M043 · Work and energy

Conservative Forces & Path Dependence

Does a force do the same work along any path between the same two points? Only if the force is conservative. In a swirling force field the work depends on the path, and running a loop backward flips its sign. Friction never flips: it takes energy whichever way you go. One loop that happens to give zero work does not prove a potential energy exists; that needs zero work around every loop.

Edit two paths that share the same endpoints. Separate the work done by the potential part of the force, the circulation of its rotational part and the loss to sliding. Then compare loop work computed from recorded force samples with the value predicted from the loop's geometry.

Analytic field classification $\text{Ready}$

First route field work $\text{Ready}$

Second route field work $\text{Ready}$

First route with resistance $\text{Ready}$

Second route with resistance $\text{Ready}$

First route length $\text{Ready}$

Second route length $\text{Ready}$

Closed-loop field work $\text{Ready}$

Closed-loop resistance work $\text{Ready}$

Closed-loop total work $\text{Ready}$

Oriented loop area $\text{Ready}$

Curl times oriented area $\text{Ready}$

Length change at double panels $\text{Ready}$

Selected route potential-part work $\text{Ready}$

Selected tangent $\text{Ready}$

Acquired force stations $\text{Ready}$

Acquired closed-loop field work $\text{Ready}$

Acquired minus reference loop $\text{Ready}$

Work along a path and around a loop

Keep the endpoints; change the journey

Radial spring coefficient

k

0.70\,\frac{\mathrm N}{\mathrm m}

Uniform downward force

G

1.00\,N

Rotational field coefficient

\gamma

1.00\,\frac{\mathrm N}{\mathrm m}

Sliding resistance magnitude

f

0.00\,N

Route to edit; first or second

j

0.00\,

Reverse both routes

\sigma

0.00\,

Length quadrature panels

n_q

8.00\,

First path control 1 horizontal

P^{()}_{1,x}

-2.00\,m

First path control 2 horizontal

P^{()}_{2,x}

2.00\,m

First path control 1 height

P^{()}_{1,y}

3.00\,m

First path control 2 height

P^{()}_{2,y}

3.00\,m

Second path control 1 horizontal

P^{()}_{1,x}

-2.00\,m

Second path control 2 horizontal

P^{()}_{2,x}

2.00\,m

Second path control 1 height

P^{()}_{1,y}

-3.00\,m

Second path control 2 height

P^{()}_{2,y}

-3.00\,m

Replay fraction

\tau

0.45\,

Force scan intervals

n_s

24.00\,

Force noise bound

\eta_F

0.00\,N

Two planar cubic Bézier paths share fixed endpoints at minus two, minus one and at two, one, in metres. Each path has two inner points you can move freely. The coordinates are right-handed: horizontal points right, vertical points up and depth points out of the path plane. The 3D paths and every integral use these same physical coordinates.

The spring part and the uniform downward part each have a known potential. When its coefficient is not zero, the rotational part has a constant, nonzero curl. Zero work around one loop, or equal work along two paths, does not prove that a potential exists everywhere. The field is classified from its given formula.

An external drive moves the object along each path; the paths are not trajectories solved from the plotted forces. No kinetic energy is inferred from the path parameter. Sliding resistance has a fixed size and points against the local direction of travel; static friction, normal contact loads and the drive's dynamics are left out. Where the tangent vanishes, the resistance has no defined direction and its arrow is not drawn.

The directed loop goes forward along the first path and back along the second. Going backward flips the sign of the work done by a position-dependent field, but resistance work stays negative. Curl times oriented area applies only to the field part of the work. A loop that crosses itself uses signed winding area.

Composite eight-point Gauss quadrature integrates the smooth field work and the path length. The field-work integrands are polynomials, but the length is a numerical result, especially near points where the tangent vanishes. The change when the panels are doubled is a rough check, not an error bound. The running plots use a separate, finer partition.

A raw acquisition records exact positions and noisy field-force components on both paths. The trapezoidal force–displacement integral uses only these records. Resistance work is calculated separately from the model and is not presented as measured. Replaying, choosing a route, or changing the resistance or quadrature keeps the field record; changing the geometry, direction, field or sensor clears it.

The paths are prescribed. Field work and sliding dissipation are computed separately. The drive and the contact dynamics are not modeled.

Start with the essentials

Focus question: Does the same destination require the same work?
One-sentence intuition: Zero work around one loop does not prove that a potential exists everywhere.

Core mathematical model

A force with conservative and rotational parts

\begin{aligned}\mathbf F&=(-kx-\gamma y,\,-G-ky+\gamma x)\\U_c&=\tfrac12k(x^2+y^2)+Gy\end{aligned}

Only the spring force and the uniform downward force come from this potential.

Work as an integral along the path

W_C=\int_0^1\mathbf F(\mathbf c(q))\cdot\mathbf c^{\prime}(q)\,\mathrm dq

Only the force component along the path's tangent contributes. The path parameter is not time; the apparatus is driven along the selected path.

Loop orientation and curl

\begin{aligned}A_{\mathrm{or}}&=\tfrac12\oint(x\,\mathrm dy-y\,\mathrm dx)\\\oint\mathbf F\cdot\mathrm d\mathbf r&=2\gamma A_{\mathrm{or}}\end{aligned}

Going forward along the first path and back along the second sets the orientation. Where the loop crosses itself, each part counts with the sign of its winding.

Dissipation does not reverse sign

\begin{aligned}W_f&=-fL\\W_{f,\mathrm{loop}}&=-f(L_A+L_B)\end{aligned}

Resistance depends on the direction of travel, so its loop work is not the curl integral of any field that depends only on position.

Work estimated from recorded forces

\widehat W_C=\sum_i\frac{\mathbf F_i^{\mathrm{obs}}+\mathbf F_{i+1}^{\mathrm{obs}}}{2}\cdot(\mathbf r_{i+1}-\mathbf r_i)

In this simulated acquisition the positions are exact, and the field-force components carry bounded deterministic noise.

Common difficulties

Reversing a loop

Typical misconceptionIf I go around the loop the other way, every work term flips sign.

Better mental modelWork by a field that depends only on position does flip. Sliding resistance turns to oppose the new direction of motion, so it still takes energy out.

Run the experiment

01
Keep the endpoints fixed
Edit both internal points on each route.
What to observe: The work of the potential part stays the same, even when the path length changes.
02
Add circulation to the field
Choose Rotational field and change its sign.
What to observe: The field work around the closed loop follows the oriented area and changes sign when the loop is reversed.
03
Add resistance
Set a sliding resistance, then press Reverse both routes.
What to observe: The field term changes sign, while the resistance term stays negative.
04
Measure the loop work
Acquire the field forces, increase the scan intervals and compare with the model.
What to observe: The measured loop work is a sum of force times displacement. The resistance is still calculated separately.