Drag Work from a Given Motion

Why Drag Work Depends on the Motion

Drag resists motion: its direction is opposite the object's velocity. Consequently, its work is generally negative. To calculate that work, however, it is not enough to know the geometric path alone. The force depends on velocity, and velocity depends on how the object moves along the path as time passes.

For a prescribed motion \(\mathbf{r}(t)\), the velocity is \(\mathbf{v}(t)=\mathbf{r}'(t)\). The work done by a force \(\mathbf{F}\) over the time interval \([a,b]\) is computed by integrating its component along the displacement. Since \(d\mathbf{r}=\mathbf{r}'(t)\,dt\), this gives

This expression is particularly useful for drag: the prescribed motion supplies the velocity, and the drag law supplies the force.

Two Common Drag Models

A simple model assumes that drag is proportional to velocity. In three dimensions, with a constant \(c\geq 0\), this force is \(\mathbf{F}_{d}=-c\mathbf{v}\). The minus sign makes it oppose the velocity. The constant \(c\) has units of force divided by speed, such as \(\mathrm{N\,s/m}\).

Another common model makes drag proportional to the square of speed. For a constant \(k\geq 0\), write \(\mathbf{F}_{d}=-k\|\mathbf{v}\|\mathbf{v}\). This force points opposite \(\mathbf{v}\), and its magnitude is \(k\|\mathbf{v}\|^2\). The units of \(k\) are \(\mathrm{kg/m}\) when force is measured in newtons and speed in metres per second.

These are different assumptions about the physical force, not interchangeable formulas. Use the model specified in the problem. In either model, if the object is momentarily at rest, the stated drag force is zero.

Why Drag Work Cannot Be Positive

The sign follows directly from the direction of the force. For linear drag, substitute \(\mathbf{F}_{d}=-c\mathbf{v}\) into the work integral and use \(\mathbf{v}\cdot\mathbf{v}=\|\mathbf{v}\|^2\):

Here \(c\geq0\), and a squared norm cannot be negative. If \(c>0\) and the object moves with nonzero speed over a time interval of positive length, the integral is positive before multiplication by \(-c\), so \(W_d<0\). If the object remains at rest, or if \(c=0\), the drag work is zero.

For quadratic drag, the same reasoning gives

This is a general feature of a resisting force: at every instant when the object moves, the force has a nonpositive component along its velocity. Integrating those instantaneous contributions cannot produce positive work.

Worked Example: Linear Drag Along a Parabolic Motion

When the Curve Parameter Is Not Time

The parameter used to describe a curve does not automatically represent time. If the position is \(\mathbf{r}(u)\) and physical time is \(t=t(u)\), with \(dt/du>0\), then the chain rule gives

Therefore the physical velocity is \(\mathbf{r}'(u)/(dt/du)\), not simply \(\mathbf{r}'(u)\). The displacement over a parameter change is still \(d\mathbf{r}=\mathbf{r}'(u)\,du\), so work may be integrated directly with respect to \(u\):

For linear drag, this becomes \(-c\int \|\mathbf{v}(u)\|^2(dt/du)\,du\). The factor \(dt/du\) accounts for how much physical time passes during a parameter change. Keeping it is essential: drag depends on velocity as a function of time, not just on the drawn shape of the path.

Worked Example: A Non-Time Parameter

Worked Example: Quadratic Drag at Constant Speed

A Common Error: Using the Path Derivative as Velocity

Suppose a motion is given by \(\mathbf{r}(u)\), but \(u\) is not time. A common mistake is to call \(d\mathbf{r}/du\) the velocity and substitute it into the drag law. For the preceding non-time-parameter example, that mistake would use \((1,2u)\) instead of the physical velocity \((1/2,u)\). It would produce the incorrect force \((-2,-4u)\) for the linear model, and the incorrect work

rather than the correct \(-\frac73\ \mathrm{J}\). The correction is to divide \(d\mathbf{r}/du\) by \(dt/du\) to find velocity, then form the force from that velocity. The work dot product uses the displacement derivative \(d\mathbf{r}/du\), not velocity a second time.

A Reliable Calculation Sequence

For a prescribed trajectory, keep the roles of position, velocity, force, and displacement distinct. The following sequence prevents the most common errors.

What the Work Tells Us

Drag removes mechanical energy from the motion: its work is negative whenever the object moves against a nonzero drag force. The work-energy theorem relates the total work of all forces to the change in kinetic energy, but the drag work alone is only one contribution. Other forces may do positive or negative work as well.

The formulas also show why timing matters. For linear drag, the work depends on the integral of speed squared over time; for quadratic drag, it depends on the integral of speed cubed over time. Two objects may trace the same geometric curve but experience different drag work if their speeds along it differ. A correct calculation therefore starts with the actual motion, not merely a sketch of its path.