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Work & Kinetic Energy

Source lecture(s): SC133 Lec 8

Intuition

Newton's laws track forces instant by instant. Energy methods take a shortcut: instead of following the whole trajectory, compare before and after. Work is the currency — force succeeding at pushing something through a distance — and kinetic energy is the account balance of motion. The exchange rate is exact: net work in, kinetic energy up, joule for joule.

Definitions

Work by a constant force \(\vec F\) over displacement \(\vec d\):

\[W = \vec F\cdot\vec d = Fd\cos\theta\]

Only the force component along the motion counts (dot product). Perpendicular forces (the normal force on a sliding block, tension in a circular swing) do zero work. For varying forces:

\[W = \int \vec F\cdot d\vec s \qquad\text{e.g. spring: } W = \int_0^x(-kx')dx' = -\tfrac12 kx^2\]

Kinetic energy:

\[K = \tfrac12 m v^2\]

Power — rate of doing work:

\[P = \frac{dW}{dt} = \vec F\cdot\vec v \qquad [1\,\text{W} = 1\,\text{J/s}]\]

The work–energy theorem

\[\boxed{\,W_\text{net} = \Delta K = \tfrac12 mv_f^2 - \tfrac12 mv_i^2\,}\]

Derivation (1-D, from Newton II): \(W = \int F\,dx = \int m\frac{dv}{dt}dx = \int m\frac{dv}{dt}v\,dt = \int mv\,dv = \tfrac12 mv_f^2 - \tfrac12 mv_i^2\). ∎ (Full version on the equation page.)

The theorem is Newton's second law integrated over distance — no new physics, but a scalar equation that skips all the trajectory details.

Worked example: braking distance

A car at speed \(v\) locks its brakes; kinetic friction \(\mu_k mg\) acts over distance \(d\). The theorem: \(-\mu_k mg\,d = 0 - \tfrac12 mv^2\), so

\[d = \frac{v^2}{2\mu_k g}\]

Mass cancels; distance grows with the square of speed. Doubling your speed quadruples the skid — the single most life-relevant equation in this course.

Signs of work

Situation Sign of \(W\) Effect on \(K\)
Force along motion (engine) + speeds up
Force against motion (friction, braking) − slows down
Force ⊥ motion (normal, centripetal) 0 speed unchanged (direction may change!)

The zero-work case explains why magnetic forces never change a particle's speed.

Common mistakes

  • "I pushed hard, so I did work." Pushing a wall does zero work — no displacement. Holding a suitcase stationary: zero work (tiring ≠ work).
  • Forgetting work is signed. Friction does negative work; the theorem needs the net, signed total.
  • Confusing power with energy. A 1000 W kettle running for one hour uses energy \(= P t = 3.6\) MJ; watts measure the rate.
  • Using \(W = Fd\) with the total force in circular motion — the centripetal force does no work at all.

Knowledge graph position

Prerequisites: Newton's laws, Vectors. Leads to: Potential energy → Conservation of energy.

Quiz

Q1 (computational). A 2 kg block accelerates from 3 m/s to 7 m/s. Net work done?

Answer

\(W = \Delta K = \tfrac12(2)(49 - 9) = 40\,\text{J}\) — regardless of how the force varied along the way.

Q2 (conceptual). A satellite in circular orbit: how much work does gravity do per revolution?

Answer

Zero. Gravity is exactly centripetal (⊥ velocity) at every instant, so \(\vec F\cdot d\vec s = 0\) throughout — consistent with constant orbital speed.

Q3 (computational). What steady power must a 1200 kg car deliver to climb a 5% grade at 20 m/s (ignore drag)?

Answer

Force along slope \(\approx mg\times 0.05 = 588\,\text{N}\); \(P = Fv = 588\times20 \approx 11.8\,\text{kW}\) (≈16 hp) — hills, not flat cruising, are what engines are sized for.

Q4 (multiple choice). Two objects with equal momentum but different masses: which has more kinetic energy? (a) heavier (b) lighter (c) equal

Answer

(b). \(K = p^2/2m\) — at fixed \(p\), smaller mass means larger \(K\). (A bullet beats a truck at equal momentum.)