Work Against Combined Forces

Why Combine Forces Before Calculating Work?

Moving an object at constant speed can require work even when its speed does not change. If gravity, a spring, and resistance all oppose the motion, an external force must balance their combined effect. The work done by that external force depends on the total force along the path, not on any one contribution in isolation.

The central task is to keep the signs consistent. A force that points in the direction of motion contributes positively to the signed force component; one that points against the motion contributes negatively. Once those components are identified, they can be added and integrated over the distance traveled.

Key idea: For straight-line motion at constant speed, the net force along the direction of motion is zero. The required external force is the negative of the sum of the other signed force components.

Signed Forces and External Work

Choose a coordinate $x$ that increases in the direction the object moves, from $x=a$ to $x=b$, where $b>a$. Write each force component along this coordinate with its sign. For example, a force that resists the motion has a negative component, while a force that assists the motion has a positive component.

Definition: If the external force component is $F_{\text{ext}}(x)$, the work done by that force as the object moves from $a$ to $b$ is $$ W_{\text{ext}}=\int_a^b F_{\text{ext}}(x)\,dx. $$ The integral gives signed work: positive work transfers energy to the object, and negative work removes energy from it.

Suppose the other forces along the path have signed components $F_1(x),F_2(x),\ldots,F_n(x)$. For straight-line motion at constant speed, the acceleration along the path is zero. By Newton’s second law, the sum of the force components is therefore zero:

F_{\text{ext}}(x)+F_1(x)+F_2(x)+\cdots+F_n(x)=0.

Thus the required external force is $F_{\text{ext}}(x)=-\sum_{i=1}^{n}F_i(x)$. This force balance is pointwise: it must hold at each position, not merely on average over the whole trip.

Why the Work Contributions Add

The work integral respects sums of forces. This follows from the linearity of integration, provided the force components are integrable over the interval. The force balance gives a direct derivation of the total external work:

Balance the forces.
At each position, $F_{\text{ext}}(x)=-\sum_{i=1}^{n}F_i(x)$.

Integrate over the path.
$W_{\text{ext}}=\int_a^b F_{\text{ext}}(x)\,dx=\int_a^b\left(-\sum_{i=1}^{n}F_i(x)\right)\,dx$.

Separate the contributions.
By linearity, $W_{\text{ext}}=-\sum_{i=1}^{n}\int_a^b F_i(x)\,dx$. Each other force contributes its own signed work, and the external work is the negative of their sum.

In a common special case, every non-external force opposes the motion. If $f_i(x)\geq 0$ denotes the magnitude of the $i$-th opposing force, then its signed component is $F_i(x)=-f_i(x)$. The external force is $F_{\text{ext}}(x)=\sum_i f_i(x)$, and the external work is $\int_a^b\sum_i f_i(x)\,dx$. This magnitude version is convenient, but it should be used only after checking that the forces really do oppose the motion throughout the interval.

Worked Examples

Worked Example: Lifting Against Gravity and Position-Dependent Resistance

A $2\text{ kg}$ object is lifted vertically upward from $y=0\text{ m}$ to $y=3\text{ m}$ at constant speed. Take $g=9.8\text{ m/s}^2$. A resistive force has magnitude $0.4y\text{ N}$, where $y$ is measured in metres. Find the external work.

Choose upward as the positive direction. The gravitational force is constant and points downward:

F_g=-mg=-(2\text{ kg})(9.8\text{ m/s}^2)=-19.6\text{ N}.

The resistance also points downward during the upward motion, so its signed component is $F_r(y)=-0.4y\text{ N}$. Constant speed requires the external force to balance both:

F_{\text{ext}}(y)=-\bigl(F_g+F_r(y)\bigr) =-\bigl(-19.6-0.4y\bigr) =19.6+0.4y\text{ N}.

Integrate from $0\text{ m}$ to $3\text{ m}$. The constant term has units of newtons, and the coefficient $0.4$ has units $\text{N/m}$:

\begin{aligned} W_{\text{ext}} &=\int_0^3(19.6+0.4y)\,dy\\ &=\left[19.6y+0.2y^2\right]_0^3\\ &=(19.6\cdot3+0.2\cdot9)-(0+0)\\ &=58.8+1.8\\ &=60.6\text{ J}. \end{aligned}

Check the force balance: $F_{\text{ext}}+F_g+F_r=(19.6+0.4y)-19.6-0.4y=0$ at every height. Check the antiderivative: differentiating $19.6y+0.2y^2$ gives $19.6+0.4y$, the external force. The total, $60.6\text{ J}$, is positive, as expected for lifting against two downward forces.

Worked Example: Moving Against a Spring and Resistance

An object moves horizontally from $x=0\text{ m}$ to $x=2\text{ m}$ at constant speed. A spring exerts the force $-4x\text{ N}$, and resistance exerts the force $-(3+2x)\text{ N}$, with $x$ measured in metres. Find the external work.

Both given force components are negative for $x\geq0$, so both oppose the positive direction of motion. Their sum and the balancing external force are

\begin{aligned} F_{\text{spring}}(x)+F_r(x) &=-4x-(3+2x)\\ &=-3-6x\text{ N},\\ F_{\text{ext}}(x)&=3+6x\text{ N}. \end{aligned}

The spring coefficient $4$ and the resistance coefficient $2$ have units $\text{N/m}$, so each force term has units of newtons. Integrating the total external force gives

\begin{aligned} W_{\text{ext}} &=\int_0^2(3+6x)\,dx\\ &=\left[3x+3x^2\right]_0^2\\ &=(3\cdot2+3\cdot4)-0\\ &=6+12\\ &=18\text{ J}. \end{aligned}

As a separate check, the work against resistance is $\int_0^2(3+2x)\,dx=[3x+x^2]_0^2=6+4=10\text{ J}$. The work against the spring is $\int_0^2 4x\,dx=[2x^2]_0^2=8\text{ J}$. Their sum is $10+8=18\text{ J}$, matching the combined-force calculation. Differentiating $3x+3x^2$ also gives $3+6x$, the external force.

Worked Example: Moving Outward Against Inverse-Square Gravity and Resistance

An object moves radially outward from $r=2\text{ m}$ to $r=4\text{ m}$ at constant speed. The gravitational force is $-12/r^2\text{ N}$, where the coefficient $12$ has units $\text{N}\cdot\text{m}^2$. A constant resistive force of $2\text{ N}$ also points inward. Find the external work.

Choose outward as positive. Both non-external forces have negative components:

F_g(r)=-\frac{12}{r^2}\text{ N}, \qquad F_r(r)=-2\text{ N}.

The force needed for constant-speed outward motion is therefore

F_{\text{ext}}(r) =-\bigl(F_g(r)+F_r(r)\bigr) =\frac{12}{r^2}+2\text{ N}.

Integrate each contribution over the radial distance:

\begin{aligned} W_{\text{ext}} &=\int_2^4\left(\frac{12}{r^2}+2\right)\,dr\\ &=\left[-\frac{12}{r}+2r\right]_2^4\\ &=\left(-\frac{12}{4}+2\cdot4\right) -\left(-\frac{12}{2}+2\cdot2\right)\\ &=(-3+8)-(-6+4)\\ &=5-(-2)\\ &=7\text{ J}. \end{aligned}

The gravity contribution is $\int_2^4 12/r^2\,dr=[-12/r]_2^4=-3-(-6)=3\text{ J}$. The resistance contribution is $\int_2^4 2\,dr=[2r]_2^4=8-4=4\text{ J}$. Their sum is $3+4=7\text{ J}$. Differentiating $-12/r+2r$ gives $12/r^2+2$, confirming the antiderivative and the external force. The force balance also checks directly: $(12/r^2+2)-12/r^2-2=0$.

A Sign Mistake to Avoid

A common error is to subtract the magnitude of a resisting force even after writing that force with a negative sign. In the lifting example, the signed forces other than the external force are $-19.6$ and $-0.4y$. The correct balance is

F_{\text{ext}}=-(-19.6-0.4y)=19.6+0.4y.

A mistaken calculation might use $F_{\text{ext}}=19.6-0.4y$, treating the resistance as if it helped lift the object. At $y=1\text{ m}$, that gives $19.2\text{ N}$, but the actual downward forces total $19.6+0.4=20.0\text{ N}$; a $19.2\text{ N}$ upward force cannot balance them. The correction is to choose either signed force components and add them algebraically, or positive magnitudes for forces known to oppose the motion. Do not mix the two conventions.

When the Result Applies

The force-balance method here assumes motion along a straight path at constant speed, with the external force chosen so that the total force along the path is zero at each position. If forces have components perpendicular to the path, those components do not contribute to the work integral along that path; use the force components parallel to the displacement. If an opposing force changes direction or begins assisting the motion, retain its signed component rather than treating its magnitude as an always-positive resistance.

Combining forces before integrating is useful because it makes the physical balance explicit and often shortens the calculation. Integrating each contribution separately is also valid, as the examples show, and can provide an independent check. In either approach, the signed force model determines whether the external work is positive, zero, or negative.

Check Your Understanding

For each question, use signed force components and state the direction chosen as positive.

An object moves right at constant speed while two forces on it are $-5\text{ N}$ and $-2x\text{ N}$. What external force is required at position $x$?
For the forces in Question 1, write an integral for the external work from $x=0\text{ m}$ to $x=3\text{ m}$.
If a force assists the motion rather than opposing it, should its signed component be added or subtracted in the force balance?
Why must the force balance hold at each position in a constant-speed, straight-line model?
A force component is $-f(x)$, where $f(x)\geq0$. What is its contribution to external work when the external force balances it?

What You'll Learn