Visighta

Work and Kinetic Energy

7.4 Conservative Forces and Potential Energy

Why Doesn't the Path Matter?

A roller coaster starts at the top of a hill and winds through loops, dips, and corkscrews before returning to the same height. If friction is negligible, what is its final speed? Exactly the same as its initial speed. It does not matter whether the track was a gentle slope, a vertical loop, or a serpentine nightmare of twists and turns. The path is irrelevant.

This should feel strange. The roller coaster experienced different forces at every point along each track --- different normal forces, different directions of acceleration. The detailed force history was completely different for different track shapes. And yet the final speed came out the same.

Something about the gravitational force makes the path not matter. That "something" is the central idea of this section, and it will transform the way you solve mechanics problems. Instead of tracking forces along entire trajectories, you will learn to keep a simple ledger of stored energy. The result is one of the most powerful tools in all of physics: energy conservation.

Before you read on: A ball rolls from point A to point B. Two paths are available --- one is a straight ramp, the other is a long, curved slide. Only gravity does work on the ball (the surfaces are frictionless, and the normal force is always perpendicular to the motion).

Is the work done by gravity the same on both paths, or does the longer path mean more work?

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. The student selects "same work on both paths" or "more work on the longer path." Their choice is locked in and revisited after the exploration.]

Exploration: Path Independence Under Gravity

The following interactive lets you test the prediction directly.

[Interactive: Path Independence Lab. Two points, A and B, are fixed in a uniform gravitational field (A is higher than B). Three or four different paths connect A to B --- a straight diagonal ramp, a curved slide, a staircase path, and a path that dips below B before rising back up. For each path, the simulation computes the work done by gravity step-by-step along the trajectory, showing $W = \int \vec{F}_g \cdot d\vec{r}$ accumulating as the ball moves.

The student can: - Select which path(s) to display - Watch the ball traverse each path at adjustable speed - See the running work total for each path update in real time - Compare final work values side by side

A toggle switch labeled "Add friction" allows the student to introduce a kinetic friction force along each path.

Guided prompts:]

Step 1: Run the ball along the straight ramp from A to B. Note the total work done by gravity.

Step 2: Now run the ball along the curved slide. Compare the total work done by gravity to the straight-ramp value.

Step 3: Try the staircase path and the dip-below path. How does the work done by gravity compare across all four paths?

Step 4: Now turn on friction. Run the ball along the straight ramp and the curved slide again. Is the work done by friction the same on both paths?

If you worked through these steps, you discovered two things:

  1. Gravity: The work done by gravity is the same on every path between A and B. It depends only on the height difference $\Delta h = h_A - h_B$, not on the shape of the route. The work is $W_{\text{grav}} = mg\Delta h$ regardless of the path taken.

  2. Friction: The work done by friction is not the same. The longer, more winding paths accumulate more friction work. Friction cares about every meter of the journey.

This is the fundamental distinction that the rest of this section builds on.

The Concept: Conservative Forces and Path Independence

Here is the idea, stated precisely.

A conservative force is one for which the work done on an object depends only on the starting and ending positions --- not on the path taken between them. Equivalently, the work done by a conservative force around any closed path (one that returns to its starting point) is zero.

$$W_{\text{conservative}} = \text{function of endpoints only}$$

$$\oint \vec{F}_{\text{conservative}} \cdot d\vec{r} = 0$$

These two statements are logically equivalent. If the work depends only on endpoints, then any round trip (same start and end) gives zero net work. Conversely, if every round trip gives zero net work, then the work between two points cannot depend on which path you take (because you could combine two different paths into a round trip, and the total must be zero).

A nonconservative force is one for which the work does depend on the path. Friction is the classic example. A longer path means more friction work. A round trip with friction does not return zero --- the friction removes energy on every leg of the trip.

Examples of conservative forces: - Gravity (near Earth's surface): $W = mg\Delta h$. Only the height difference matters. - The spring force: $W = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$. Only the initial and final compressions matter. - The gravitational force (general, $1/r^2$ law): the work depends only on initial and final distances from the center.

Examples of nonconservative forces: - Kinetic friction: the work depends on the length of the path. Longer path, more energy removed. - Air resistance: same issue. The work depends on how far the object actually travels through the air. - A hand pushing an object along a specified route: the work depends on the path you choose to push.

Pause and think: A block slides from point A to point B on a frictionless surface, then back from B to A along the same path. The only horizontal force is gravity (the surface is tilted). What is the total work done by gravity for the round trip?

Now imagine the same round trip, but with friction. What is the total work done by friction?

Check your answer **Gravity:** The total work for the round trip is zero. On the way from A to B, gravity does work $W_{AB}$. On the way back from B to A, gravity does work $-W_{AB}$ (same magnitude, opposite sign, because the displacement is reversed). The sum is zero. This is exactly what "conservative" means. **Friction:** The total work for the round trip is *negative* and *nonzero*. On the way from A to B, friction opposes the motion and does negative work. On the way back from B to A, friction *still* opposes the motion (it reversed direction!) and does negative work again. The friction work on each leg is negative, so the total is negative. Energy is lost on both legs --- friction never gives it back. This is why friction is called "nonconservative." A round trip with friction always costs energy.

From Path Independence to Potential Energy

Path independence is not just a curiosity. It unlocks a powerful conceptual and computational tool: potential energy.

Here is the logic. If the work done by a conservative force depends only on the starting and ending positions, then we can define a function of position --- call it $U(\vec{r})$ --- that captures all the information about the work. Specifically, we define the potential energy $U$ so that:

$$W_{\text{conservative}} = -\Delta U = -(U_f - U_i) = U_i - U_f$$

The work done by the conservative force equals the negative of the change in potential energy. When the object moves to a position where $U$ is lower, the conservative force does positive work (it pushes the object in the direction it moves). When $U$ increases, the force does negative work (it opposes the motion).

Why the negative sign? Because potential energy represents stored energy. When gravity does positive work on a falling ball (the ball speeds up), the gravitational potential energy decreases --- the stored energy is being released and converted to kinetic energy. The decrease in $U$ equals the work done by the force. The negative sign keeps the bookkeeping consistent.

For gravity near Earth's surface:

$$U_{\text{grav}} = mgy$$

where $y$ is the height above some chosen reference level. The work done by gravity is:

$$W_{\text{grav}} = -\Delta U = -(mgy_f - mgy_i) = mg(y_i - y_f) = mg\Delta h$$

This matches what you already know: gravity does positive work when the object descends ($y_i > y_f$) and negative work when it ascends.

For a spring (Hooke's law):

$$U_{\text{spring}} = \frac{1}{2}kx^2$$

where $x$ is the displacement from the spring's natural length. The work done by the spring is:

$$W_{\text{spring}} = -\Delta U = -\left(\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2\right) = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$$

The spring does positive work when it returns toward its natural length ($|x_f| < |x_i|$) and negative work when it is stretched or compressed further.

Notice the critical requirement: defining $U$ is only possible because the work is path-independent. If the work between two points depended on the route, you could not assign a single number $U$ to each position. The function $U(\vec{r})$ would be ambiguous --- its value would depend on how you got there. Path independence guarantees that $U$ at each position is well-defined.

This is why you cannot define a "friction potential energy." Friction work depends on the path, so no single function of position can capture it.

The Payoff: Energy Conservation

Now combine potential energy with the work-energy theorem from Section 7.3.

The work-energy theorem says:

$$W_{\text{net}} = \Delta KE$$

Split the net work into work done by conservative forces and work done by nonconservative forces:

$$W_{\text{conservative}} + W_{\text{nonconservative}} = \Delta KE$$

For the conservative part, substitute $W_{\text{conservative}} = -\Delta U$:

$$-\Delta U + W_{\text{nonconservative}} = \Delta KE$$

Rearrange:

$$\Delta KE + \Delta U = W_{\text{nonconservative}}$$

If there are no nonconservative forces (or they do no work --- for example, the normal force, which is always perpendicular to the motion), then:

$$\Delta KE + \Delta U = 0$$

$$KE_f + U_f = KE_i + U_i$$

$$\frac{1}{2}mv_f^2 + U_f = \frac{1}{2}mv_i^2 + U_i$$

This is conservation of mechanical energy. The total mechanical energy $E = KE + U$ stays constant throughout the motion, as long as only conservative forces do work.

This single equation replaces the entire force-analysis workflow for a huge class of problems. You do not need to know the forces at every point. You do not need to solve a differential equation. You set up two snapshots --- initial and final --- write down the kinetic and potential energies in each, and equate the totals.

Return to the roller coaster. The coaster starts at height $h$ with speed $v_0$. It returns to height $h$ at the end. By energy conservation:

$$\frac{1}{2}mv_f^2 + mgh = \frac{1}{2}mv_0^2 + mgh$$

The $mgh$ terms cancel:

$$v_f = v_0$$

The final speed equals the initial speed. The track shape is irrelevant --- the equation never mentioned it. The entire complex trajectory collapsed into a comparison of two states.

Historical Context

The concept of "stored energy" --- potential energy --- was a 19th-century breakthrough. Before it, every mechanics problem required a force-by-force analysis. You had to trace the object's path, compute forces at each point, integrate, and solve differential equations. There was no shortcut.

The realization that certain forces could be described by an energy function, and that total energy was conserved, collapsed entire classes of problems into one equation. The work of scientists like Leibniz, Lagrange, and Helmholtz transformed mechanics from a problem-by-problem exercise into a systematic framework.

Energy conservation was not discovered by watching objects move. It was discovered by noticing that something did not change --- that a particular combination of quantities ($KE + U$) remained constant. The ability to see what stays the same while everything else changes is one of the deepest skills in physics.

Connection to Previous Sections

The work-energy theorem (Section 7.3) was already a step up from Newton's laws: it traded a vector equation for a scalar one, connecting force, distance, and speed without needing time. But it still required computing net work --- summing up contributions from every force along the path.

The move to energy conservation is the next leap. By splitting forces into conservative (bookkeepable) and nonconservative (path-dependent), you can absorb the conservative forces into a potential energy function. The work-energy theorem becomes:

$$\Delta KE + \Delta U = W_{\text{nc}}$$

where $W_{\text{nc}}$ is the work done by nonconservative forces only. When $W_{\text{nc}} = 0$, energy is conserved. When $W_{\text{nc}} \neq 0$, the equation tells you exactly how much mechanical energy was added or removed by nonconservative forces.

This is not a new law of physics. It is the work-energy theorem, reorganized. But the reorganization is transformative. It turns a calculation (integrate forces along a path) into bookkeeping (compare energies at two points).

Worked Example: A Ball Launched Vertically

A ball of mass 0.5 kg is thrown straight up with an initial speed of 10 m/s from a height of 2 m above the ground. Find the maximum height it reaches. Ignore air resistance.

Using energy conservation:

Choose the ground as the reference level ($y = 0$). At the initial point, the ball has kinetic energy and gravitational potential energy. At the maximum height, the ball is momentarily at rest ($v_f = 0$).

$$KE_i + U_i = KE_f + U_f$$

$$\frac{1}{2}(0.5)(10)^2 + (0.5)(9.8)(2) = 0 + (0.5)(9.8)(y_{\max})$$

$$25 + 9.8 = 4.9\, y_{\max}$$

$$y_{\max} = \frac{34.8}{4.9} \approx 7.1 \text{ m}$$

Notice: we never needed to think about the force of gravity acting along the path. We compared two snapshots --- start and top --- and the equation did the rest.

Faded Example: Mass on a Spring

A 0.3 kg block is pressed against a horizontal spring ($k = 400$ N/m), compressing it by 0.15 m. The block is released and slides along a frictionless surface. Find the speed of the block when it leaves the spring (at the spring's natural length).

Step 1: Identify the initial and final states. What are the kinetic and potential energies in each?

Check your answer **Initial state:** The block is at rest ($v_i = 0$), and the spring is compressed by $x_i = 0.15$ m. So $KE_i = 0$ and $U_i = \frac{1}{2}(400)(0.15)^2 = 4.5$ J. **Final state:** The spring is at its natural length ($x_f = 0$), and the block has speed $v_f$. So $KE_f = \frac{1}{2}(0.3)v_f^2$ and $U_f = 0$.

Step 2: Apply conservation of mechanical energy to find $v_f$.

Check your answer $$KE_i + U_i = KE_f + U_f$$ $$0 + 4.5 = \frac{1}{2}(0.3)v_f^2 + 0$$ $$v_f^2 = \frac{4.5}{0.15} = 30$$ $$v_f = \sqrt{30} \approx 5.48 \text{ m/s}$$ All the energy stored in the spring's compression was converted to kinetic energy. Compare this to the Debug problem in Section 6.4, where a student tried to solve this same type of problem using constant-acceleration kinematics and got the wrong answer. Energy conservation handles the variable spring force effortlessly.

Practice

Layer 1: Concrete

Problem 1. A 2 kg book is lifted from a table (height 0.8 m above the floor) to a shelf (height 2.5 m above the floor). Taking the floor as the reference level, find:

(a) The gravitational potential energy of the book on the table and on the shelf.

(b) The change in gravitational potential energy.

(c) The work done by gravity during the lift.

Check your answer **(a)** On the table: $U = mgy = (2)(9.8)(0.8) = 15.68$ J. On the shelf: $U = (2)(9.8)(2.5) = 49.0$ J. **(b)** $\Delta U = U_f - U_i = 49.0 - 15.68 = 33.32$ J. The potential energy increased. **(c)** $W_{\text{grav}} = -\Delta U = -33.32$ J. Gravity did negative work because the book moved upward, against the direction of the gravitational force.

Problem 2. A spring ($k = 250$ N/m) is stretched from its natural length to 0.12 m beyond its natural length. Then it is stretched further to 0.20 m beyond its natural length.

(a) Find the elastic potential energy at each position.

(b) Find the change in elastic potential energy.

(c) Find the work done by the spring force during this stretch.

Check your answer **(a)** At $x = 0.12$ m: $U = \frac{1}{2}(250)(0.12)^2 = 1.8$ J. At $x = 0.20$ m: $U = \frac{1}{2}(250)(0.20)^2 = 5.0$ J. **(b)** $\Delta U = 5.0 - 1.8 = 3.2$ J. The elastic potential energy increased. **(c)** $W_{\text{spring}} = -\Delta U = -3.2$ J. The spring did negative work because the spring was being stretched further (the displacement was in the direction opposite to the spring force).

Layer 2: Pattern

Problem 3. For each force below, determine whether it is conservative or nonconservative. Justify your answer using path independence.

(a) The gravitational force on an object near Earth's surface.

(b) Kinetic friction on a block sliding across a floor.

(c) The spring force (Hooke's law).

(d) A constant horizontal wind that always blows east with force $F_0$.

Check your answer **(a)** **Conservative.** The work done by gravity is $W = mg\Delta h$, which depends only on the change in height (the endpoints). Two paths between the same starting and ending heights produce the same work. The round-trip work is zero. **(b)** **Nonconservative.** The work done by kinetic friction is $W = -f_k d$, where $d$ is the total distance traveled. A longer path means more negative work. The round-trip work is not zero --- friction removes energy on every leg. **(c)** **Conservative.** The work done by the spring is $W = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$. This depends only on the initial and final displacements (the endpoints), not on the path. If the mass oscillates back and forth before arriving at $x_f$, the spring work is the same as if it went there directly. **(d)** **Conservative.** A constant force does work $W = F_0 \Delta x$, which depends only on the net displacement in the east-west direction. It does not matter whether the object took a direct route or a winding one; only the east-west component of the displacement matters. (This might be surprising --- a constant force in a fixed direction is conservative, even though it does not look like gravity or a spring.)

Layer 3: Structure

Problem 4. Why can't we define a "friction potential energy" --- a function $U_f(x)$ such that the work done by friction equals $-\Delta U_f$?

Check your answer Defining a potential energy function requires that the work done by the force depends only on the starting and ending positions. Then $U$ at each position has a single, well-defined value, and $W = U_i - U_f$ regardless of the path. Friction violates this requirement. The work done by friction depends on the total distance traveled, not just the endpoints. If a block slides from A to B along a short path, the friction work is $W_1 = -f_k d_1$. If it slides from A to B along a longer path, the friction work is $W_2 = -f_k d_2 \neq W_1$. Since $W_1 \neq W_2$, you cannot find a single function $U_f$ such that $W = U_f(A) - U_f(B)$ for both paths. The function would need to give two different values for the same pair of points, which is a contradiction. In short: potential energy requires path independence, and friction is path-dependent. The concept does not apply.

Layer 4: Debug

Problem 5. A student is solving the following problem: "A 5 kg block slides down a rough incline (height 3 m, kinetic friction coefficient $\mu_k = 0.2$, incline angle 30 degrees). Find the speed at the bottom."

The student writes:

$$KE_i + U_i = KE_f + U_f$$

$$0 + mgh = \frac{1}{2}mv_f^2 + 0$$

$$v_f = \sqrt{2gh} = \sqrt{2(9.8)(3)} = 7.67 \text{ m/s}$$

(a) What is wrong with this solution?

(b) How should the problem be set up correctly?

Check your answer **(a)** The student applied conservation of mechanical energy ($KE + U = \text{constant}$), but this equation is only valid when no nonconservative forces do work. There *is* a nonconservative force here --- kinetic friction --- and it *does* do work. The student ignored friction entirely. Friction removes mechanical energy from the system. The actual final speed will be less than 7.67 m/s. **(b)** The correct equation includes the work done by nonconservative forces: $$\Delta KE + \Delta U = W_{\text{nc}}$$ $$\left(\frac{1}{2}mv_f^2 - 0\right) + (0 - mgh) = W_{\text{friction}}$$ The work done by friction is $W_{\text{friction}} = -f_k d = -\mu_k mg\cos\theta \cdot d$, where $d = h / \sin\theta = 3/\sin 30° = 6$ m is the length of the incline. $$W_{\text{friction}} = -(0.2)(5)(9.8)\cos 30° \cdot 6 = -(0.2)(5)(9.8)(0.866)(6) \approx -50.9 \text{ J}$$ So: $$\frac{1}{2}(5)v_f^2 - (5)(9.8)(3) = -50.9$$ $$2.5\,v_f^2 = 147 - 50.9 = 96.1$$ $$v_f = \sqrt{38.4} \approx 6.2 \text{ m/s}$$ The correct speed is 6.2 m/s, significantly less than the frictionless value of 7.67 m/s. The friction work term is not optional --- omitting it violates the physics.

Reflection

Think about the central idea of this section.

What makes conservative forces "special" --- and why is that specialness so useful?

Conservative forces are special because their work depends only on where you start and where you end, not on how you get there. This path independence means you can assign a single number --- the potential energy --- to every position. And that means you can track energy as a ledger: kinetic energy in one column, potential energy in another, the total always balancing.

Nonconservative forces like friction break this accounting. They care about every step of the journey, so no single position-based number can capture their effect. You have to compute their work explicitly along whatever path the object actually takes.

The power of the distinction is practical: for systems with only conservative forces, you replace the entire force-along-the-path analysis with a comparison of two snapshots. One equation. Two states. No path. That is why energy conservation is one of the most used tools in all of physics.

Looking Ahead

You now have the machinery of energy conservation: identify conservative forces, write down potential energies, compare initial and final states, and solve. This is already a powerful problem-solving technique, and the next section makes it even more visual.

In Section 7.5, you will learn to read energy diagrams --- graphs of $U(x)$ that reveal, at a glance, where an object speeds up, where it slows down, where it turns around, and where it is trapped. All without solving a single equation. The potential energy function you defined here becomes a landscape, and the object's motion becomes a story told by the shape of that landscape.