Classical-Mechanics · Unit 14 · Video 3 · Interactive Practice

Positive, Negative, or Zero: The Work Done by Constant Forces

IKey Formulas

Formula	Name	What you need
$W = F_x\,\Delta x$	Work done by a constant force	Force component along the motion, and $\Delta x$ of the point of application
$W^f = -\mu_k N\,\Delta x$	Work done by kinetic friction	$\mu_k$ , the normal force $N$ , and the slide distance
$N = mg - F\sin\theta$	Normal force under an angled push	Vertical force balance: $F\sin\theta + N - mg = 0$
$W^g = -mg\,\Delta y$	Work done by gravity (near Earth, $+y$ up)	$\Delta y = y_f - y_i$

Key Insight: The sign is physical, not a convention. $W > 0$ feeds energy in, $W < 0$ drains it out, and $W = 0$ — a perpendicular force, or an application point that never moves — leaves the energy account untouched.

IIVisualization 1 — One Force, Two Effects

Tilting the push upward costs driving force but lightens the press on the table — and the friction follows.

Push angle θ (°) 30 Push magnitude F (N) 2.0

💡 Challenge: With $F = 2.0\ \text{N}$ , find the angle at which $F\sin\theta$ reaches $mg = 1.96\ \text{N}$ — the cup lifts off and friction vanishes entirely.

IIIVisualization 2 — Whose Displacement Counts?

Work tracks the point where the force acts, not the body that force belongs to.

Contact Motion 0.00 m

💡 The walker still gains kinetic energy — it comes from chemical energy in the muscles, not from the ground, which hands over exactly zero joules.

IVVisualization 3 — Gravity, and the Axis That Cancels

A falling body gains energy from gravity, a rising body loses it — and calling down positive changes nothing.

Final height (m) 1.2

VQuiz Questions

Problem 1 · Work Done by a Constant Push

Given: A constant horizontal force of $12\ \text{N}$ pushes a box $3.0\ \text{m}$ across a floor, in the same direction as the force — find the work $W$ done by that force.

✅ Correct! Force component and displacement are both positive, so the push feeds

36\ \text{J}

into the box.

❌ Close, but check the sign. The force points along the motion, so

F_x > 0

and

\Delta x > 0

— their product is positive.

❌ Not quite. Work is force times displacement,

W = F_x\,\Delta x

— not force divided by displacement.

❌ Not quite. Apply

W = F_x\,\Delta x

with

F_x = +12\ \text{N}

and

\Delta x = +3.0\ \text{m}

— a force with a component along the motion, acting through a real displacement, does nonzero work.

Show solution

Point the $+x$ axis along the motion. The force lies along $+x$ , so $F_x = +12\ \text{N}$ , and the displacement is $\Delta x = +3.0\ \text{m}$ :

W = F_x\,\Delta x = (12\ \text{N})(3.0\ \text{m}) = 36\ \text{J}

Check the units: $1\ \text{N}\cdot\text{m} = 1\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2} = 1\ \text{J}$ — the same combination as $K = \tfrac{1}{2}mv^2$ .

$W > 0$ : the push feeds energy into the box.

Problem 2 · The Force That Accelerates You For Free

Given: You start from rest and walk forward. Your planted foot does not slip, and static friction from the ground on that foot has magnitude $140\ \text{N}$ while your body advances $0.80\ \text{m}$ — find the work static friction does on you.

✅ Correct! The planted foot never moves, so the application point has

\Delta x = 0

and the ground transfers no energy at all.

❌ Close, but that is the wrong displacement.

\Delta x

W = F_x\,\Delta x

belongs to the point of application, and the non-slipping foot stays put.

❌ Not quite. Ask where the force acts and whether that spot moves — the foot is planted, so its displacement is zero.

Show solution

Static friction is the external force that accelerates you, but the work formula uses the displacement of the point where the force acts:

W = f_s\,\Delta x_{\text{application}} = (140\ \text{N})(0\ \text{m}) = 0\ \text{J}

The foot does not slip and does not slide, so its displacement is zero for as long as it is planted. Your body's $0.80\ \text{m}$ belongs to your center of mass, not to the contact point.

Where does the kinetic energy come from, then? From inside: chemical energy stored in muscle converts into kinetic energy plus some thermal energy. Friction redirects your push into forward acceleration — it does not supply the joules.

Problem 3 · Angled Push — The Normal Force Trap

Given: A $2.5\ \text{kg}$ block is pushed $1.5\ \text{m}$ across a horizontal floor by a constant $20\ \text{N}$ force directed $37\degree$ above the horizontal. Take $\mu_k = 0.20$ , $g = 9.8\ \text{m}\cdot\text{s}^{-2}$ , $\cos 37\degree = 0.80$ , $\sin 37\degree = 0.60$ — find the work done by kinetic friction.

✅ Correct! The upward component

F\sin\theta = 12\ \text{N}

carries half the load, so

N = 12.5\ \text{N}

and friction drains only

3.75\ \text{J}

❌ Close — but you used $N = mg$ . The push has an upward component of

12\ \text{N}

, so the table supports far less than the full weight.

❌ Not quite. Friction's work is

-\mu_k N\,\Delta x

, and it is negative: the force opposes the motion while

\Delta x > 0

Show solution

Step 1 — Vertical force balance. The block does not accelerate vertically:

F\sin\theta + N - mg = 0 \quad\Longrightarrow\quad N = mg - F\sin\theta

N = (2.5)(9.8) - (20)(0.60) = 24.5\ \text{N} - 12\ \text{N} = 12.5\ \text{N}

Step 2 — Friction force.

f_k = \mu_k N = (0.20)(12.5\ \text{N}) = 2.5\ \text{N}

Step 3 — Its work. Friction points opposite the motion, so $f_{k,x} = -2.5\ \text{N}$ while $\Delta x = +1.5\ \text{m}$ :

W^f = -\mu_k N\,\Delta x = -(2.5\ \text{N})(1.5\ \text{m}) = -3.75\ \text{J}

Why the distractors are tempting: assuming $N = mg$ gives $-(0.20)(24.5)(1.5) = -7.35\ \text{J}$ , roughly double; and $-24\ \text{J}$ is the push's work $F\cos\theta\,\Delta x = (16)(1.5) = 24\ \text{J}$ with a sign flipped onto it.

Problem 4 · Gravity Up and Back Down

Given: A $0.50\ \text{kg}$ ball is thrown straight up, rises $6.0\ \text{m}$ to the top of its flight, then falls the same $6.0\ \text{m}$ back into the thrower's hand. Take $g = 9.8\ \text{m}\cdot\text{s}^{-2}$ and $+y$ upward.

Work done by gravity during the rise?

Work done by gravity over the whole up-and-down trip?

✅ Excellent! Gravity drains

29.4\ \text{J}

on the way up and feeds the same

29.4\ \text{J}

back on the way down — net zero, because

\Delta y = 0

❌ Check the rise. Force down, motion up:

\Delta y = +6.0\ \text{m}

and

F^g_y = -mg

, so the product is negative.

❌ Check the round trip.

W^g = -mg\,\Delta y

uses only the net change in height, not the distance travelled.

Show solution

With $+y$ up, gravity has the single component $F^g_y = -mg = -(0.50)(9.8) = -4.9\ \text{N}$ , and $W^g = F^g_y\,\Delta y = -mg\,\Delta y$ .

Rise: $\Delta y = +6.0\ \text{m}$ , so

W^g = -(4.9\ \text{N})(+6.0\ \text{m}) = -29.4\ \text{J}

Force down, motion up — gravity drains energy and the ball slows.

Round trip: the ball returns to the hand, so $y_f = y_i$ and $\Delta y = 0$ :

W^g = -(4.9\ \text{N})(0\ \text{m}) = 0\ \text{J}

The $-29.4\ \text{J}$ of the rise is exactly repaid by $+29.4\ \text{J}$ on the fall, where $\Delta y = -6.0\ \text{m}$ .

Axis check: call downward positive instead. On the rise, $F^g_y = +4.9\ \text{N}$ and $\Delta y = -6.0\ \text{m}$ , giving $W^g = -29.4\ \text{J}$ again — the choice of axis cancels out of a physical quantity.

Solved: 0 / 4