Classical-Mechanics · Unit 15 · Video 2 · Interactive Practice

Chop, Dot, Add: Work Along Any Curved Path

IKey Formulas

Formula	Name	What it gives
$W = \int_i^f \vec{F} \cdot d\vec{r}$	Work as a line integral	Any force, any path
$d\vec{r} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}$	Cartesian line element	$dW = F_x\,dx + F_y\,dy + F_z\,dz$
$d\vec{r} = dr\,\hat{r} + r\,d\theta\,\hat{\theta} + dz\,\hat{k}$	Cylindrical line element	$dW = F_r\,dr + F_\theta\,r\,d\theta + F_z\,dz$
$W = -mg\,(y_f - y_0)$ $W_{\text{spring}} = -\tfrac{1}{2}k\,(x_f^2 - x_0^2)$	The two worked results	Endpoints only — the route drops out

Key Insight: Every term in the sum is a scalar product, so work is a plain number, not a vector — and once the limit is taken the time parameter has cancelled, leaving only the geometry of the path.

IIChop, Dot, Add — Then Take the Limit

Cut the path into chords, dot each with the force acting there, add — and the sum settles.

Chords N 1

IIIDoes the Route Matter?

Two legs joining $i = (0,0)$ to $f = (1,1)$ — the force decides whether the route matters.

Force field Corner a 1.00 Corner b 0.00

💡 Forces that return the same number for every route between two points earn a name of their own — conservative forces — a few chapters ahead.

IVA Force That Genuinely Varies

Hooke's force grows with displacement, so the work is a signed area, never a rectangle.

Start x₀ (m) 0.00 End x_f (m) 0.30

💡 The combination $\tfrac{1}{2}kx^2$ is fixed by where the block is, not by how it got there — that is what lets a spring store and return work, and it earns a name in the chapters ahead.

VQuiz Questions

Problem 1 · Gravity Over a Curved Flight

Given: A ball of mass $m = 0.50\ \text{kg}$ flies along a parabolic arc from $(x_0, y_0) = (0,\ 1.50\ \text{m})$ to $(x_f, y_f) = (12.0\ \text{m},\ 0.30\ \text{m})$ , with $g = 9.8\ \text{m/s}^2$ — find the work done by gravity.

✅ Correct! The ball ends

1.20\ \text{m}

lower, so

y_f - y_0

is negative, the two minus signs cancel, and gravity's work comes out positive.

❌ Check the sign.

y_f - y_0 = 0.30 - 1.50 = -1.20\ \text{m}

, and

-mg

multiplied by a negative height change is positive.

❌ That is the horizontal run. Since

\hat j \cdot \hat i = 0

, the

12.0\ \text{m}

of sideways travel contributes nothing at all.

❌ Not quite. A curved path does not make the work vanish — only a net height change of zero would do that.

Show solution

Gravity is uniform, so choose Cartesian axes with $\hat j$ along the force and expand the line element:

\vec F \cdot d\vec r = (-mg\,\hat j)\cdot(dx\,\hat i + dy\,\hat j) = -mg\,dy

The $dx$ term dies because $\hat j \cdot \hat i = 0$ , so the line integral collapses to one ordinary integral over height:

W = \int_{y_0}^{y_f} (-mg)\,dy = -mg\,(y_f - y_0)

W = -(0.50)(9.8)\big(0.30 - 1.50\big) = -(4.9)(-1.20) = +5.88\ \text{J}

The parabola never enters the answer: $W \approx +5.9\ \text{J}$ for any route between those two heights.

Problem 2 · The Factor Hiding in the Arc Step

Given: A puck slides along a circular groove of radius $r = 0.40\ \text{m}$ through an angle $\Delta\theta = \pi/2$ , driven by a constant tangential force $F_\theta = 6.0\ \text{N}$ along $\hat\theta$ (with $F_r = F_z = 0$ ) — find the work done by that force.

✅ Correct! The arc step is

r\,d\theta

, so

W = F_\theta\, r\, \Delta\theta = (6.0)(0.40)(\pi/2)

❌ You integrated $F_\theta\,d\theta$ . An angle is not a length: sweeping through

d\theta

at radius

r

covers an arc of length

r\,d\theta

, and that factor of

r = 0.40\ \text{m}

is the one genuinely new feature of the cylindrical line element.

❌ Not quite. A tangential force points along the motion. Here it is the radial term that vanishes, because

dr = 0

on a circle.

❌ That is only $F_\theta r$ . It is the coefficient of

d\theta

in the integrand — you still have to integrate it over

\Delta\theta = \pi/2

Show solution

Dot the force into the cylindrical line element:

dW = F_r\,dr + F_\theta\,r\,d\theta + F_z\,dz = F_\theta\,r\,d\theta

The radial and vertical terms drop out because $F_r = F_z = 0$ (and $dr = dz = 0$ on the groove anyway). Both $F_\theta$ and $r$ are constants, so they come outside:

W = F_\theta\, r \int_0^{\pi/2} d\theta = (6.0)(0.40)\left(\frac{\pi}{2}\right) = (2.4)(1.5708) = 3.77\ \text{J}

So $W \approx 3.8\ \text{J}$ . Dropping the $r$ would have given $6.0 \times \pi/2 = 9.4$ , which is not even a work: $F_\theta\,d\theta$ has units of newtons, not joules.

Problem 3 · Spring Work on Both Sides of Equilibrium

Given: A block attached to a spring of stiffness $k = 200\ \text{N/m}$ moves from $x_0 = -0.10\ \text{m}$ (compressed) to $x_f = +0.30\ \text{m}$ (stretched), with $x$ measured from the relaxed position.

What work does the spring do?

Which move gives exactly zero spring work?

✅ Correct! Only

x_f^2 - x_0^2

enters, so a move that ends the same distance from equilibrium as it started costs the spring nothing.

❌ You squared the displacement. The antiderivative of

-kx

-\tfrac12 kx^2

evaluated at each limit — that gives

x_f^2 - x_0^2

, never

(x_f - x_0)^2

❌ Watch the compressed side.

x_0 = -0.10\ \text{m}

still contributes

x_0^2 = +0.010\ \text{m}^2

, and it is subtracted:

0.090 - 0.010 = 0.080

, not

0.090 + 0.010

❌ Check the sign. The block ends farther from equilibrium than it began, so the spring resisted the whole way and its work must be negative.

❌ Not quite. Integrate

F_x = -kx

over

x

from

x_0

x_f

: the integrand is not constant, so the answer is not

F\,\Delta x

❌ Not quite.

W_{\text{spring}} = 0

needs

x_f^2 = x_0^2

— equal distances from equilibrium, on either side.

Show solution

Part 1. The block moves along $\hat i$ , so only the $x$ term of the work integral survives, and Hooke's law makes the integrand vary with position:

W_{\text{spring}} = \int_{x_0}^{x_f} (-kx)\,dx = -\tfrac{1}{2}k\big(x_f^2 - x_0^2\big)

W_{\text{spring}} = -\tfrac{1}{2}(200)\big((0.30)^2 - (-0.10)^2\big) = -100\,(0.090 - 0.010) = -8.0\ \text{J}

Negative, because the block finishes farther from equilibrium than it started: the spring pulled backwards the whole way.

Part 2. Setting $W_{\text{spring}} = 0$ requires $x_f^2 = x_0^2$ , i.e. $|x_f| = |x_0|$ . Only $-0.30\ \text{m} \to +0.30\ \text{m}$ satisfies that:

W_{\text{spring}} = -100\big((0.30)^2 - (-0.30)^2\big) = -100\,(0.090 - 0.090) = 0

The block crosses equilibrium: the spring does $+9.0\ \text{J}$ on the way in and $-9.0\ \text{J}$ on the way out. The side of the origin is irrelevant — only the square of the distance counts.

Problem 4 · Two Routes, One Pair of Endpoints

Given: The circulating force $\vec F = c(-y\,\hat i + x\,\hat j)$ with $c = 3.0\ \text{N/m}$ acts on a bead travelling from $i = (0,0)$ to $f = (2.0\ \text{m},\ 2.0\ \text{m})$ . Route 1 runs along the $x$ -axis to $(2.0, 0)$ and then straight up; route 2 runs up the $y$ -axis to $(0, 2.0)$ and then straight across — find $W_1$ and $W_2$ .

✅ Correct! On route 1 the force pushes along the climb; on route 2 it opposes the crossing. Same endpoints, opposite answers — the route is part of the answer.

❌ That assumes the endpoints settle it. This force changes from point to point, so the two routes sample different force values — that is exactly what path dependence means.

❌ Check the integrand. On the second leg of route 1,

\vec F \cdot d\vec r = cx\,dy

with

x

held at

2.0\ \text{m}

, so the integrand is

(3.0)(2.0) = 6.0\ \text{N}

and the leg is

2.0\ \text{m}

long.

Show solution

In Cartesian components $\vec F \cdot d\vec r = F_x\,dx + F_y\,dy = -cy\,dx + cx\,dy$ . Evaluate leg by leg.

Route 1 — along the $x$ -axis, then up:

$(0,0) \to (2,0)$ : here $y = 0$ and $dy = 0$ , so $\vec F \cdot d\vec r = 0$ . Contribution $0$ .
$(2,0) \to (2,2)$ : here $x = 2.0\ \text{m}$ and $dx = 0$ , so $\vec F \cdot d\vec r = cx\,dy = 6.0\,dy$ , giving $\int_0^{2} 6.0\,dy = +12\ \text{J}$ .

W_1 = 0 + 12 = +12\ \text{J}

Route 2 — up the $y$ -axis, then across:

$(0,0) \to (0,2)$ : here $x = 0$ and $dx = 0$ , so $\vec F \cdot d\vec r = cx\,dy = 0$ . Contribution $0$ .
$(0,2) \to (2,2)$ : here $y = 2.0\ \text{m}$ and $dy = 0$ , so $\vec F \cdot d\vec r = -cy\,dx = -6.0\,dx$ , giving $\int_0^{2} (-6.0)\,dx = -12\ \text{J}$ .

W_2 = 0 - 12 = -12\ \text{J}

Same $i$ , same $f$ , and $W_1 \neq W_2$ . Compare with gravity in Problem 1, where the shape of the path cancelled completely — the difference is that gravity is the same vector everywhere, while this force turns as you move.

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