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Conservative Fields and Path Independence · Tutorial 791 of 999

Work and Change in Potential Energy

Learn why conservative-force work is the negative change in potential energy, and use that relationship to solve gravitational and spring examples.

Intermediate 9 min read

What You'll Learn

  • Define potential energy for a conservative force field and distinguish it from a scalar potential.
  • Derive the relationship between work and the change in potential energy along a curve.
  • Calculate gravitational work and potential-energy changes using a chosen reference height.
  • Use the spring-force law to calculate work and elastic potential-energy changes.
  • Check signs, units, and results by direct line integration.

Work and Potential Energy

A force can transfer energy as an object moves. The amount transferred depends on the force and the motion, and in general it may depend on the path taken. For a conservative force, however, the work is determined by the starting and ending positions. Potential energy packages that endpoint information into a scalar quantity: the work done by the force is the decrease in potential energy.

The sign matters. If a conservative force does positive work, the potential energy decreases. If it does negative work, the potential energy increases. This relationship connects the line integrals of vector calculus with familiar examples such as gravity and springs.

Core relationship: If \(\mathbf{F}\) is a conservative force with potential energy \(U\), so that \(\mathbf{F}=-\nabla U\), then the work done by \(\mathbf{F}\) in moving from \(A\) to \(B\) is \(U(A)-U(B)\). Equivalently, the work is the negative of the change in potential energy.

Force Potential and Potential Energy

A vector field \(\mathbf{F}\) is conservative on a domain if there is a scalar function \(\phi\) such that \(\mathbf{F}=\nabla\phi\). The fundamental theorem for line integrals then gives the work from \(A\) to \(B\) as \(\phi(B)-\phi(A)\). In mechanics, the conventional potential energy \(U\) is the negative of this scalar potential, up to an arbitrary additive constant. Thus \(\mathbf{F}=-\nabla U\).

Definition: A potential energy function for a conservative force \(\mathbf{F}\) is a differentiable scalar function \(U\) satisfying \(\mathbf{F}=-\nabla U\). The potential energy change from \(A\) to \(B\) is \(\Delta U=U(B)-U(A)\), and the work by the force along a curve \(C\) from \(A\) to \(B\) is \(W_C=\int_C\mathbf{F}\cdot d\mathbf{r}\).

Potential energy is measured in joules when force is measured in newtons and position in metres, since a line integral of force over displacement has units \(\mathrm{N\,m}=\mathrm{J}\). The zero of \(U\) is a choice: adding a constant changes potential-energy values but not their differences or the force \(-\nabla U\).

Why Work Equals the Decrease in Potential Energy

Let \(\mathbf{r}(t)\), for \(a\le t\le b\), be a piecewise smooth path from \(A=\mathbf{r}(a)\) to \(B=\mathbf{r}(b)\). Assume \(U\) is differentiable along the path and \(\mathbf{F}=-\nabla U\). On each smooth segment, the chain rule gives the rate of change of \(U\) along the path.

1
Write work as a parameter integral.
By the definition of a line integral, \(W_C=\int_a^b\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt\).
2
Substitute the potential-energy relation.
Since \(\mathbf{F}=-\nabla U\), the integrand is \(-\nabla U(\mathbf{r}(t))\cdot\mathbf{r}'(t)\).
3
Apply the chain rule and integrate.
The chain rule identifies \(\nabla U(\mathbf{r}(t))\cdot\mathbf{r}'(t)\) with \(\frac{d}{dt}U(\mathbf{r}(t))\). Therefore the integral is \(-[U(\mathbf{r}(b))-U(\mathbf{r}(a))]\).
$$ W_C = -\int_a^b \frac{d}{dt}U(\mathbf{r}(t))\,dt = U(A)-U(B) = -\Delta U. $$

The calculation applies segment by segment if the path has finitely many smooth pieces; the intermediate endpoint terms cancel when the pieces are added. This proves the result for any piecewise smooth path on which the potential is defined. It also explains path independence: the final expression contains only the endpoint values.

Worked Example: Work Done by Gravity During a Descent

Worked Example: Work Done by Gravity During a Descent

A \(2\,\mathrm{kg}\) object moves from height \(12\,\mathrm{m}\) to height \(4\,\mathrm{m}\) in a uniform gravitational field. Let the vertical coordinate \(y\) increase upward, and use \(g=9.8\,\mathrm{m/s^2}\). Find the work done by gravity and the change in gravitational potential energy.

The gravitational force is \(\mathbf{F}=(0,-mg)=(0,-19.6)\,\mathrm{N}\). The corresponding potential energy, choosing zero at \(y=0\), is \(U(y)=mgy\). To check the work directly, take the straight path \(\mathbf{r}(t)=(3t,12-8t)\,\mathrm{m}\), \(0\le t\le1\). It has the required endpoint heights, and \(\mathbf{r}'(t)=(3,-8)\,\mathrm{m}\).

$$ W_C = \int_0^1 (0,-19.6)\cdot(3,-8)\,dt = \int_0^1 156.8\,dt = 156.8\,\mathrm{J}. $$

The endpoint energies are \(U(A)=(2)(9.8)(12)=235.2\,\mathrm{J}\) and \(U(B)=(2)(9.8)(4)=78.4\,\mathrm{J}\). Hence \(\Delta U=78.4-235.2=-156.8\,\mathrm{J}\), so \(-\Delta U=156.8\,\mathrm{J}\), agreeing with the line integral. Gravity does positive work as the object descends, while its potential energy decreases by the same amount.

Worked Example: A One-Dimensional Spring

Worked Example: Work Done by a Spring

A spring with spring constant \(k=8\,\mathrm{N/m}\) exerts force \(F(x)=-kx\), where \(x\) is displacement in metres from its equilibrium position. Find the work done by the spring as the spring moves from \(x=2\,\mathrm{m}\) to \(x=0.5\,\mathrm{m}\).

The potential energy is \(U(x)=\frac12kx^2\), because \(-\frac{dU}{dx}=-kx=F(x)\). For a direct calculation, parameterize the motion by \(x(t)=2-1.5t\), \(0\le t\le1\), so \(x'(t)=-1.5\,\mathrm{m}\). Along the path, \(F(x(t))=-8(2-1.5t)\,\mathrm{N}\).

$$ \begin{aligned} W_C &=\int_0^1 -8(2-1.5t)(-1.5)\,dt\\ &=12\int_0^1(2-1.5t)\,dt\\ &=12[2t-0.75t^2]_0^1\\ &=12(2-0.75)=15\,\mathrm{J}. \end{aligned} $$

At the initial position, \(U(2)=\frac12(8)(2^2)=16\,\mathrm{J}\). At the final position, \(U(0.5)=\frac12(8)(0.5^2)=1\,\mathrm{J}\). Therefore \(U(A)-U(B)=16-1=15\,\mathrm{J}\), matching the direct calculation. The spring does positive work while returning toward equilibrium, and the stored elastic potential energy falls by \(15\,\mathrm{J}\).

Worked Example: A Spring in Three Dimensions

Worked Example: Work Along a Three-Dimensional Spring Path

A spring centered at the origin exerts the three-dimensional force \(\mathbf{F}(\mathbf{r})=-3\mathbf{r}\), with position in metres and force in newtons. Find its work as the endpoint changes from \(A=(1,2,0)\,\mathrm{m}\) to \(B=(2,0,2)\,\mathrm{m}\) along the straight segment.

This force has potential energy \(U(\mathbf{r})=\frac12(3)\|\mathbf{r}\|^2\). Parameterize the segment by \(\mathbf{r}(t)=(1+t,2-2t,2t)\,\mathrm{m}\), \(0\le t\le1\), with \(\mathbf{r}'(t)=(1,-2,2)\,\mathrm{m}\). Thus \(\mathbf{F}(\mathbf{r}(t))=(-3(1+t),-3(2-2t),-6t)\,\mathrm{N}\).

$$ \begin{aligned} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) &=-3(1+t)+6(2-2t)-12t\\ &=9-27t,\\ W_C&=\int_0^1(9-27t)\,dt =[9t-13.5t^2]_0^1 =-4.5\,\mathrm{J}. \end{aligned} $$

The endpoint squared lengths are \(\|A\|^2=1^2+2^2+0^2=5\,\mathrm{m^2}\) and \(\|B\|^2=2^2+0^2+2^2=8\,\mathrm{m^2}\). Therefore \(U(A)=1.5(5)=7.5\,\mathrm{J}\) and \(U(B)=1.5(8)=12\,\mathrm{J}\). The endpoint formula gives \(U(A)-U(B)=7.5-12=-4.5\,\mathrm{J}\), confirming the line-integral result. The negative work is consistent with the spring being stretched farther from its equilibrium position.

From Potential Energy to Mechanical Energy

Potential energy describes the energy associated with position in a conservative force field. For a particle of mass \(m\) moving with velocity \(\mathbf{v}\), its kinetic energy is \(K=\frac12m\|\mathbf{v}\|^2\). The work-energy principle says that the net work on the particle equals its change in kinetic energy. Consequently, if the conservative force is the only force doing work, then

$$ \Delta K=W_C=-\Delta U, \qquad \Delta(K+U)=0. $$

This conclusion requires that no other force contribute work over the motion. If friction or another nonconservative force does work, its contribution must also be included; in that case \(K+U\) need not remain constant. The work-potential relation for the conservative force itself remains valid.

A Common Sign Mistake

A frequent error is to report the work by a conservative force as \(U(B)-U(A)\). That quantity is the change in potential energy, not the work. For the descending object above, this mistaken calculation would say \(W=78.4-235.2=-156.8\,\mathrm{J}\). But gravity points downward and the object moves downward, so its work must be positive. The correction is

$$ W_C=U(A)-U(B) =235.2-78.4 =156.8\,\mathrm{J} =-\Delta U. $$

A reliable check is to compare the directions of force and displacement locally: when they align overall, positive work is plausible; when they oppose, negative work is plausible. Then verify the sign using \(-\Delta U\). This check does not replace the calculation, but it can reveal a reversed endpoint subtraction.

Why the Relationship Is Useful

A line integral may require tracking a force along an entire curve. Once a potential energy is known, the work can instead be found from two endpoint evaluations. This is especially useful when a path is complicated but the force is conservative. It also provides a physical interpretation for endpoint potentials: the decrease in stored potential energy is the energy transferred by the force as work.

For uniform gravity near Earth, \(U=mgy\) after choosing a reference height. For an ideal spring, \(U=\frac12kx^2\) relative to equilibrium. In both cases, the potential formula must be paired with the force convention and coordinate choice used to define it. The central sign rule remains the same: work by a conservative force equals initial potential energy minus final potential energy.

Check Your Understanding

Use the definitions and sign convention in this tutorial to answer the following questions.

  1. If a conservative force does \(-6\,\mathrm{J}\) of work, what is the change in potential energy?
  2. For a mass moving downward in a uniform gravitational field with upward-positive height coordinate, does gravitational potential energy increase or decrease?
  3. Why can adding a constant to a potential energy function not change the force or work?
  4. A spring has \(k=4\,\mathrm{N/m}\). Find its potential energy at \(x=3\,\mathrm{m}\) relative to equilibrium.
  5. What additional condition allows the conclusion that \(K+U\) stays constant?

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