Classical-Mechanics · Unit 16 · Video 1 · Interactive Practice

Conservative vs Non-Conservative Forces: Does Work Depend on the Path?

IKey Formulas

Formula	Name	What the answer contains
$W = \displaystyle\int_{\text{path}} \vec{F}\cdot d\vec{r}$	Work, by definition	The force and — in principle — the route
$W_{\text{grav}} = -mg\,(y_f - y_i)$	Work by gravity near the earth	The two heights, nothing else
$W^{f} = -\mu_k N s$	Work by kinetic friction	The full distance $s$ actually travelled
$\Delta E_{\text{system}} + \Delta E_{\text{surroundings}} = 0$	Conservation of energy	A closed system and its two states

Key Insight: A force is conservative when $W$ between two points comes out the same for every path joining them. Gravity's answer holds only $y_f - y_i$ ; friction's holds $s$ — and $s$ is the one quantity the route is free to change.

IIVisualization 1 — Five Routes, One Answer

A $2.0\ \text{kg}$ ball falls $4.0\ \text{m}$ and travels $6.0\ \text{m}$ sideways; does the route change gravity's work?

Route Progress 1.00

IIIVisualization 2 — Friction Bills Every Metre

Same block, same two endpoints, $\mu_k = 0.25$ : an overshoot of $d$ adds $2d$ of path that friction charges for.

Overshoot d (m) 2.00 Progress 1.00

💡 Gravity does $0$ on both of these trips: the floor is horizontal, so $y_f = y_i$ and every step has $\vec{F}\perp d\vec{r}$ . Path-dependence is friction's alone.

IVVisualization 3 — Why Only the Height Change Is Billed

Approximate any route by horizontal and vertical steps, then see which of them survive the sum.

Step 1

Step 1 — Replace the route by a staircase

Every step is either purely horizontal or purely vertical:

d\vec{r} = dx\,\hat{\imath} \qquad \text{or} \qquad d\vec{r} = dy\,\hat{\jmath}

Step 2 — Every tread contributes nothing

\vec{F}\cdot d\vec{r} = -mg\,\hat{\jmath}\cdot dx\,\hat{\imath} = -mg\,dx\,(\hat{\jmath}\cdot\hat{\imath}) = 0

Extra horizontal wandering is free, however much of it there is.

Step 3 — Risers cancel in pairs

\vec{F}\cdot d\vec{r} = -mg\,dy

A riser climbing

h

gives

-mgh

; the matching riser back down gives

+mgh

-mgh + mgh = 0

Step 4 — Only the net drop survives

W = -mg\sum \Delta y = -mg\,(y_f - y_i)

W = 19.6\ \text{N} \times 4.0\ \text{m} = 78.4\ \text{J}

The sum never learned how long the route was — so no route can change it.

VQuiz Questions

Problem 1 · Gravity Down a Curved Ramp

Given: A $2.0\ \text{kg}$ ball leaves a shelf $4.0\ \text{m}$ above the floor and lands $6.0\ \text{m}$ away horizontally, following a long curved ramp. Take $g = 9.8\ \text{m/s}^2$ . Find the work done by gravity.

✅ Correct!

W = -mg(y_f - y_i) = -(2.0)(9.8)(0 - 4.0) = +78.4\ \text{J}

— positive, because the ball ends lower and gravity helped.

❌ Check the sign. The ball moves down, in the same direction as the force, so gravity does positive work.

y_f - y_i = -4.0\ \text{m}

, and the minus sign in

-mg(y_f-y_i)

cancels it.

❌ Wrong displacement.

19.6 \times 6.0

bills the horizontal run, but

\hat{\jmath}\cdot\hat{\imath} = 0

kills every horizontal contribution. Only

\Delta y

appears.

❌ That is exactly what the video disproves. Gravity's work is path-independent: the ramp's shape drops out and only

y_f - y_i

survives.

❌ Not quite. Use

W = -mg(y_f - y_i)

with

m = 2.0\ \text{kg}

g = 9.8\ \text{m/s}^2

y_i = 4.0\ \text{m}

y_f = 0

Show solution

Along any route the force is $\vec{F} = -mg\,\hat{\jmath}$ , so a step $d\vec{r} = dx\,\hat{\imath} + dy\,\hat{\jmath}$ contributes

\vec{F}\cdot d\vec{r} = -mg\,dy

The horizontal part of every step is annihilated by $\hat{\jmath}\cdot\hat{\imath} = 0$ . Integrating from $y_i$ to $y_f$ :

W = \int_{y_i}^{y_f} -mg\,dy = -mg\,(y_f - y_i) = -(2.0)(9.8)(0 - 4.0) = 78.4\ \text{J}

The $6.0\ \text{m}$ of horizontal travel and the curvature of the ramp never enter the calculation — that is the whole content of "gravity is conservative".

Problem 2 · Friction on the Overshoot Route

Given: The same $2.0\ \text{kg}$ block slides on a horizontal floor with $\mu_k = 0.25$ and $N = mg = 19.6\ \text{N}$ . It is sent from $x_i$ toward $x_f$ , a displacement of $\Delta x = 6.0\ \text{m}$ , but it overshoots by $d = 2.0\ \text{m}$ and slides back to $x_f$ . Find the work done by friction over the whole trip.

✅ Correct!

s_2 = \Delta x + 2d = 10.0\ \text{m}

, so

W^{f} = -\mu_k N s_2 = -(4.90)(10.0) = -49.0\ \text{J}

— worse than the straight route's

-29.4\ \text{J}

, with identical endpoints.

❌ That is the straight route's answer. Friction is charged on distance travelled, not on displacement: the overshoot adds

2d = 4.0\ \text{m}

of extra path that still costs

\mu_k N

per metre.

❌ You paid for the overshoot only once. The block covers

d

going out and

d

coming back, so

s_2 = \Delta x + 2d = 10.0\ \text{m}

, not

8.0\ \text{m}

❌ Check the sign. Friction reverses when the block reverses, so it opposes the motion on both legs — each contributes negative work and nothing cancels.

❌ Not quite. First get

\mu_k N = 0.25 \times 19.6 = 4.90\ \text{N}

, then the total distance travelled

s_2

Show solution

Step 1 — the friction force:

|\vec{f}_k| = \mu_k N = 0.25 \times 19.6\ \text{N} = 4.90\ \text{N}

Step 2 — the distance actually travelled. Out to the turning point is $\Delta x + d = 8.0\ \text{m}$ ; back to $x_f$ is another $d = 2.0\ \text{m}$ :

s_2 = (\Delta x + d) + d = \Delta x + 2d = 6.0 + 4.0 = 10.0\ \text{m}

Step 3 — the work. Friction points backward on the way out and forward on the way back, but the motion reverses too, so $\vec{f}_k\cdot d\vec{r} < 0$ on every leg:

W^{f} = -\mu_k N s_2 = -(4.90)(10.0) = -49.0\ \text{J}

The straight route gives $-(4.90)(6.0) = -29.4\ \text{J}$ . Same endpoints, different work — kinetic friction is non-conservative.

Problem 3 · Both Forces on the Same Two Routes

Given: A $2.0\ \text{kg}$ block is dragged across a horizontal floor from $x_i$ to $x_f$ , with $\Delta x = 6.0\ \text{m}$ , $\mu_k = 0.25$ and $N = 19.6\ \text{N}$ . Route A is straight; route B overshoots $x_f$ by $d = 1.5\ \text{m}$ and returns. Find what each force does on each route.

Work done by gravity

Work done by friction

✅ Correct! Gravity is perpendicular to every step, so it does

0

on both. Friction charges

4.90\ \text{N}

per metre of path:

6.0\ \text{m}

against

9.0\ \text{m}

, giving

-29.4\ \text{J}

against

-44.1\ \text{J}

❌ Check gravity. The floor is horizontal, so

y_f = y_i

and every step satisfies

-mg\,\hat{\jmath}\cdot dx\,\hat{\imath} = 0

— a force perpendicular to the motion does no work, on either route.

❌ Check the distance travelled. Route B covers

\Delta x + 2d = 6.0 + 3.0 = 9.0\ \text{m}

— the overshoot is paid for twice, once out and once back.

Show solution

Gravity. Motion is horizontal, so $d\vec{r} = dx\,\hat{\imath}$ everywhere while $\vec{F} = -mg\,\hat{\jmath}$ :

\vec{F}\cdot d\vec{r} = -mg\,(\hat{\jmath}\cdot\hat{\imath})\,dx = 0

Equivalently $W = -mg(y_f - y_i) = 0$ since the height never changes. Both routes: $0\ \text{J}$ .

Friction. $\mu_k N = 0.25 \times 19.6 = 4.90\ \text{N}$ .

s_A = 6.0\ \text{m} \quad\Rightarrow\quad W_A^{f} = -(4.90)(6.0) = -29.4\ \text{J}

s_B = \Delta x + 2d = 6.0 + 3.0 = 9.0\ \text{m} \quad\Rightarrow\quad W_B^{f} = -(4.90)(9.0) = -44.1\ \text{J}

The two routes differ by $-\mu_k N (2d) = -(4.90)(3.0) = -14.7\ \text{J}$ . Gravity did not notice the detour at all; friction charged for every centimetre of it.

Problem 4 · The Round-Trip Test

Given: A $2.0\ \text{kg}$ puck is pushed around a closed loop on a rough ramp, starting and finishing at the same point $A$ . Along the way it climbs $0.60\ \text{m}$ at the loop's high point before coming back down. The total path length around the loop is $s = 8.0\ \text{m}$ , the friction force has constant magnitude $\mu_k N = 4.90\ \text{N}$ , and $mg = 19.6\ \text{N}$ . Find the work each force does around the loop.

Work done by gravity around the loop

Work done by friction around the loop

✅ Correct! A closed loop has

y_f = y_i

, so gravity's path-independent answer is forced to

0

; friction, charged on

s = 8.0\ \text{m}

, gives

-39.2\ \text{J}

. Zero work around every closed loop is the round-trip form of "conservative".

❌ Check gravity.

W = -mg(y_f - y_i)

, and the puck returns to

A

, so

y_f - y_i = 0

exactly. The climb's

-11.76\ \text{J}

is undone by the descent's

+11.76\ \text{J}

, and the

8.0\ \text{m}

of path length never appears.

❌ Check friction. It opposes the motion at every instant, so it never contributes positively and nothing cancels on the return:

W^{f} = -\mu_k N s = -(4.90)(8.0)

Show solution

Gravity. Its work depends only on the endpoints, and here they coincide:

W_{\text{grav}} = -mg\,(y_f - y_i) = -(19.6)(0) = 0\ \text{J}

Broken into pieces, the climb costs $-(19.6)(0.60) = -11.76\ \text{J}$ and the matching descent returns $+11.76\ \text{J}$ — the pair-cancellation of the staircase argument, applied to a whole loop.

Friction. Its magnitude is constant and it always points opposite $d\vec{r}$ , so every element of the loop contributes $-\mu_k N\,ds$ :

W^{f} = -\mu_k N \oint ds = -\mu_k N s = -(4.90)(8.0) = -39.2\ \text{J}

This is the sharpest form of the test: a conservative force does zero work around any closed path, because the "two paths" from $A$ back to $A$ must agree. Friction fails it by $39.2\ \text{J}$ every lap — energy that leaves as heat and never comes back.

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