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Work and Variable Forces · Tutorial 471 of 998

Solving Work Problems for Unknown Parameters

Use work integrals as equations to determine unknown quantities, then check that each answer fits the model and its physical conditions.

Intermediate 9 min read

What You'll Learn

  • Rearrange the constant-force work equation to find an unknown distance or force.
  • Determine a spring constant from the work required for a known extension.
  • Solve for an unknown spring extension while respecting its nonnegative physical domain.
  • Find an unknown coefficient in a polynomial force law from a measured work value.
  • Use a work integral to solve for distance and reject algebraic roots outside the interval conditions.
  • Verify solutions by substituting them into the original work equation.

Using Work as an Equation

Work problems do not always ask for the work done. Sometimes the work is already specified, and the unknown is a distance, a force constant, or a parameter in a force model. The integral that represents work then becomes an equation to solve.

The central task is to preserve the physical meaning of the model while doing the algebra. A candidate answer must satisfy the work equation, have the correct units, and belong to the allowed domain. For example, a spring extension is nonnegative, and a distance endpoint may be required to lie beyond the starting position.

Key idea: Write the work integral for the stated force and interval, set it equal to the specified work, and solve for the unknown. Then check the result in the original integral and against the physical conditions.

Work Equations and Unknowns

For a force \(F(x)\) acting along a line as an object moves from \(x=a\) to \(x=b\), the work done by that force is

$$ W=\int_a^b F(x)\,dx. $$

If \(F\) is constant, this reduces to \(W=F(b-a)\). The sign depends on the direction of the force relative to the displacement: a force opposing the motion does negative work. In the examples below, positive work means work supplied in the direction of the stated motion. For a spring being stretched slowly from its natural length, we use the positive applied-force magnitude \(F(x)=kx\); the work done by the spring itself has the opposite sign.

Once the integral is evaluated, the remaining equation may be linear, quadratic, or another algebraic equation. Solving the equation is only part of the task. The original interval, force model, and units determine which algebraic solutions are admissible.

Worked Examples

Worked Example: Find a Distance Under a Constant Force

A constant force of \(12\text{ N}\) does \(180\text{ J}\) of work while moving an object in the force's direction. Find the distance \(d\).

For constant force, \(W=Fd\). Substitute the known values:

$$ 180\text{ J}=(12\text{ N})d. $$

Divide both sides by \(12\text{ N}\):

$$ d=\frac{180\text{ J}}{12\text{ N}} =15\text{ m}, $$

since \(1\text{ J}=1\text{ N}\cdot\text{m}\). Check by substituting into the work equation:

$$ W=(12\text{ N})(15\text{ m})=180\text{ J}. $$

The distance is \(15\text{ m}\), and it reproduces the specified work.

Worked Example: Find a Spring Constant

An applied force stretches a spring from its natural length to an extension of \(3\text{ m}\). The required work is \(72\text{ J}\). Assuming Hooke's law \(F(x)=kx\), find the spring constant \(k\).

The extension coordinate runs from \(0\) to \(3\text{ m}\). Set the work integral equal to \(72\text{ J}\):

$$ 72\text{ J} =\int_0^3 kx\,dx =k\left[\frac{x^2}{2}\right]_0^3 =k\left(\frac{9}{2}\right). $$

Therefore,

$$ k=\frac{72}{9/2} =72\cdot\frac{2}{9} =16\text{ N/m}. $$

The units follow because \(W=\frac12 kx^2\), so \(k\) has units \(\text{J}/\text{m}^2=\text{N/m}\). Check the value in the original integral:

$$ \int_0^3 (16x)\,dx =16\left[\frac{x^2}{2}\right]_0^3 =8(9) =72\text{ J}. $$

Thus the spring constant is \(16\text{ N/m}\).

Worked Example: Find a Spring Extension

A spring has constant \(k=8\text{ N/m}\). How far must it be stretched from its natural length for the applied work to equal \(32\text{ J}\)?

Let \(x\) be the extension. The work equation is

$$ 32\text{ J} =\int_0^x 8u\,du =4x^2. $$

Solving gives

$$ x^2=8\text{ m}^2, \qquad x=\pm 2\sqrt{2}\text{ m}. $$

The negative root is not an admissible extension in this setup: \(x\) measures stretching from the natural length and must satisfy \(x\ge 0\). Hence \(x=2\sqrt{2}\text{ m}\). Check:

$$ \int_0^{2\sqrt{2}} 8u\,du =4(2\sqrt{2})^2 =4(8) =32\text{ J}. $$

The required extension is \(2\sqrt{2}\text{ m}\), approximately \(2.83\text{ m}\) if rounded to two decimal places.

Worked Example: Find a Polynomial Force Parameter

Suppose a force along a line is modeled, for \(0\le x\le 2\text{ m}\), by \(F(x)=ax^2+4\text{ N}\), where \(x\) is measured in metres and \(a\) is an unknown coefficient. The work from \(0\) to \(2\text{ m}\) is \(16\text{ J}\). Find \(a\).

In SI units, the coefficient \(a\) has units \(\text{N}/\text{m}^2\). Evaluate the work integral and equate it to the specified value:

$$ 16 =\int_0^2 (ax^2+4)\,dx =\left[\frac{ax^3}{3}+4x\right]_0^2 =\frac{8a}{3}+8. $$

Subtract \(8\) from both sides and solve:

$$ 8=\frac{8a}{3}, \qquad a=3\text{ N/m}^2. $$

Check by substituting the coefficient into the force model and integrating:

$$ \int_0^2 (3x^2+4)\,dx =\left[x^3+4x\right]_0^2 =(8+8)-0 =16\text{ J}. $$

The coefficient \(a=3\text{ N/m}^2\) gives the required work.

Worked Example: Find an Endpoint and Select the Physical Root

A force \(F(x)=3x+2\text{ N}\), with \(x\) measured in metres, acts from position \(x=1\text{ m}\) to an unknown endpoint \(x=d\). The work is \(24\text{ J}\), and the motion proceeds forward, so \(d\ge 1\text{ m}\). Find \(d\).

Using the work integral and evaluating in SI units:

$$ 24 =\int_1^d (3x+2)\,dx =\left[\frac{3x^2}{2}+2x\right]_1^d =\frac{3d^2}{2}+2d-\frac{7}{2}. $$

Multiply by \(2\) and rearrange:

$$ 48=3d^2+4d-7, \qquad 3d^2+4d-55=0. $$

Factor the quadratic:

$$ (3d-11)(d+5)=0. $$

The algebraic candidates are \(d=11/3\text{ m}\) and \(d=-5\text{ m}\). Only \(11/3\text{ m}\) satisfies the required condition \(d\ge1\text{ m}\). Check it in the original integral:

$$ \int_1^{11/3}(3x+2)\,dx =\left[\frac{3x^2}{2}+2x\right]_1^{11/3} =\left(\frac{55}{2}\right)-\left(\frac{7}{2}\right) =24\text{ J}. $$

The physically admissible endpoint is \(d=11/3\text{ m}\).

Why the Distance Solution Is Often Unique

When the force is continuous and positive throughout the interval of motion, increasing the endpoint increases the accumulated work. This explains why a specified work value determines at most one endpoint in that range.

1
Define accumulated work.
For a fixed starting point \(a\), let \(W(d)=\int_a^d F(x)\,dx\).
2
Compare two endpoints.
If \(d_2>d_1\ge a\), then \(W(d_2)-W(d_1)=\int_{d_1}^{d_2}F(x)\,dx\).
3
Use positivity.
If \(F(x)>0\) on that interval, then \(\int_{d_1}^{d_2}F(x)\,dx>0\), so \(W(d_2)>W(d_1)\). Therefore the same work value cannot correspond to two different endpoints in the allowed range.

This argument requires positivity on the interval being considered. If a force changes sign, accumulated work need not increase with the endpoint, and an equation for distance may have multiple admissible solutions. In every case, the original interval restrictions must be applied after solving the algebraic equation.

A Common Mistake: Ignoring the Domain

A frequent error is to keep every algebraic root without checking what the variable represents. In the spring-extension example, solving \(x^2=8\) gives two real numbers. But \(x\) was defined as an extension measured outward from the natural length, so the model requires \(x\ge0\). The negative root is an algebraic solution of the squared equation, not a valid extension for this situation.

Another useful check is dimensional consistency. If a calculation for an unknown distance produces units of newtons, or a spring constant is reported in joules, something has gone wrong in the setup or algebra. Units do not replace substitution, but they can reveal errors before the final check.

Check each solution: Substitute it into the original work equation, verify its units, and confirm it lies in the stated physical domain. Do not rely only on the rearranged equation.

Why Parameter Solving Matters

Work measurements can be used to infer properties of a system. A known force and measured work can reveal a distance; a known extension and work can determine a spring constant; and a specified force model can be calibrated by solving for an unknown coefficient. The common mathematical structure is an integral equation in which one quantity is unknown.

Keeping the model and its assumptions visible makes these calculations more reliable. In particular, distinguish the force doing the work from a force opposing the motion, include the correct endpoints, carry units through the calculation, and use the original equation to verify the final result.

Check Your Understanding

For each problem, identify the work equation, solve for the requested quantity, and check any domain restrictions.

  1. A constant force of \(9\text{ N}\) does \(63\text{ J}\) of work in its direction. What distance does the object move?
  2. A spring is stretched \(2\text{ m}\) from its natural length using \(20\text{ J}\) of work. What is its spring constant?
  3. A spring with constant \(5\text{ N/m}\) is stretched from its natural length until the applied work is \(10\text{ J}\). What nonnegative extension is required?
  4. For \(F(x)=ax+2\text{ N}\), where \(x\) is measured in metres, the work from \(0\) to \(2\text{ m}\) is \(12\text{ J}\). Find \(a\), including its units.
  5. Why must a solution of an algebraic equation for an endpoint be checked against the interval conditions in the problem?

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