Classical-Mechanics · Unit 18 · Video 4 · Interactive Practice

Friction in One Substitution: Cart, Spring, and the Inclined Plane

IKey Formulas

Formula	Name	What it assumes
$\tfrac{1}{2}kx^2 - xmg\sin\theta = mgd\sin\theta$	Conservation of mechanical energy	Frictionless slide; both $U^g = 0$ and $U^s = 0$ at the unstretched spring end
$x = \dfrac{mg}{k}\left(\sin\theta + \sqrt{\sin^2\theta + \dfrac{2kd}{mg}\sin\theta}\right)$	Maximum compression	Positive root only: compression drives the spring end forward
$\sin\theta \;\longrightarrow\; \sin\theta - \mu_k\cos\theta$	The friction substitution	Turns the frictionless answer into the friction answer
$E_{\text{lost}} = \|W_{nc}\| = \mu_k mg\cos\theta\,(d + x)$	Energy converted to thermal energy	$x$ is the compression of the case you are actually in

Key Insight: Friction never starts a fresh derivation. Collecting its terms returns the frictionless quadratic with $\sin\theta$ demoted to $\sin\theta - \mu_k\cos\theta$ , so gravity's driving component is simply weakened and the spring stores less.

IIVisualization 1 — Where the Energy Sits

Released $1.00\ \text{m}$ up a $30^\circ$ ramp, the cart's energy only ever moves between three accounts.

Time t (s) 0.000

IIIVisualization 2 — Two Roots, One Motion

The energy line meets the potential curve twice, but only one crossing is a place this cart reaches.

Contact Elapsed (s) 0.000

💡 The unattached cart does not stop at $x = 0$ : it leaves the spring with all $9.80\ \text{J}$ back as kinetic energy and coasts a further $1.00\ \text{m}$ up the slope, arriving exactly at its release point — off the left of this frame.

IVVisualization 3 — Friction in One Substitution

Friction changes exactly one factor in the answer: $\sin\theta$ becomes $\sin\theta - \mu_k\cos\theta$ .

Friction μk 0.20

Part (a) — frictionless

0 = \tfrac{1}{2}kx^2 - xmg\sin\theta - dmg\sin\theta

Part (b) — collect the friction terms of $-\mu_k mg\cos\theta\,(d+x) = E_f - E_i$

0 = \tfrac{1}{2}kx^2 - xmg(\sin\theta - \mu_k\cos\theta) - dmg(\sin\theta - \mu_k\cos\theta)

The same quadratic, one factor changed

\sin\theta \;\longrightarrow\; R \equiv \sin\theta - \mu_k\cos\theta \qquad\Longrightarrow\qquad x = \frac{mg}{k}\left(R + \sqrt{R^2 + \frac{2kd}{mg}\,R}\right)

💡 At $\mu_k = \tan\theta \approx 0.577$ the reduced factor reaches zero: friction alone balances the gravitational component along the slope, and a cart released from rest never slides at all.

VQuiz Questions

Problem 1 · Maximum Compression

Given: A $2.0\ \text{kg}$ cart is released from rest on a frictionless $30^\circ$ incline, a distance $d = 1.00\ \text{m}$ along the slope from the free end of an unstretched spring with $k = 400\ \text{N/m}$ ; take $g = 9.8\ \text{m/s}^2$ . Find the maximum compression $x$ .

✅ Correct! With

\frac{mg}{k} = 0.049\ \text{m}

and

\frac{2kd}{mg}\sin\theta = 20.41

, the root is

0.049\left(0.5 + \sqrt{0.25 + 20.41}\right) = 0.247\ \text{m}

❌ That solves $\tfrac{1}{2}kx^2 = mgd\sin\theta$ . The cart keeps descending while it compresses the spring, dropping a further

x\sin\theta

and contributing another

xmg\sin\theta

to the energy account.

❌ That is the rejected root. The quadratic also gives

x_- = -0.198\ \text{m}

; a compression must move the spring end forward, so the plus sign is the physical one.

❌ That is the equilibrium compression.

mg\sin\theta/k = 0.0245\ \text{m}

is where the spring force balances the gravity component — the cart is moving fastest there, not stopped.

Show solution

Put $U^g = 0$ and $U^s = 0$ both at the unstretched spring end. Released from rest a distance $d$ up the slope, the cart starts a height $d\sin\theta$ above that line:

E_i = K_i + U_i = 0 + mgd\sin\theta = (19.6)(1.00)(0.5) = 9.80\ \text{J}

At maximum compression the cart is momentarily at rest, the spring holds $\tfrac{1}{2}kx^2$ , and the cart sits a height $x\sin\theta$ below the zero line:

E_f = \tfrac{1}{2}kx^2 - xmg\sin\theta

Nothing non-conservative acts, so $E_f = E_i$ :

\tfrac{1}{2}kx^2 - xmg\sin\theta - mgd\sin\theta = 0 \quad\Longrightarrow\quad 200x^2 - 9.8x - 9.8 = 0

x = \frac{mg}{k}\left(\sin\theta + \sqrt{\sin^2\theta + \frac{2kd}{mg}\sin\theta}\right) = 0.049\left(0.5 + \sqrt{0.25 + 20.41}\right) = \mathbf{0.247\ \text{m}}

Check: $\tfrac{1}{2}kx^2 = 12.22\ \text{J}$ and $xmg\sin\theta = 2.42\ \text{J}$ , so $E_f = 12.22 - 2.42 = 9.80\ \text{J} = E_i$ ✓

Problem 2 · Factoring the Root

Given: the quadratic formula applied to $\tfrac{1}{2}kx^2 - xmg\sin\theta - dmg\sin\theta = 0$ returns $x = \dfrac{mg\sin\theta}{k} + \sqrt{\left(\dfrac{mg\sin\theta}{k}\right)^{2} + \dfrac{2dmg\sin\theta}{k}}$ . Find the equivalent form after factoring $\dfrac{mg}{k}$ out front.

✅ Correct! Dividing the radicand by

\left(\frac{mg}{k}\right)^2

sends

\left(\frac{mg\sin\theta}{k}\right)^2 \to \sin^2\theta

and

\frac{2dmg\sin\theta}{k} \to \frac{2kd}{mg}\sin\theta

❌ The first term under the root is not $1$ .

\left(\frac{mg\sin\theta}{k}\right)^{2} \div \left(\frac{mg}{k}\right)^{2} = \sin^2\theta

. This is the misprint carried by the printed textbook — the algebra gives

\sin^2\theta

❌ Check what pulling out $\sin^2\theta$ leaves behind.

\sqrt{\sin^2\theta + \frac{2kd}{mg}\sin\theta} = \sin\theta\sqrt{1 + \frac{2kd}{mg\sin\theta}}

— a second factor of

\sin\theta

moves into the denominator.

❌ The leading term is missing. Only the square root was factored; the

\frac{mg\sin\theta}{k}

sitting outside the radical becomes the

\sin\theta

in front.

Show solution

Factor $\frac{mg}{k}$ out of both terms. Outside the root it is immediate; inside, pulling out a factor means dividing the radicand by its square:

\left(\frac{mg\sin\theta}{k}\right)^{2} = \left(\frac{mg}{k}\right)^{2}\sin^2\theta, \qquad \frac{2dmg\sin\theta}{k} = \left(\frac{mg}{k}\right)^{2}\cdot\frac{2kd}{mg}\sin\theta

x = \frac{mg}{k}\left(\sin\theta + \sqrt{\sin^2\theta + \frac{2kd}{mg}\sin\theta}\right)

Why the $1$ is wrong. Some printings show a $1$ under this root. Substituting that expression back into $\tfrac{1}{2}kx^2 - xmg\sin\theta - dmg\sin\theta$ does not give zero — it leaves a residue of $\frac{(mg\cos\theta)^2}{2k}$ , which for the numbers of Problem 1 is $0.36\ \text{J}$ . The correct root, with $\sin^2\theta$ , satisfies the equation exactly.

Numerically (Problem 1 values, $\frac{mg}{k} = 0.049$ , $\frac{2kd}{mg} = 40.82$ ): the correct form gives $0.247\ \text{m}$ , the $1$ -version gives $0.251\ \text{m}$ .

Problem 3 · Turning Friction On

Given: the same $2.0\ \text{kg}$ cart, $30^\circ$ ramp, $d = 1.00\ \text{m}$ and $k = 400\ \text{N/m}$ , now with kinetic friction $\mu_k = 0.20$ acting over the whole slide. Find the reduced factor and the new maximum compression.

What is $\sin\theta - \mu_k\cos\theta$ ?

What is the maximum compression $x$ ?

✅ Correct!

R = 0.500 - 0.20(0.866) = 0.327

, and the same formula with

\sin\theta

replaced by

R

gives

x = 0.049\left(0.327 + \sqrt{0.107 + 13.34}\right) = 0.196\ \text{m}

❌ The normal force carries a $\cos\theta$ .

N = mg\cos\theta

, so

f_k = \mu_k mg\cos\theta

and the subtracted term is

\mu_k\cos\theta = 0.173

, not

\mu_k

❌ Check the sign. Friction opposes the motion, so its work is negative and it reduces the driving factor:

\sin\theta - \mu_k\cos\theta

❌ That used $\mu_k\tan\theta$ . The friction work is

-\mu_k mg\cos\theta\,(d+x)

and the gravitational work carries

\sin\theta

; dividing both by

mg

leaves

\sin\theta - \mu_k\cos\theta

❌ Friction acts over $d + x$ , not $d$ . The cart keeps sliding while the spring compresses, so the unknown itself sets the path length — that is why

x

appears inside

W_{nc}

❌ The substitution is everywhere or nowhere. Both the

x

term and the

d

term collect the same factor, so

R

must replace

\sin\theta

inside the

\frac{2kd}{mg}

term as well.

❌ That is the frictionless answer. With friction the cart arrives with less kinetic energy, so the spring stores less and stops it sooner.

Show solution

Step 1 — the friction force. Perpendicular to the incline the cart does not accelerate, so $N = mg\cos\theta = 16.97\ \text{N}$ and

f_k = \mu_k mg\cos\theta = 0.20(16.97) = 3.39\ \text{N}, \qquad W_{nc} = -\mu_k mg\cos\theta\,(d + x)

Step 2 — collect. Setting $W_{nc} = E_f - E_i$ and moving everything to one side:

0 = \tfrac{1}{2}kx^2 - xmg(\sin\theta - \mu_k\cos\theta) - dmg(\sin\theta - \mu_k\cos\theta)

This is the Problem 1 equation with $\sin\theta$ replaced by $R = \sin\theta - \mu_k\cos\theta = 0.500 - 0.173 = \mathbf{0.327}$ .

Step 3 — substitute into the old answer. With $\frac{mg}{k} = 0.049$ and $\frac{2kd}{mg} = 40.82$ :

x = \frac{mg}{k}\left(R + \sqrt{R^2 + \frac{2kd}{mg}R}\right) = 0.049\left(0.327 + \sqrt{0.107 + 13.34}\right) = \mathbf{0.196\ \text{m}}

Check: $\tfrac{1}{2}kx^2 = 7.66\ \text{J}$ , while gravity releases $mg\sin\theta(d+x) = 11.72\ \text{J}$ and friction takes $\mu_k mg\cos\theta(d+x) = 4.06\ \text{J}$ ; $7.66 + 4.06 = 11.72$ ✓

Problem 4 · The Energy Bill

Given: the friction case of Problem 3 — $\mu_k = 0.20$ , $mg\cos\theta = 16.97\ \text{N}$ , $d = 1.00\ \text{m}$ , and the compression it produced, $x = 0.196\ \text{m}$ . Find the mechanical energy converted to thermal energy.

✅ Correct!

E_{\text{lost}} = \mu_k mg\cos\theta\,(d+x) = 0.20(16.97)(1.196) = 4.06\ \text{J}

, which is exactly

E_i - E_f = 9.80 - 5.74

❌ The path is $d + x$ . Friction rubs for the whole slide, including the

0.196\ \text{m}

during which the spring is being compressed.

❌ That used the frictionless compression.

d + x

must be the distance actually travelled,

1.00 + 0.196 = 1.196\ \text{m}

, not

1.247\ \text{m}

— friction shortens the slide, so it also shortens its own path.

❌ Friction is set by the normal force.

N = mg\cos\theta

, so

f_k = \mu_k mg\cos\theta

;

\sin\theta

belongs to the gravitational component along the slope, not to friction.

Show solution

The lost energy is the magnitude of the non-conservative work, evaluated over the distance the cart actually slid:

E_{\text{lost}} = |W_{nc}| = \mu_k mg\cos\theta\,(d + x) = 0.20(19.6)(0.866)(1.00 + 0.196) = \mathbf{4.06\ \text{J}}

Cross-check with the energy ledger. Starting energy $E_i = mgd\sin\theta = 9.80\ \text{J}$ . At maximum compression,

E_f = \tfrac{1}{2}kx^2 - xmg\sin\theta = 7.66 - 1.92 = 5.74\ \text{J}

E_i - E_f = 9.80 - 5.74 = 4.06\ \text{J}\ \checkmark

Those joules are now thermal energy inside the cart — for a wheeled cart, mostly in the bearings. Using the part (a) compression instead would report $4.23\ \text{J}$ , overstating both the slide and the loss.

Solved: 0 / 4