Theory

Work and energy

How to compute the work of a constant or variable force, what power measures, how the work-energy theorem and the conservation of mechanical energy give speeds without solving the equations of motion, and how a potential energy curve describes motion and equilibrium.

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Work of a constant force
Work of a variable force

Work of a constant force

Throughout this topic the body is treated as a point particle and the reference frame is inertial.

Consider a point particle that undergoes a straight-line displacement $s$ while a constant force $F$ acts on it. The work done by the force is the scalar

L = F \cdot s = F s cos θ,

where $θ$ is the angle between $F$ and $s$ . In the International System work is measured in joules, $1 J = 1 N \cdot m$ . In Cartesian components $L = F_{x} Δ x + F_{y} Δ y + F_{z} Δ z$ , where $Δ x$ , $Δ y$ , $Δ z$ are the components of the displacement.

The work depends only on the component $F cos θ$ of the force parallel to the displacement, because the perpendicular component has zero dot product with $s$ . For example, a cart pulled for $10 m$ on a horizontal plane with a force of $50 N$ inclined at $3 0^{\circ}$ receives from the force the work $L = 50 \cdot 10 \cdot cos 3 0^{\circ} J \approx 433 J$ .

A constant force

F

inclined at

θ

to the displacement

s

does the work

F s cos θ

, which depends only on the projection of

F

onto the direction of

s

Sign of the work

The sign of the work is the sign of $cos θ$ .

If $0 \leq θ < \frac{π}{2}$ , the work is positive and the force is called driving; for example, the weight does positive work on a body moving down.
If $θ = \frac{π}{2}$ , the work is zero; the normal force of a plane on which the body slides and the centripetal force in uniform circular motion are perpendicular to the displacement and do not do work.
If $\frac{π}{2} < θ \leq π$ , the work is negative and the force is called resistive; kinetic friction, opposite to the displacement, has $θ = π$ and does the work $L = - f s$ .

Therefore a force can act on a moving body without doing work, when it is perpendicular to the displacement.

Work of the net force

If the constant forces $F_{1}, \dots, F_{n}$ act on the point particle, with net force $R = F_{1} + \dots + F_{n}$ , the total work during the displacement $s$ is

L_{tot} = L_{1} + \dots + L_{n} = R \cdot s .

The equality follows from the distributive property of the dot product over addition. Therefore the total work can be computed in two equivalent ways, by adding the works of the individual forces or by computing the work of the net force; the first is more convenient when the contribution of each force is also required, and the total work is the quantity that appears in the work-energy theorem.

Work of a variable force

Consider a point particle moving along the $x$ -axis from $x_{A}$ to $x_{B}$ , subject to a force whose component $F_{x} (x)$ depends on the position. The interval is divided into pieces of width $Δ x_{i}$ so small that on each of them the force can be considered constant; the work on the $i$ -th piece is approximately $F_{x} (x_{i}) Δ x_{i}$ , and the sum of these contributions tends to the integral as the pieces shrink. We therefore set

L = \int_{x_{A}}^{x_{B}} F_{x} (x) d x .

The work is the signed area under the graph of $F_{x}$ as a function of $x$ , and for a constant force the definition gives back $L = F_{x} (x_{B} - x_{A})$ . For example, with $F_{x} (x) = 4 + 2 x$ , where $x$ is in meters and $F_{x}$ in newtons, the work between $x = 0$ and $x = 3 m$ is $[4 x + x^{2}]_{0}^{3} = 21 J$ .

The work of

F_{x} (x) = 4 + 2 x

between

x = 0

and

x = 3

is the area of the shaded trapezoid,

21 J

Work along a curved path

If the particle travels along a curved path $γ$ from $A$ to $B$ and the force depends on the position, the work is

L = \int_{γ} F \cdot d r,

where $d r$ is the infinitesimal displacement along the path. The definition repeats the previous reasoning, adding the contributions $F \cdot Δ r$ of the small pieces into which the curve is divided. It is a line integral, for which see the topic on curves and line integrals in Mathematical Analysis 2. In problems of particle mechanics the path is often straight or the force is conservative, and the computation reduces to an integral in one variable or to a difference of potential energies.

Work of the elastic force

Consider an ideal spring with spring constant $k$ , with one end fixed and the other attached to a body moving along the $x$ -axis, with the origin at the rest position; by Hooke’s law the elastic force is $F_{x} = - k x$ . When the deformation goes from $x_{A}$ to $x_{B}$ the elastic force does the work

L = \frac{1}{2} k x_{A}^{2} - \frac{1}{2} k x_{B}^{2} .

The work is negative when the deformation increases in absolute value, because the force opposes the displacement, and positive when the spring returns toward its rest length. For a spring with $k = 200 N/m$ that returns to rest from an extension of $0.10 m$ we have $L = \frac{1}{2} \cdot 200 \cdot 0.1 0^{2} J = 1.0 J$ .

The elastic force

F_{x} = - k x

: when the spring stretches from

0

x

the work is the signed area of the triangle,

- \frac{1}{2} k x^{2}

Proof

By the definition of the work of a variable force,

L = \int_{x_{A}}^{x_{B}} (- k x) d x = [- \frac{1}{2} k x^{2}]_{x_{A}}^{x_{B}} = \frac{1}{2} k x_{A}^{2} - \frac{1}{2} k x_{B}^{2} .

The result depends only on the initial and final deformations.

The theory continues with 7 more sections (power, kinetic energy and the work energy theorem, conservative forces…), plus all worked examples, the solutions and the PDF.

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