Conservative Forces and Potential Energy
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You should be comfortable calculating line integrals of vector fields and applying the work-energy theorem. Familiarity with the gradient operator \(\nabla\) and basic vector calculus identities is assumed.
Why this matters
Conservative forces allow us to replace complex path-dependent work calculations with simple state-function differences.
In many physical systems, the work done by a force depends only on the starting and ending positions, not the specific trajectory taken. This property defines conservative forces and introduces the concept of potential energy. By identifying conservative forces, we can store the work done in a scalar potential field, simplifying the analysis of motion and laying the groundwork for the conservation of mechanical energy.
The Idea of Path Independence
Imagine pushing a box across a floor. If the floor is rough, the work you do depends on how far you push it and the path you take. However, if you lift a book from a table to a shelf, the work done against gravity depends only on the vertical height change, not on whether you lift it straight up or move it sideways first. This distinction is the core of conservative forces. A force is conservative if the work it does moving a particle between two points is independent of the path taken. This implies that if you move a particle around a closed loop and return it to its starting point, the net work done by the conservative force is zero. This path independence allows us to define a scalar quantity, potential energy, associated with the position of the particle.
Mathematical Definition and Equivalence
Formally, a force field \(\vec{F}(\vec{r})\) is conservative if the line integral around any closed loop \(\mathcal{C}\) vanishes:
For a force field that is continuously differentiable on a simply connected region, this condition is equivalent to the field being curl-free:
Furthermore, a conservative force can always be expressed as the negative gradient of a scalar potential energy function \(U(\vec{r})\):
These three statements are equivalent. The work done by a conservative force moving a particle from point \(A\) to point \(B\) is given by the difference in potential energy:
Note the negative sign in the force-potential relationship. This ensures that the force points in the direction of decreasing potential energy, which is the direction in which the force pushes the particle.
Examples of Conservative Forces
Gravity near the Earth's surface is a classic example. The force is \(\vec{F} = -mg\hat{z}\). We can identify a potential energy \(U(z) = mgz\) such that \(\vec{F} = -\nabla U\). The work done by gravity as an object of mass \(m\) rises from height \(z_1\) to \(z_2\) is \(W = mgz_1 - mgz_2 = -mg(z_2 - z_1)\). This result is independent of the horizontal path taken.
Another example is the spring force in Hooke's law, \(\vec{F} = -k\vec{x}\). The potential energy is \(U(x) = \frac{1}{2}kx^2\). The work done by the spring moving from \(x_1\) to \(x_2\) is \(W = \frac{1}{2}kx_1^2 - \frac{1}{2}kx_2^2\). Again, the work depends only on the endpoints.
Electrostatic forces are also conservative. The electric field \(\vec{E}\) from a static charge distribution satisfies \(\nabla\times\vec{E}=\vec{0}\), allowing the definition of an electric potential \(V\) such that \(\vec{E} = -\nabla V\). The potential energy of a charge \(q\) is \(U = qV\).
Non-Conservative Forces
Not all forces are conservative. Friction is the most common example. The force of kinetic friction always opposes the direction of motion. If you slide a block around a closed loop, the friction force does negative work along every segment of the path. The total work done by friction around the closed loop is not zero; it is negative and proportional to the total distance traveled. Therefore, friction is non-conservative. We cannot define a potential energy function for friction. The work done by friction depends on the path length, making it path-dependent.
Other non-conservative forces include air resistance and driving forces in engines. When non-conservative forces are present, the work-energy theorem must be modified to account for the work done by these forces separately, as they do not contribute to a potential energy term.
The derivation
Conservative Forces and Potential Energy
For a force field \(\vec{F}(\vec{r})\) that depends on position alone and is continuously differentiable on a simply connected region of \(\mathbb{R}^3\), the following three statements are equivalent: (i) the work \(\int_{\mathcal{C}}\vec{F}\cdot d\vec{r}\) depends only on the endpoints of the path \(\mathcal{C}\), equivalently \(\oint\vec{F}\cdot d\vec{r}=0\) around every closed loop; (ii) the field is curl-free, \(\nabla\times\vec{F}=\vec{0}\) everywhere; and (iii) there exists a single-valued scalar field \(U(\vec{r})\), the potential energy, with \(\vec{F}=-\nabla U\). A force satisfying any (hence all) of these is called conservative, and the work it does is \(W_{A\to B}=U(A)-U(B)\).
The equivalence of path independence and curl-free conditions holds only for simply connected regions. In regions with holes (for example the region outside a long straight current-carrying wire, where the magnetic field is curl-free but circulates around the wire), a field can be curl-free locally but not conservative globally because the line integral around a loop encircling the hole may not vanish. This subtlety is crucial in advanced electromagnetism and fluid dynamics. C
Worked example
Problem. A particle of mass \(m = 2.0\ \mathrm{kg}\) moves in the \(xy\)-plane under the influence of a force field \(\vec{F}(x,y) = (3x^2 + 2y)\hat{i} + (2x)\hat{j}\ \mathrm{N}\). Determine if the force is conservative. If so, find the potential energy function \(U(x,y)\) given that \(U(0,0) = 0\). Calculate the work done by the force moving the particle from \((0,0)\) to \((2,1)\ \mathrm{m}\).
- Check if the force is conservative by verifying \(\nabla\times\vec{F}=\vec{0}\). For a 2D field, this requires \(\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}\). Here, \(\frac{\partial F_y}{\partial x} = \frac{\partial (2x)}{\partial x} = 2\) and \(\frac{\partial F_x}{\partial y} = \frac{\partial (3x^2 + 2y)}{\partial y} = 2\). Since they are equal, the force is conservative.
- Find the potential energy \(U(x,y)\) such that \(\vec{F} = -\nabla U\). This gives \(\frac{\partial U}{\partial x} = -F_x = -(3x^2 + 2y)\) and \(\frac{\partial U}{\partial y} = -F_y = -2x\). Integrate \(\frac{\partial U}{\partial x}\) with respect to \(x\): \(U(x,y) = -x^3 - 2xy + f(y)\). Differentiate with respect to \(y\): \(\frac{\partial U}{\partial y} = -2x + f'(y)\). Set equal to \(-2x\): \(f'(y) = 0\), so \(f(y) = C\). Given \(U(0,0)=0\), we have \(C=0\). Thus, \(U(x,y) = -x^3 - 2xy\).
- Calculate the work done from \((0,0)\) to \((2,1)\). \(W = U(0,0) - U(2,1)\). \(U(0,0) = 0\). \(U(2,1) = -(2)^3 - 2(2)(1) = -8 - 4 = -12\ \mathrm{J}\). Therefore, \(W = 0 - (-12) = 12\ \mathrm{J}\).
Units check. The force is in Newtons, distance in meters, so work is in Joules. The potential energy terms \(x^3\) and \(xy\) must have units of Joules. This implies the coefficients in the force expression have appropriate units to make the integral consistent. The calculation is dimensionally consistent.
Three quick tests
- Is the force \(\vec{F} = y\hat{i} - x\hat{j}\) conservative?
Answer
No. \(\frac{\partial F_y}{\partial x} = -1\) and \(\frac{\partial F_x}{\partial y} = 1\). They are not equal, so \(\nabla\times\vec{F} \neq \vec{0}\). - If a conservative force does \(50\ \mathrm{J}\) of work moving a particle from A to B, what is the change in potential energy \(\Delta U\)?
Answer
\(\Delta U = U_B - U_A = -W = -50\ \mathrm{J}\). - Can you define a potential energy for the force of kinetic friction?
Answer
No, because the work done by friction depends on the path length, not just the endpoints. It is non-conservative.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.