The Timpson Lyceum · University Physics I (2025)

Work and Kinetic Energy – Lecture 6 (University Physics I)

Overview

This lecture builds up the concept of work from a simple, oversimplified picture to its full vector definition, then connects work to motion through the work–energy theorem. We start with work as force times displacement, refine it to a dot product (only the component of force along the displacement counts), and finally write it as a path integral $\int \vec F \cdot d\vec r$ for forces that vary in both magnitude and direction. Along the way we define kinetic energy and prove $W_\text{total} = \Delta K$ three times (constant force, varying force in 1-D, varying force in 3-D). We apply these ideas to pushing problems, a pile-driver, springs (Hooke's law) with and without friction, and a swing, and we close with power as the time rate of doing work. Mastering this material lets you solve dynamics problems through energy bookkeeping instead of tracking forces and accelerations moment by moment.

Key Concepts

Work — the building-up of the definition

• Simplest (force along motion): $W = Fs$, where $s$ is the displacement and $F$ acts in the same direction. Units are newton-meters, called joules (J): $1\text{ J} = 1\text{ N}\cdot\text{m}$.
• Force at an angle (dot product): only the component of force along the displacement does work. In two equivalent forms, $W = \vec F \cdot \vec s = |\vec F|\,|\vec s|\cos\theta = F_x s_x + F_y s_y$ where $\theta$ is the angle between $\vec F$ and $\vec s$. Work is a scalar (the dot product is also called the scalar product) — it has no direction.
• Sign of work, from $\cos\theta$: if $\theta = 0$, $\cos\theta = 1$ (maximum positive work); if $\theta = 90^\circ$, $\cos\theta = 0$ (no work — e.g. a force perpendicular to motion, like holding a book steady or the tension in a swing chain); if $\theta = 180^\circ$, $\cos\theta = -1$ (negative work — e.g. friction opposing the motion).
• Full (general) definition: for a force that varies in magnitude and direction along a path, $W = \int_{\vec r_1}^{\vec r_2} \vec F \cdot d\vec r$ This sums $\vec F \cdot d\vec r$ over infinitely many infinitesimal steps $d\vec r = (dx,\,dy,\,dz)$. In one dimension it reduces to $W = \int_{x_1}^{x_2} F_x\,dx$, and if $F$ is constant it collapses back to $W = F(x_2 - x_1) = Fs$. Each simpler formula is a special case of this one.

Kinetic energy

• Definition: $K = \tfrac{1}{2}mv^2$ — the energy associated with motion. Intuitively, it is the work needed to bring an object from rest up to speed $v$ (or, with the opposite sign, to bring it to rest).

Work–energy theorem

• The net (total) work done on an object equals its change in kinetic energy: $W_\text{total} = \Delta K = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2$
• $W_\text{total}$ uses the net force (vector sum of all forces) — or equivalently the sum of the works done by each individual force.
• Consequence: positive net work speeds an object up; negative net work slows it down. If the net force (and net work) is zero, the speed is unchanged.
• The derivation (1-D, constant force) starts from $W_\text{total} = F\!\cdot\! s = ma(x_2 - x_1)$ and uses the kinematics relation $v_2^2 = v_1^2 + 2a(x_2 - x_1)$ to replace $a(x_2 - x_1)$ with $\tfrac12 v_2^2 - \tfrac12 v_1^2$. For varying forces one instead integrates $F_x = m\,dv/dt$ and changes variables to get $\int_{v_1}^{v_2} mv\,dv = \tfrac12 mv_2^2 - \tfrac12 mv_1^2$. The 3-D proof splits $\vec F \cdot d\vec r$ into $x$, $y$, $z$ integrals and recombines them via $v^2 = v_x^2 + v_y^2 + v_z^2$.

Springs and Hooke's law

• A spring exerts a variable force proportional to its stretch or compression: $F = kx$, where $k$ is the force constant in N/m. A large $k$ is a stiff spring (car shock); a small $k$ is a soft spring (slinky).
• Work to stretch/compress a spring from $x_1$ to $x_2$ (integrating $kx$, equal to the area under the force–displacement line): $W = \int_{x_1}^{x_2} kx\,dx = \tfrac{1}{2}kx_2^2 - \tfrac{1}{2}kx_1^2$
• You cannot use $W = Fs$ for a spring because the force is not constant; $Fs$ only works if you use the average force.

Power

• Power is the time rate of doing work. Average: $P_\text{avg} = \dfrac{\Delta W}{\Delta t}$; instantaneous: $P = \dfrac{dW}{dt}$. Units: joules per second, called watts (W).
• An equivalent form, since $dW = \vec F \cdot d\vec r$ and $d\vec r/dt = \vec v$: $P = \vec F \cdot \vec v = |\vec F|\,|\vec v|\cos\theta$
• Useful conversions: $1\text{ hp} = 746\text{ W}$; a kilowatt-hour is a unit of energy (power × time), not power: $1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6\times10^6\text{ J}$.

Worked Example

Tractor pulling a sled — final speed via the work–energy theorem. A tractor pulls a sled ($m = 1500\text{ kg}$) a distance $s = 20\text{ m}$ with a force $F = 5000\text{ N}$ at $\theta = 36.9^\circ$ above horizontal, against friction $f_k = 3500\text{ N}$. The sled starts at $v_1 = 2\text{ m/s}$. Find its final speed.

First find the work done by each force. The weight and normal force are perpendicular to the (horizontal) motion, so they do zero work. Friction opposes motion ($\cos 180^\circ = -1$):

Now apply the work–energy theorem, $W_\text{total} = \tfrac12 mv_2^2 - \tfrac12 mv_1^2$, and solve for $v_2$:

Power of a jet engine. Each engine produces a forward thrust $F = 322{,}000\text{ N}$ while the plane flies at $v = 250\text{ m/s}$ in the same direction ($\cos 0 = 1$):

Summary

• Work is the dot product of force and displacement, $W = \vec F\cdot\vec s = |\vec F||\vec s|\cos\theta$; in general it is the path integral $\int \vec F\cdot d\vec r$. It is a scalar measured in joules.
• Only the force component along the displacement does work: perpendicular forces do none, opposing forces do negative work.
• Kinetic energy is $K = \tfrac12 mv^2$, and the work–energy theorem states $W_\text{total} = \Delta K$ — net work changes an object's speed.
• For a spring, force varies as $F = kx$ (Hooke's law), so the work is $\tfrac12 kx_2^2 - \tfrac12 kx_1^2$; you cannot use $Fs$ for a variable force.
• Power is $P = dW/dt = \vec F\cdot\vec v$, measured in watts; remember $1\text{ hp} = 746\text{ W}$ and that a kilowatt-hour is energy, not power.
• Energy methods let you find final speeds, forces, or distances directly from work and kinetic energy, sidestepping detailed force-and-acceleration analysis.