From Work to Potential Energy
A force can transfer energy as an object moves. When the force depends on position, the work it does is found by integrating the force along the motion. For a conservative force, that work can be recorded instead as a change in potential energy: the energy associated with the object’s position in the force field.
The sign is essential. If a conservative force does positive work, the potential energy decreases. If the force does negative work, the potential energy increases. This tutorial derives that relationship and applies it to spring and gravitational force models in one dimension.
Definition and Sign Convention
Let an object move along a line, with position coordinate \(x\), and let \(F(x)\) denote the component of the force in the positive \(x\)-direction. The work done by this force as the object moves from \(a\) to \(b\) is \(W_F=\int_a^b F(x)\,dx\). The integral is signed: a force pointing opposite the direction of motion contributes negative work.
In one dimension, motion is along an interval, so there is no alternative route between the same two positions that could change the work. In particular, if \(F\) is continuous on an interval, the integral defines a potential-energy function there. A potential-energy function is determined only up to an additive constant: choosing a different reference changes its numerical values but not its changes between positions.
Why the Negative Integral Appears
The definition follows from energy bookkeeping: the potential-energy change is chosen to cancel the work done by the conservative force. The calculus also makes the connection precise. The following derivation assumes \(F\) is continuous on an interval and chooses a reference position \(x_0\) with assigned potential energy \(U(x_0)\).
For any position \(x\) in the interval, set \(U(x)=U(x_0)-\int_{x_0}^{x}F(s)\,ds\). The variable \(s\) is a dummy integration variable.
By the Fundamental Theorem of Calculus, \(\frac{d}{dx}\int_{x_0}^{x}F(s)\,ds=F(x)\). Therefore \(U'(x)=-F(x)\), or equivalently \(F(x)=-U'(x)\).
Subtract the values at \(a\) and \(b\): \(U(b)-U(a)=-\int_{x_0}^{b}F(s)\,ds+\int_{x_0}^{a}F(s)\,ds=-\int_a^b F(s)\,ds\). This is the required change formula.
Thus, the force is the negative derivative of potential energy. A potential-energy function that rises as \(x\) increases corresponds to a force pointing toward smaller \(x\). A potential that falls corresponds to a force pointing toward larger \(x\).
The change formula is often more useful than an absolute value of \(U\), because physical calculations commonly compare two positions. If the potential is specified relative to a reference position \(x_0\), then the value at any other position follows from \(U(x)=U(x_0)-\int_{x_0}^{x}F(s)\,ds\).
Worked Example: Potential Energy of a Spring
Worked Example: Potential Energy of a Spring
A spring obeys Hooke’s law, with force \(F(x)=-kx\), where \(x\) is its displacement from its natural length. Let \(k=50\ \mathrm{N/m}\). Find the change in spring potential energy as the spring moves from \(x=0.10\ \mathrm{m}\) to \(x=0.30\ \mathrm{m}\), and take \(U(0)=0\).
Step 1: Write the change formula. Since the spring force is \(F(x)=-50x\ \mathrm{N}\),
Step 2: Integrate and substitute both limits.
The squared positions are \(0.30^2=0.09\) and \(0.10^2=0.01\), so their difference is \(0.08\ \mathrm{m^2}\). Therefore, \(\Delta U=25\ \mathrm{N/m}\times0.08\ \mathrm{m^2}=2.0\ \mathrm{J}\).
Step 3: Check using the potential function. Because \(U(0)=0\), integrate from zero to \(x\): \(U(x)=-\int_0^x(-50s)\,ds=25x^2\ \mathrm{J}\), with \(x\) measured in metres. At the two positions, \(U(0.30)=25(0.09)=2.25\ \mathrm{J}\) and \(U(0.10)=25(0.01)=0.25\ \mathrm{J}\). Their difference is \(2.25-0.25=2.0\ \mathrm{J}\), as calculated. Differentiating gives \(U'(x)=50x\ \mathrm{N}\), so \(-U'(x)=-50x=F(x)\).
The energy increases as the spring is stretched farther. The spring force points back toward \(x=0\), so while the spring is stretched in the positive direction, that force does negative work.
Constant Gravitational Force Near Earth
For vertical motion near Earth’s surface, take the upward direction as positive and let \(y\) measure height. If the mass is \(m\) and gravitational acceleration has magnitude \(g\), the force component is \(F(y)=-mg\). The force is constant in this model. Taking a change from \(y_i\) to \(y_f\), the potential-energy change is \(mg(y_f-y_i)\).
If upward motion increases the height, then \(y_f-y_i>0\) and gravitational potential energy increases. The formula does not require the initial height to be zero; only the height difference matters. The familiar choice \(U(y)=mgy\) corresponds to setting the zero of potential energy at \(y=0\).
Worked Example: Lifting an Object Against Constant Gravity
A \(2.5\ \mathrm{kg}\) object is lifted from \(y_i=1.0\ \mathrm{m}\) to \(y_f=4.0\ \mathrm{m}\). Use \(g=9.8\ \mathrm{m/s^2}\). Find the change in gravitational potential energy and check it against the work done by gravity.
Step 1: Identify the force and displacement. With upward positive, \(F(y)=-mg=-(2.5\ \mathrm{kg})(9.8\ \mathrm{m/s^2})=-24.5\ \mathrm{N}\). The height change is \(4.0-1.0=3.0\ \mathrm{m}\).
Step 2: Calculate the potential-energy change.
Step 3: Check by integrating the gravitational force. Its work is \(W_F=\int_{1.0}^{4.0}(-24.5)\,dy=(-24.5)(4.0-1.0)\ \mathrm{J}=-73.5\ \mathrm{J}\). Hence \(-W_F=73.5\ \mathrm{J}=\Delta U\), as required.
Inverse-Square Gravitational Force
At distances where the gravitational force cannot be treated as constant, a common one-dimensional model uses the distance \(r>0\) between two masses. Let \(C=GMm\), where \(G\) is the gravitational constant and \(M,m\) are the masses. Taking increasing \(r\) as the positive direction, the force on the smaller mass is \(F(r)=-C/r^2\): its direction is toward smaller separation.
The potential-energy change from \(r_i\) to \(r_f\) is obtained by integrating this force. If the reference is chosen so that \(U\) tends to zero as \(r\) tends to infinity, the resulting potential is \(U(r)=-C/r\). This choice is convenient, but other choices differing by a constant describe the same energy changes.
Worked Example: Change in Inverse-Square Gravitational Potential Energy
Suppose an object is subject to the radial force \(F(r)=-12/r^2\ \mathrm{N}\), with \(r\) measured in metres. Find the potential-energy change as the separation increases from \(2\ \mathrm{m}\) to \(4\ \mathrm{m}\). Here the force coefficient is \(C=12\ \mathrm{N\,m^2}\).
Step 1: Use the negative-work integral.
Step 2: Evaluate the integral with its limits. Since an antiderivative of \(r^{-2}\) is \(-r^{-1}\),
Step 3: Check with the potential function. Taking zero potential at infinity gives \(U(r)=-12/r\), with the coefficient carrying units \(\mathrm{N\,m^2}\). Thus \(U(2)=-6\ \mathrm{J}\) and \(U(4)=-3\ \mathrm{J}\). The difference is \(-3-(-6)=3\ \mathrm{J}\). Also, differentiating \(U(r)=-12/r\) gives \(U'(r)=12/r^2\), so \(-U'(r)=-12/r^2=F(r)\).
Reference Levels and a Common Sign Error
The value assigned to potential energy at a chosen position is a reference choice. Replacing \(U(x)\) by \(U(x)+K\), where \(K\) is any constant, leaves \(U(b)-U(a)\) unchanged and also leaves the force \(-U'(x)\) unchanged. Therefore, avoid treating a particular numerical value of \(U\) as absolute unless a reference has been specified.
A frequent error is to set the potential-energy change equal to the work done by the conservative force, without changing the sign. For the spring in the example, the work done by the spring from \(0.10\ \mathrm{m}\) to \(0.30\ \mathrm{m}\) is
Incorrect calculation: \(\Delta U=W_F=-2.0\ \mathrm{J}\). This would say that stretching the spring makes its potential energy decrease.
Correction: The definition gives \(\Delta U=-W_F=-(-2.0\ \mathrm{J})=+2.0\ \mathrm{J}\). The spring’s force opposes the stretching motion, so it does negative work while the spring’s stored potential energy increases.
Why the Relationship Is Useful
Potential energy lets us describe the effect of a conservative force through position rather than repeatedly evaluating the force’s work over a displacement. The derivative relationship \(F=-U'\) also gives a direct way to interpret a potential-energy graph: the force points downhill in the direction of decreasing \(U\). The sign and reference conventions remain important, especially when comparing different force models.
| Force model | Potential-energy change | One common reference choice |
|---|---|---|
| Spring, \(F(x)=-kx\) | \(\Delta U=\frac12k(x_f^2-x_i^2)\) | \(U(0)=0\), giving \(U(x)=\frac12kx^2\) |
| Constant gravity, \(F(y)=-mg\) | \(\Delta U=mg(y_f-y_i)\) | \(U(0)=0\), giving \(U(y)=mgy\) |
| Inverse-square gravity, \(F(r)=-C/r^2\) | \(\Delta U=C(1/r_i-1/r_f)\) | \(U(\infty)=0\), giving \(U(r)=-C/r\) |
In each row, the change formula follows from the same rule: integrate the force over position and take the negative. The model determines the integral; the reference choice determines only the additive constant in the potential.
Check Your Understanding
Use the force-to-potential relationship to answer these questions. Keep track of the direction of the force and the sign of work.
- If a conservative force does \(8\ \mathrm{J}\) of work while an object moves between two positions, what is the potential-energy change?
- A spring has force \(F(x)=-kx\), with \(k>0\). Starting from \(U(0)=0\), derive its potential-energy function.
- For constant gravity with upward-positive coordinate \(y\), does potential energy increase or decrease when an object is lowered?
- For \(F(r)=-C/r^2\), with \(C>0\), explain why increasing \(r\) increases potential energy when the zero is chosen at infinity.
- Why does adding a constant to \(U(x)\) leave both the force and the change in potential energy unchanged?
