Case Study 1 — The Work of a Climb: Energy on a Mountain Tram

Field: Classical mechanics (physics) Calculus used: Vector line integral $\int_C\mathbf{F}\cdot d\mathbf{r}$ (Section 35.3); Fundamental Theorem for Line Integrals (Section 35.5); closed-loop corollary and conservation of energy (Section 35.5)

A Cable Car and a Question About Energy

A mountain resort runs an aerial tram: a cabin of mass $m = 4000$ kg is hauled up a steel cable from the valley station at elevation $y = 0$ to the summit station at $y = 800$ m. The cable does not run straight. To clear a ridge, it rises steeply, dips into a saddle, then climbs again to the peak — the cabin's path through the vertical plane is a genuine curve, not a ramp. The chief engineer asks a question that sounds simple and turns out to be the heart of this chapter: how much work must the haul motor do against gravity to lift the cabin to the top, and does the wiggling shape of the cable change that number?

The honest way to answer is to set up the work as a vector line integral and compute it. Work is not "force times distance" when the force and the motion keep changing direction; it is the accumulated dot product of force and displacement along the path, $$W = \int_C \mathbf{F}\cdot d\mathbf{r}.$$ We will compute the work three ways — over a straight haul, over the wiggly real cable, and finally with a single subtraction using the Fundamental Theorem for Line Integrals — and watch the three answers agree. That agreement is the physics lesson hiding inside the mathematics.

Setting Up the Force Field

Work the problem in the vertical plane, with $x$ horizontal (in km, say) and $y$ the elevation in meters. Near the surface of the Earth, gravity is a constant downward force, $$\mathbf{F}_{\text{grav}} = \langle 0,\, -mg\rangle, \qquad g = 9.8\ \text{m/s}^2.$$ With $m = 4000$ kg, $mg = 3.92\times 10^4$ N. This is a vector field of exactly the kind Chapter 34 introduced: at every point of the plane it assigns the same downward arrow. It is also, importantly, a conservative field, the gradient of a potential. Define the gravitational potential energy per the whole cabin as $U(x,y) = mgy$. Then $$-\nabla U = -\langle 0,\, mg\rangle = \langle 0,\, -mg\rangle = \mathbf{F}_{\text{grav}}. \checkmark$$ The motor must overcome gravity, so the force the motor supplies (against gravity, ignoring friction and the cable's own weight for now) is $\mathbf{F}_{\text{motor}} = -\mathbf{F}_{\text{grav}} = \langle 0,\, mg\rangle$, pointing up. The work the cabin does against gravity is therefore $W_{\text{against}} = \int_C \mathbf{F}_{\text{motor}}\cdot d\mathbf{r} = -\int_C \mathbf{F}_{\text{grav}}\cdot d\mathbf{r}$.

Computation 1: The Straight Haul

First imagine an idealized straight cable from the valley station $(0,0)$ to the summit $(2,\,800)$ — a horizontal run of 2 km and a rise of 800 m. Parametrize the segment: $$\mathbf{r}(t) = \langle 2000\,t,\ 800\,t\rangle, \qquad 0\le t\le 1,$$ (using meters for both coordinates so the dot product carries units of joules). Then $\mathbf{r}'(t) = \langle 2000,\ 800\rangle$, and the gravitational force along the path is the constant $\mathbf{F}_{\text{grav}} = \langle 0,\ -mg\rangle$. The integrand is $$\mathbf{F}_{\text{grav}}\cdot\mathbf{r}'(t) = (0)(2000) + (-mg)(800) = -800\,mg.$$ So $$\int_C \mathbf{F}_{\text{grav}}\cdot d\mathbf{r} = \int_0^1 (-800\,mg)\,dt = -800\,mg = -800(3.92\times 10^4) = -3.136\times 10^7\ \text{J}.$$ Gravity does $-31.36$ MJ of work (negative, since the cabin rises against it). The motor must therefore do $+31.36$ MJ against gravity. Notice already what did not appear: the horizontal run of 2000 m contributed nothing, because gravity has no horizontal component. Only the vertical motion cost energy.

Computation 2: The Wiggly Real Cable

Now the realistic path. Model the cable's profile as a curve that climbs, dips into a saddle, and climbs again, but still starts at $(0,0)$ and ends at $(2000,\,800)$. A clean parametrization with that shape is $$\mathbf{r}(t) = \big\langle 2000\,t,\ \ 800\,t + 300\sin(2\pi t)\big\rangle, \qquad 0\le t\le 1.$$ The $300\sin(2\pi t)$ term makes the elevation overshoot on the way up and dip below the straight line near the saddle, returning to the same endpoints. Differentiate: $$\mathbf{r}'(t) = \big\langle 2000,\ \ 800 + 600\pi\cos(2\pi t)\big\rangle.$$ The force is still $\mathbf{F}_{\text{grav}} = \langle 0,\ -mg\rangle$, so the integrand is again only the second component: $$\mathbf{F}_{\text{grav}}\cdot\mathbf{r}'(t) = -mg\big(800 + 600\pi\cos(2\pi t)\big).$$ Integrate term by term over $[0,1]$: $$\int_C\mathbf{F}_{\text{grav}}\cdot d\mathbf{r} = -mg\int_0^1 \big(800 + 600\pi\cos(2\pi t)\big)\,dt = -mg\Big(800 + 600\pi\cdot\underbrace{\big[\tfrac{\sin 2\pi t}{2\pi}\big]_0^1}_{=\,0}\Big).$$ The cosine integrates to $\tfrac{\sin 2\pi t}{2\pi}\big|_0^1 = 0$. The entire wiggle contributes nothing. We are left with $$\int_C\mathbf{F}_{\text{grav}}\cdot d\mathbf{r} = -800\,mg = -3.136\times 10^7\ \text{J},$$ exactly as for the straight haul. The detours up and down the saddle each cost and then refund their energy; only the net 800 m of rise survives.

Computation 3: One Subtraction, No Integral

The Fundamental Theorem for Line Integrals (§35.5) tells us in advance that the two computations had to agree, and lets us skip the integral entirely. Because $\mathbf{F}_{\text{grav}} = -\nabla U$ with $U = mgy$, for any path $C$ from $A=(0,0)$ to $B=(2000,800)$, $$\int_C\mathbf{F}_{\text{grav}}\cdot d\mathbf{r} = -\big(U(B) - U(A)\big) = -\big(mg\cdot 800 - mg\cdot 0\big) = -800\,mg.$$ No parametrization, no $\cos(2\pi t)$, no integral — just the potential evaluated at two points. This is precisely the original Fundamental Theorem of Calculus (Chapter 14), $\int_a^b F'\,dx = F(b)-F(a)$, with the gradient playing the role of the derivative and the endpoints playing the role of the limits of integration. The engineer's worry about the "wiggling shape" was unfounded: gravity is conservative, so the work depends only on the change in elevation. A cabin spiraling lazily up the mountain and one yanked straight up the cliff face cost the haul motor the same $31.36$ MJ against gravity.

The Round Trip: Why the Descent Pays You Back

The closed-loop corollary completes the picture. Suppose the cabin makes a full round trip: up the wiggly cable to the summit and back down to the valley. The path is now a closed curve, with $A = B$. The Fundamental Theorem gives $$\oint_C\mathbf{F}_{\text{grav}}\cdot d\mathbf{r} = U(A) - U(A) = 0.$$ Gravity does zero net work over any round trip. The energy spent lifting the cabin is exactly recovered on the way down — which is why a well-designed tram uses a counterweight or a descending second cabin to recapture that potential energy, and why regenerative braking on the downhill run can feed power back to the grid. Conservation of energy, $K + U = \text{constant}$, is not a separate physical law bolted onto the mathematics; it is the closed-loop corollary of the Fundamental Theorem for Line Integrals, lived out on a mountainside.

Where the Conservative Story Breaks: Friction

Real cables have friction, and friction is the cautionary counterexample. The frictional force always points opposite the motion: $\mathbf{F}_{\text{fric}} = -\mu\,\big|\mathbf{N}\big|\,\hat{\mathbf{T}}$, where $\hat{\mathbf{T}} = \mathbf{r}'/|\mathbf{r}'|$ is the unit tangent. Its work is $$W_{\text{fric}} = \int_C \mathbf{F}_{\text{fric}}\cdot d\mathbf{r} = -\mu|\mathbf{N}|\int_C \hat{\mathbf{T}}\cdot d\mathbf{r} = -\mu|\mathbf{N}|\int_C ds = -\mu|\mathbf{N}|\,L,$$ because $\hat{\mathbf{T}}\cdot d\mathbf{r} = \hat{\mathbf{T}}\cdot\mathbf{T}\,ds = ds$. The frictional work is proportional to the arc length $L$ of the path — a scalar line integral $\int_C ds$ — not to the displacement between endpoints. The wiggly cable, being longer, dissipates strictly more energy to friction than the straight one. Friction is non-conservative: it has no potential, its work depends on the route taken, and around a closed loop $\oint\mathbf{F}_{\text{fric}}\cdot d\mathbf{r} \ne 0$ (it is always negative, draining mechanical energy into heat). This is the clean dividing line of the chapter: conservative forces care only about endpoints (vector line integral via a potential); dissipative forces care about the whole path (scalar line integral of arc length).

Lessons

  1. Work is a vector line integral. When force and motion change direction, $W = \int_C\mathbf{F}\cdot d\mathbf{r}$ replaces "force times distance," automatically extracting the tangential component at every instant.
  2. Gravity is conservative. With potential $U = mgy$, the work against gravity over any path is $mg\,\Delta y$ — the wiggle, the saddle, the horizontal run all drop out. The three computations agreeing is the Fundamental Theorem for Line Integrals at work.
  3. The closed loop returns zero. A round trip in a conservative field does no net work; this is conservation of mechanical energy, and it is why counterweights and regenerative braking recover the climb's energy.
  4. Friction breaks the pattern. A non-conservative force has no potential; its work is the scalar line integral of arc length, so longer paths cost more and closed loops lose energy to heat.

Discussion Questions

  1. The wiggle term $300\sin(2\pi t)$ contributed nothing to the gravitational work. Which single property of the integrand $\cos(2\pi t)$ over $[0,1]$ guaranteed this, and how does the Fundamental Theorem for Line Integrals predict it without computing the integral?
  2. If the resort doubled the horizontal run from 2 km to 4 km but kept the 800 m rise, how does the work against gravity change? How does the friction loss change? Explain the asymmetry.
  3. Why can an engineer quote a cabin's potential energy as a single number "$31.36$ MJ at the summit" without ever specifying which cable route was used — but cannot do the same for the energy lost to friction?
  4. Sketch a force field for which lifting the cabin straight up and lifting it via the saddle would give different work. What property must such a field lack?

Further Reading

  • Halliday, D., Resnick, R., and Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley. Chapters 7–8 develop work, the work–energy theorem, and conservative forces at exactly the level used here; the cable-car energy balance is a worked variant of their incline problems.
  • Taylor, J. R. (2005). Classical Mechanics. University Science Books. Section 4.2–4.4 give the cleanest undergraduate treatment of when a force has a potential, and prove the path-independence ⇔ conservative equivalence used in Computation 3.
  • Feynman, R. P., Leighton, R. B., and Sands, M. (1964). The Feynman Lectures on Physics, Vol. I, Chapters 13–14 ("Work and Potential Energy"). Feynman's discussion of why gravity "forgets the path" is the physical intuition behind the Fundamental Theorem for Line Integrals.
  • Stewart, J. (2020). Calculus: Early Transcendentals (9th ed.), Section 16.3. The textbook source for the Fundamental Theorem for Line Integrals and worked work-as-line-integral problems.