Non-conservative forces — friction, air drag: Exercises

Throughout, take $g = 9.8 m/s^{2}$ unless a problem says otherwise. Every symbol below was defined in the parent; if you meet one you don't recognise, the reminder is right there in the callout.

Level 1 — Recognition

Goal: can you tell conservative from non-conservative, and get the sign right, without heavy algebra?

For each force, say conservative (C) or non-conservative (NC), and give the one-line reason: (a) gravity, (b) kinetic friction, (c) the spring force $- k x$ , (d) air drag, (e) the normal force on a block sliding along a flat floor.

Recall Solution 1.1

The test is always: does the work around a closed loop come out zero? (the little circle on $\oint$ means "add up $F \cdot d r$ all the way around a loop back to the start").

(a) gravity — C. Lift up, come back down: PE returned exactly. Loop work $= 0$ .
(b) kinetic friction — NC. It always points opposite to velocity, so on a round trip it subtracts on the way out AND on the way back. Loop work $< 0$ .
(c) spring $- k x$ — C. It has a potential energy $U = \frac{1}{2} k x^{2}$ ; stretch and release, energy returns. Loop work $= 0$ .
(d) air drag — NC. Same reason as friction: always opposes motion, always steals.
(e) normal force — C-ish / neither. On a flat floor the normal force is perpendicular to the motion, so $N \cdot d r = 0$ : it does zero work on every segment. It steals nothing, so it can't be a "thief" — it simply sits out.

Problem 1.2

A block is pushed 3 m across a rough floor. $μ_{k} = 0.30$ , $m = 2.0 kg$ , floor flat so $N = m g$ . What is the work done by friction? State its sign before computing.

Recall Solution 1.2

Sign first: friction opposes motion, so $W_{f r i c}$ is negative — the mnemonic " $W_{f r i c}$ is always $-$ ". Normal force on a flat floor: $N = m g = 2.0 \times 9.8 = 19.6 N$ . $W_{f r i c} = - μ_{k} N L = - (0.30) (19.6) (3.0) = - 17.64 J .$ So the floor stole $17.64 J$ , turning it into heat.

Level 2 — Application

Goal: plug the parent's formulas into single-stage problems, all quadrants of sign and geometry covered.

Problem 2.1

A $4.0 kg$ box slides $L = 5.0 m$ down a rough incline of angle $θ = 3 0^{\circ}$ , starting from rest. $μ_{k} = 0.20$ . Find the final speed. (See Figure below.)

Figure — Non-conservative forces — friction, air drag

Recall Solution 2.1

On an incline the weight $m g$ splits into two perpendicular pieces (look at the figure): one along the slope, $m g sin θ$ , that drives the slide; one into the slope, $m g cos θ$ , that the surface must push back against — so $N = m g cos θ$ . Use the master equation $W_{n c} = Δ K E + Δ U$ :

Height dropped $= L sin θ$ , so $Δ U = - m g L sin θ$ .
Friction: $W_{n c} = - μ_{k} N L = - μ_{k} m g cos θ L$ . $- μ_{k} m g cos θ L = \frac{1}{2} m v^{2} - 0 + (- m g L sin θ)$ Cancel $m$ , solve for $v$ — this is exactly the parent's result: $v = 2 g L (sin θ - μ_{k} cos θ) .$ Numbers: $sin 3 0^{\circ} = 0.5$ , $cos 3 0^{\circ} = 0.866$ . $v = 2 (9.8) (5.0) (0.5 - 0.20 \times 0.866) = 98 (0.3268) = 32.02 = 5.66 m/s .$

Problem 2.2

Same box and incline, but now the incline is shallow: $θ = 1 0^{\circ}$ , $μ_{k} = 0.20$ . Does the box slide down on its own from rest? Justify with numbers.

Recall Solution 2.2

The bracket $(sin θ - μ_{k} cos θ)$ decides everything. It is positive only when $tan θ > μ_{k}$ . $tan 1 0^{\circ} = 0.176, μ_{k} = 0.20.$ Since $0.176 < 0.20$ , the bracket is negative → the formula would demand $v^{2} < 0$ , which is impossible → the box does not accelerate; it stays put (gravity's pull along the slope is too weak to beat friction). This is the same condition as the parent's $μ_{k} \geq tan θ$ stopping rule — energy and statics agree.

Problem 2.3

A skydiver of mass $80 kg$ falls with quadratic drag. Given $C_{d} ρ A = 0.25 kg/m$ (so $F_{d} = \frac{1}{2} C_{d} ρ A v^{2}$ ), find the terminal velocity.

Recall Solution 2.3

Terminal velocity is where drag exactly cancels gravity, so acceleration stops: $m g = \frac{1}{2} C_{d} ρ A v_{t}^{2} \Rightarrow v_{t} = \frac{2 m g}{C _{d} ρ A} .$ $v_{t} = \frac{2 ( 80 ) ( 9.8 )}{0.25} = 6272 = 79.2 m/s .$

Level 3 — Analysis

Goal: combine two ideas — path-dependence, or momentum then energy.

Problem 3.1 (the loop test, numerically)

A $1.5 kg$ puck is pushed $2.0 m$ east across a rough table ( $μ_{k} = 0.25$ , $N = m g$ ), then $2.0 m$ back west to its start. (a) What is the net displacement? (b) What is the total path length? (c) What total work does friction do over the round trip, and what does the nonzero answer prove?

Recall Solution 3.1

(a) Start and end are the same point ⇒ net displacement $= 0$ . (b) Path length $L = 2.0 + 2.0 = 4.0 m$ — this is the distance travelled, which is what friction cares about. (c) $N = m g = 1.5 \times 9.8 = 14.7 N$ . $W_{f r i c} = - μ_{k} N L = - (0.25) (14.7) (4.0) = - 14.7 J .$ Because the loop work is $\neq = 0$ (strictly negative), friction cannot be conservative and has no potential energy. A conservative force with zero displacement would give zero net work; friction gave $- 14.7 J$ . That $14.7 J$ became heat.

Problem 3.2 (momentum then energy)

A $12 g$ bullet at $350 m/s$ embeds in a $1.8 kg$ wooden block resting on a rough floor ( $μ_{k} = 0.35$ ). How far does the block-plus-bullet slide before stopping?

Recall Solution 3.2

Stage 1 — collision (use momentum, NOT energy). The bullet embeds: perfectly inelastic, so KE is lost inside the wood, but momentum survives (no external horizontal impulse during the brief hit). $m_{b} u = (m_{b} + M) v \Rightarrow (0.012) (350) = (1.812) v \Rightarrow v = \frac{4.2}{1.812} = 2.318 m/s .$ Stage 2 — sliding (now use energy). Friction stops the combined mass over distance $d$ . Total mass $m = 1.812 kg$ , $N = m g$ . $W_{n c} = Δ K E : - μ_{k} m g d = 0 - \frac{1}{2} m v^{2} .$ The $m$ cancels: $d = \frac{v ^{2}}{2 μ _{k} g} = \frac{( 2.318 ) ^{2}}{2 ( 0.35 ) ( 9.8 )} = \frac{5.373}{6.86} = 0.783 m .$

Level 4 — Synthesis

Goal: chain conservative + non-conservative stages into one energy ledger.

Problem 4.1 (spring → rough patch → smooth ramp)

A spring ( $k = 800 N/m$ ) is compressed $0.15 m$ and launches a $0.50 kg$ cart. The cart crosses a rough horizontal patch of length $1.2 m$ ( $μ_{k} = 0.30$ ), then meets a frictionless ramp. How high up the ramp does it rise? (See Figure.)

Recall Solution 4.1

Track the energy from start (spring loaded, cart at rest) to top of ramp (cart momentarily at rest). Master equation over the whole journey: $W_{n c} = Δ K E + Δ U_{g r a v} + Δ U_{s p r in g} .$

Start KE $= 0$ , end KE $= 0$ ⇒ $Δ K E = 0$ .
Spring releases its stored energy: $Δ U_{s p r in g} = 0 - \frac{1}{2} k x^{2} = - \frac{1}{2} (800) (0.15)^{2} = - 9.0 J$ .
Gravity: $Δ U_{g r a v} = + m g h$ (rises by $h$ ).
Non-conservative: only the rough patch, $W_{n c} = - μ_{k} m g L = - (0.30) (0.50) (9.8) (1.2) = - 1.764 J$ .

Plug in: $- 1.764 = 0 + m g h + (- 9.0) \Rightarrow m g h = 9.0 - 1.764 = 7.236 J .$ $h = \frac{7.236}{( 0.50 ) ( 9.8 )} = \frac{7.236}{4.9} = 1.477 m .$ The spring gave $9.0 J$ ; friction stole $1.764 J$ as heat; the rest ( $7.236 J$ ) climbed the ramp.

Problem 4.2 (energy audit / heat)

For Problem 4.1, how much energy became heat, and what fraction of the spring's stored energy was that?

Recall Solution 4.2

Heat $=$ energy taken by non-conservative forces $= - W_{n c} = + 1.764 J$ (from the parent's $Q = - W_{n c}$ ). Fraction of spring energy lost to heat: $\frac{1.764}{9.0} = 0.196 = 19.6%.$ So roughly one-fifth of the launch energy warmed the rough patch; four-fifths made it to gravitational PE.

Level 5 — Mastery

Goal: subtle, open-ended, degenerate cases.

Problem 5.1 (drag work needs an integral)

An object moving with linear drag $F_{d} = b v$ ( $b = 2.0 N\cdots/m$ ) is pushed at constant speed $v = 3.0 m/s$ over $L = 5.0 m$ . (a) Find the drag work. (b) Now the object decelerates from $6.0 m/s$ under drag alone (no other horizontal force) and we know it travels $4.0 m$ total while its mass is $1.0 kg$ — find the drag work here without pretending $F_{d}$ is constant.

Recall Solution 5.1

(a) Constant speed ⇒ $F_{d} = b v$ is constant $= 2.0 \times 3.0 = 6.0 N$ , and it opposes motion, so $W_{d} = - F_{d} L = - (6.0) (5.0) = - 30.0 J .$ Here the shortcut is legal because speed is constant. (b) Decelerating: $F_{d} = b v$ changes every instant, so $- F_{d} L$ is meaningless. Use energy instead — drag is the only horizontal force, so $W_{d} = Δ K E$ . It starts at $6.0 m/s$ ; where does it stop? With linear drag the object actually asymptotically slows but, taking the stated final rest (it has effectively stopped after 4 m), $Δ K E = 0 - \frac{1}{2} (1.0) (6.0)^{2} = - 18.0 J$ . $W_{d} = Δ K E = - 18.0 J .$ Notice we never used the $4.0 m$ with a constant force — that would give the wrong number. Energy differences sidestep the varying force entirely.

Problem 5.2 (degenerate incline: $μ_{k} = tan θ$ )

A block is placed on a rough incline with $μ_{k} = tan θ$ exactly, and given a gentle initial nudge downhill at $u = 1.0 m/s$ . Describe its motion and, if it stops, over what distance. Take $θ = 2 5^{\circ}$ , so $μ_{k} = tan 2 5^{\circ} = 0.466$ , $g = 9.8$ .

Recall Solution 5.2

The along-slope driving force is $m g sin θ$ ; the resisting friction while moving is $μ_{k} m g cos θ = tan θ m g cos θ = m g sin θ$ . They are equal — the net along-slope force is zero once it's sliding! So the block moves at constant velocity while it's going... except there is no force to keep it going and none to stop it, so ideally it coasts forever at $1.0 m/s$ (net force zero ⇒ zero acceleration). This is the exact knife-edge between "slides faster" ( $tan θ > μ_{k}$ ) and "won't move" ( $tan θ < μ_{k}$ ). Energetically: gravity releases $m g sin θ \cdot L$ and friction takes $μ_{k} m g cos θ \cdot L$ — exactly equal, so $Δ K E = 0$ : speed never changes. A beautiful degenerate case where the block glides at whatever speed you gave it.

Problem 5.3 (which route loses more?)

A hockey puck goes from corner A to corner B of a rough square table, side $s = 1.0 m$ , $μ_{k} = 0.10$ , $m = 0.20 kg$ . Route 1: straight along the diagonal. Route 2: along two edges (A→corner→B). Which route loses more energy to friction, and by how much?

Recall Solution 5.3

Friction cares about path length, not endpoints — this is the whole point of non-conservative forces.

Route 1 (diagonal): length $= 2 s = 1.414 m$ .
Route 2 (two edges): length $= 2 s = 2.0 m$ . $N = m g = 0.20 \times 9.8 = 1.96 N$ ; per metre friction takes $μ_{k} N = 0.196 J/m$ .
Route 1 loss: $0.196 \times 1.414 = 0.277 J$ .
Route 2 loss: $0.196 \times 2.0 = 0.392 J$ . Extra loss on Route 2: $0.392 - 0.277 = 0.115 J$ . The longer path costs more — exactly the path-dependence signature. A conservative force would charge the same for both routes.

Recall One-screen summary of the toolbox

Which tool for which job? ::: Collision/sticking → momentum; sliding/climbing → energy $W_{n c} = Δ K E + Δ U$ ; constant-force drag or friction → $- F L$ ; varying drag → energy differences only. Sign of friction/drag work? ::: Always negative — they are thieves. Path length vs displacement for friction? ::: Always total path length $L$ ; displacement is irrelevant. Heat produced? ::: $Q = - W_{n c}$ .

Connections

Parent topic — the theory these drills exercise.
Work–Energy theorem — the always-true starting point $W_{t o t a l} = Δ K E$ .
Mechanical energy conservation — the special case when $W_{n c} = 0$ .
Friction — static & kinetic — where $f_{k} = μ_{k} N$ comes from.
Inelastic collisions — why Problem 3.2 needs momentum first.
Terminal velocity & projectile with drag — Problem 2.3's home.
Heat & first law of thermodynamics — where the stolen energy goes.

Exercises — Non-conservative forces — friction, air drag