/40 = 5000 N — found in one step with no need for acceleration.

Potential Energy, Conservation, and Power

Potential energy comes in two forms tested on the FE exam:

• Gravitational: PE = mgh, with h measured from a chosen datum.
• Elastic (spring): PE = ½kx², with x the deformation from the spring's free length.

When only conservative forces (gravity, springs) do work, mechanical energy is conserved:

$KE_1 + PE_1 = KE_2 + PE_2$ $\tfrac{1}{2}mv_1^2 + mgh_1 + \tfrac{1}{2}kx_1^2 = \tfrac{1}{2}mv_2^2 + mgh_2 + \tfrac{1}{2}kx_2^2$

If friction or another non-conservative force acts, add its work as an energy loss: KE₁ + PE₁ − W_friction = KE₂ + PE₂.

Power

Power is the rate of doing work:

$P = \frac{dW}{dt} = \vec{F}\cdot\vec{v} = Fv\cos\theta, \qquad P_{rot} = T\omega$

Trap: Use conservation of energy only when no friction or applied non-conservative force does work. A block sliding down a rough ramp loses energy to friction, so KE₁ + PE₁ does not equal KE₂ + PE₂ — you must subtract the friction work μN·d.

Impulse, Momentum, and Collisions

Linear momentum is p = mv. The impulse-momentum principle integrates force over time:

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

For a constant force, F·Δt = mΔv. This is the right tool when a force acts over a known time interval — impacts, jet thrust, a bat striking a ball.

Conservation of momentum holds when no external impulse acts on the system:

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$

Coefficient of Restitution

For collisions, the coefficient of restitution e relates separation speed to approach speed:

$e = \frac{v_2' - v_1'}{v_1 - v_2} = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$

For a perfectly plastic collision (e = 0) the bodies move together: m₁v₁ + m₂v₂ = (m₁ + m₂)v′.

Worked Example — Plastic Collision

A 2 kg ball at 10 m/s strikes a 3 kg ball at rest and they stick:

$v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{2(10) + 3(0)}{5} = \frac{20}{5} = 4\text{ m/s}$

Momentum is conserved even though kinetic energy drops from 100 J to 40 J — the 60 J difference becomes heat and deformation.

Vibrations — Natural Frequency

For an undamped spring-mass system, the natural frequency is:

$\omega_n = \sqrt{\frac{k}{m}}\;(\text{rad/s}), \quad T = 2\pi\sqrt{\frac{m}{k}}, \quad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\;(\text{Hz})$

For a simple pendulum, ωn = √(g/L). Note ωn (rad/s) and f (Hz) differ by 2π — confusing them is a frequent error.

Choosing Energy vs. Momentum vs. F = ma

The three kinetics methods each fit a different problem type, and picking the right one saves time:

For a body falling and compressing a spring, energy methods are fastest (KE + PE balance). For a collision, momentum is essential because the impact force and duration are unknown. For finding a car's acceleration on a grade, ΣF = ma is direct.

Worked Example — Energy Method on a Spring

A 4 kg block slides down a frictionless ramp, dropping 2 m, then compresses a spring (k = 8000 N/m). Find the maximum compression x. Energy conservation: mgh = ½kx² → 4(9.81)(2) = ½(8000)x² → 78.5 = 4000x² → x² = 0.0196 → x = 0.140 m. The block's gravitational PE converts entirely into spring PE at maximum compression, where its velocity is momentarily zero. No time or acceleration calculation is required — the hallmark advantage of the energy method.

Trap — restitution direction: in the e equation, the numerator is the separation velocity (after) and the denominator is the approach velocity (before). Reversing them, or forgetting that e ranges only from 0 to 1, produces impossible results.