Lesson 2 · 30 min
The Work of a Force
Work measures what a force contributes to a particle's speed as the particle moves. Only the part of the force along the path counts. A force pushing forward does positive work, one holding back does negative work, and one pushing sideways does none at all.
Learning objectives
- Define the work of a force, \(dU = \Fvec\cdot d\rvec = F\cos\theta\,ds\), and give it in joules.
- Decide the sign of the work from the angle \(\theta\) between the force and the motion.
- Calculate the work of a constant force, along a straight path and along any path.
- Read the work of a varying force as the area under its force–displacement graph.
- Recognize the forces that do no work.
Definition
A particle moves along its path through a small displacement \(d\rvec\), of length \(ds\). The work done on it by a force \(\Fvec\) is the dot product of the force and the displacement:
Work of a force
\[ dU = \Fvec\cdot d\rvec = F\cos\theta\,ds \qquad\qquad U_{1\to2} = \int_{\rvec_1}^{\rvec_2}\Fvec\cdot d\rvec = \int_{s_1}^{s_2} F\cos\theta\,ds \]\(\theta\) is the angle between \(\Fvec\) and the direction of motion. \(F\cos\theta\) is the force's component along the path, the tangential component \(F_t\) of Lesson 1.
In rectangular components the dot product is \(dU = F_x\,dx + F_y\,dy + F_z\,dz\). That form is handy when you know the force's components and the path's coordinates.
The sign of the work
| Angle \(\theta\) | Work | Example |
|---|---|---|
| \(0 \le \theta \lt 90^\circ\) | positive: the force helps the motion | The weight of a block sliding downhill |
| \(\theta = 90^\circ\) | zero | The push of a smooth fixed surface |
| \(90^\circ \lt \theta \le 180^\circ\) | negative: the force resists the motion | Kinetic friction, always at \(180^\circ\) |
Work is a scalar: it has a sign but no direction. Its unit is the joule, \(1\ \text{J} = 1\ \text{N}\cdot\text{m}\), the work of a \(1\ \text{N}\) force acting through \(1\ \text{m}\) along its own line. In US customary units, work is in ft·lb. Write work in joules: a moment is also a newton times a meter, but it is a different quantity.
The work of a constant force
On a straight path, a force of constant magnitude \(F_c\) at a constant angle \(\theta\) to the path does work
\[ U_{1\to2} = F_c\cos\theta\,(s_2 - s_1) \]More generally, a force that is constant in magnitude and direction can come out of the integral, whatever the path:
\[ U_{1\to2} = \int_{\rvec_1}^{\rvec_2}\Fvec\cdot d\rvec = \Fvec\cdot(\rvec_2 - \rvec_1) = F_x\,\Delta x + F_y\,\Delta y + F_z\,\Delta z \]Only the displacement between the end points matters, not the shape of the path in between. Lesson 3 uses this for the most important constant force of all, the weight.
Example 2.1 — Pulling a sled
A child pulls a sled \(15\ \text{m}\) across level snow with a rope at \(30^\circ\) above the horizontal. The tension in the rope is \(120\ \text{N}\). Find the work done by the rope, and say which other forces on the sled do no work.
Show solution
The rope's tension is constant in magnitude and at a constant \(\theta = 30^\circ\) to the straight path:
\[ \colP{U_P} = F_c\cos\theta\,\Delta s = (120)\cos 30^\circ\,(15) = 1559\ \text{J} \]Only the horizontal component of the tension, \(103.9\ \text{N}\), does work. The vertical component, \(60\ \text{N}\), is perpendicular to the motion. The weight and the normal force of the snow are vertical too, so they do no work on a level path. Friction from the snow, if any, does negative work.
Example 2.2 — A constant force along a curved ramp
A box is pushed from \(A\) to \(B\) along a curved ramp by a horizontal force of constant magnitude \(P = 50\ \text{N}\). \(B\) is \(3\ \text{m}\) to the right of \(A\) and \(1.2\ \text{m}\) higher. Find the work of \(P\).
Show solution
\(\colP{\mathbf{P}} = 50\,\ihat\ \text{N}\) is constant in magnitude and direction, so only the displacement matters, \(\rvec_B - \rvec_A = 3\,\ihat + 1.2\,\jhat\ \text{m}\):
\[ \colP{U_P} = \mathbf{P}\cdot(\rvec_B - \rvec_A) = (50)(3) + (0)(1.2) = 150\ \text{J} \]The shape of the ramp never entered. Along a steep part of the ramp the angle between \(\mathbf{P}\) and the motion is large, along a flat part it is small, but the integral of \(P\cos\theta\,ds\) is always \(P\,\Delta x\).
Work as an area
When the force changes along the path, plot its tangential component \(F\cos\theta\) against \(s\). The work \(\int F\cos\theta\,ds\) is the area under the graph between \(s_1\) and \(s_2\): area above the axis counts as positive work, area below it as negative.
Example 2.3 — A force that grows with distance
A horizontal force \(F = (20 + 10x)\ \text{N}\), with \(x\) in meters, pushes a block along a level floor from \(x = 0\) to \(x = 2\ \text{m}\). Find its work.
Show solution
The graph is a trapezoid with parallel sides \(20\) and \(40\ \text{N}\) and width \(2\ \text{m}\): area \(\tfrac12(20 + 40)(2) = 60\ \text{J}\). Figure 2.2 shows this force as "Growing push".
Forces that do no work
A force does no work when it is perpendicular to the motion, or when the point where it acts does not move. The ones you will meet most:
- The normal force of a fixed, smooth or rough surface: it is perpendicular to the sliding.
- The reaction at a fixed pin, such as the pivot of a pendulum: its point of application does not move.
- The tension in a cord that swings a particle around a fixed point: the cord stays perpendicular to the circular path.
- The weight during horizontal motion.
These forces still matter. The normal force sets the size of the friction force, \(F_f = \mu_k N\), and Lesson 8 needs the normal force on curved paths.
Check your understanding
Key takeaways
- \(dU = \Fvec\cdot d\rvec = F\cos\theta\,ds\). Only the component of the force along the path does work.
- Work is a scalar in joules: positive when the force helps the motion, negative when it resists, zero when it is perpendicular.
- A constant force does work \(\Fvec\cdot(\rvec_2 - \rvec_1)\): only the end points of the path matter.
- The work of a varying force is the area under its \(F\cos\theta\)–\(s\) graph.
- Next, Lesson 3 works out the weight, springs and friction.