Work Against Position-Dependent Resistance

Introduction

When an object moves through a medium, across a surface, or along a track, it may experience a force that opposes its motion. The size of that resistance need not remain constant. A rougher section of a track, for example, may create more resistance farther along the path.

To find the work required to overcome such resistance, we must keep track of both its changing magnitude and its direction. The resistance force points against the motion, while an applied force that overcomes it points along the motion. They therefore do opposite signed work during the same displacement.

Key distinction: Work done against resistance is the positive amount of work supplied to oppose it. Work done by the resisting force is negative when the object moves in the direction under consideration.

Representing Position-Dependent Resistance

Let an object move along a straight line in the positive \(x\)-direction, from position \(x=a\) to position \(x=b\), where \(a<b\). Suppose the resistance magnitude at position \(x\) is \(R(x)\). A resistance magnitude is nonnegative: \(R(x)\geq 0\). Its unit is a force unit, such as newtons.

Because resistance points opposite to the positive direction of motion, the signed force exerted by resistance is \(-R(x)\). If an applied force balances resistance so that the object moves at constant speed, the applied force is \(+R(x)\). The work integral for each force is therefore different in sign.

W_{\text{resistance}} = \int_a^b -R(x)\,dx = -\int_a^b R(x)\,dx, \qquad W_{\text{against resistance}} = \int_a^b R(x)\,dx.

The second expression is the positive work supplied to balance resistance over the path. It describes work required solely to overcome resistance; it is not automatically the total work of an applied force if that force also changes the object's speed or counters other forces.

Definition: For motion from \(x=a\) to \(x=b\) in the positive direction, the work against a position-dependent resistance \(R(x)\) is \(\int_a^b R(x)\,dx\), provided \(R\) is integrable on the interval. The signed work done by the resisting force is the negative of this quantity.

If \(R(x)\) is measured in newtons and \(x\) in metres, the work is measured in newton-metres, equivalent to joules. A resistance law may be continuous or piecewise continuous; for a piecewise law, calculate the integral on each interval and add the results.

Why the Integral Gives the Work

Over a short displacement \(\Delta x\), a force that is nearly constant does work approximately equal to force times displacement. Here the applied force needed to balance resistance changes with position, so the approximation on a small interval uses the resistance magnitude at a point in that interval. Adding these small contributions and taking the limit gives the integral.

Divide the path into small intervals.
On the interval from \(x_{i-1}\) to \(x_i\), let its length be \(\Delta x_i=x_i-x_{i-1}\), and choose a sample point \(\xi_i\) within it.

Approximate the work on each interval.
The applied force balancing resistance is approximately \(R(\xi_i)\), so the small amount of work supplied is approximately \(R(\xi_i)\Delta x_i\).

Add the contributions and take the limit.
As the interval widths shrink, the sums approach the definite integral, which is the work against resistance.

W_{\text{against resistance}} = \lim_{\max \Delta x_i\to 0} \sum_{i=1}^{n} R(\xi_i)\Delta x_i = \int_a^b R(x)\,dx.

This derivation also explains why the sign matters. During each positive displacement, resistance acts in the negative direction, so its contribution is approximately \(-R(\xi_i)\Delta x_i\). Its signed work is consequently \(-\int_a^b R(x)\,dx\), whereas the applied balancing force supplies \(+\int_a^b R(x)\,dx\).

Worked Examples with Linear Resistance

Worked Example: Resistance Increasing Linearly

An object moves from \(x=0\) m to \(x=3\) m. The resistance magnitude is \(R(x)=4+2x\) N. Find the work required to overcome resistance, assuming the applied force balances it throughout the motion.

Since \(R(x)\geq 0\) on this interval, the work against resistance is the integral of \(R\):

W_{\text{against resistance}} = \int_0^3 (4+2x)\,dx.

An antiderivative is \(4x+x^2\), because its derivative is \(4+2x\). Evaluating at the endpoints gives

\begin{aligned} W_{\text{against resistance}} &= \left[4x+x^2\right]_0^3\\ &=(4\cdot 3+3^2)-(4\cdot 0+0^2)\\ &=(12+9)-0\\ &=21\ \text{J}. \end{aligned}

Check: Differentiating \(4x+x^2\) gives \(4+2x\), the resistance law. The positive answer has units \(\text{N}\cdot\text{m}=\text{J}\). The signed work done by resistance is \(-21\) J.

Worked Example: Linear Resistance Over a Longer Displacement

A cart moves from \(x=1\) m to \(x=5\) m against resistance \(R(x)=3x+2\) N. Find the work required to overcome the resistance.

The endpoints are the actual positions of the motion, so integrate over \([1,5]\):

\begin{aligned} W_{\text{against resistance}} &=\int_1^5 (3x+2)\,dx\\ &=\left[\frac{3}{2}x^2+2x\right]_1^5\\ &=\left(\frac{3}{2}\cdot 25+2\cdot 5\right) -\left(\frac{3}{2}\cdot 1^2+2\cdot 1\right)\\ &=\left(\frac{75}{2}+10\right)-\left(\frac{3}{2}+2\right)\\ &=\frac{95}{2}-\frac{7}{2}\\ &=44\ \text{J}. \end{aligned}

Check: The derivative of \(\frac{3}{2}x^2+2x\) is \(3x+2\). Also, \(R(x)\) is positive throughout \([1,5]\), so positive work against resistance is consistent with the model. The resistance itself does \(-44\) J of work.

Piecewise Resistance Laws

Resistance can change according to the part of the path the object occupies. In that case, use the law that applies on each subinterval. The total work is the sum of the work over those subintervals. A change in the resistance law does not require a new work principle; it changes only how the integral is evaluated.

Worked Example: Resistance Changes at a Position

An object moves from \(x=0\) m to \(x=5\) m. Its resistance magnitude is

R(x)= \begin{cases} 2+x, & 0\leq x\leq 2,\\ 4+2(x-2), & 2<x\leq 5, \end{cases} \qquad \text{in newtons}.

On the first interval, integrate \(2+x\). On the second, integrate \(4+2(x-2)\). Split the work integral at \(x=2\):

\begin{aligned} W_{\text{against resistance}} &=\int_0^2 (2+x)\,dx +\int_2^5 \bigl(4+2(x-2)\bigr)\,dx\\ &=\left[2x+\frac{x^2}{2}\right]_0^2 +\left[4(x-2)+(x-2)^2\right]_2^5\\ &=\left(4+\frac{4}{2}\right)-0 +\bigl(4(5-2)+(5-2)^2\bigr) -\bigl(4(2-2)+(2-2)^2\bigr)\\ &=6+(12+9)-0\\ &=27\ \text{J}. \end{aligned}

Check: On the first interval, the derivative of \(2x+\frac{x^2}{2}\) is \(2+x\). On the second interval, the derivative of \(4(x-2)+(x-2)^2\) is \(4+2(x-2)\). Both resistance magnitudes are nonnegative on their intervals, and the total work is \(6+21=27\) J. The signed work by resistance is \(-27\) J.

A Sign Error to Avoid

A common mistake is to use the positive resistance magnitude as though it were the signed resisting force. For the first example, that can lead to the following incorrect statement:

W_{\text{done by resistance}} = \int_0^3 (4+2x)\,dx = 21\ \text{J}.

The integral is indeed \(21\) J, but it is the positive work supplied against resistance, not the work done by the resisting force. The signed resisting force is \(-(4+2x)\) N, so the correction is

W_{\text{done by resistance}} = \int_0^3 -(4+2x)\,dx = -21\ \text{J}, \qquad W_{\text{against resistance}}=21\ \text{J}.

Before integrating, identify which quantity is being requested: the work supplied against resistance or the signed work done by resistance. The distinction prevents a correct integral from being assigned the wrong physical meaning.

Interpreting the Result

The integral measures the accumulated effect of resistance across the whole displacement. It is not generally correct to take the resistance at the starting point or ending point and multiply by the distance; that shortcut applies only when the force magnitude is constant. For position-dependent resistance, the integral accounts for the values along the entire path.

The setup also gives useful checks. If \(R(x)\geq 0\) throughout a forward displacement, work against resistance cannot be negative. If a calculation gives a negative value for that quantity, check the limits, the sign, and whether the signed force has been confused with its magnitude. If resistance is piecewise-defined, check that every part of the displacement has been included exactly once.

Takeaway: For forward motion from \(a\) to \(b\), integrate the resistance magnitude to find positive work against it. Attach a minus sign when finding the signed work done by resistance itself.

Check Your Understanding

Use the distinction between a resistance magnitude and a signed force to answer these questions.

If \(R(x)=5+x\) N on \(0\leq x\leq 2\) m, what definite integral gives the work against resistance?
For the motion and resistance in Question 1, what is the signed work done by resistance, and why does its sign differ?
Why must the work integral be split if a resistance law changes at an interior position?
For forward motion over an interval where \(R(x)\geq 0\), what sign should work against resistance have?
What units result when resistance is in newtons and position is in metres?