Classical-Mechanics · Unit 18 · Video 3 · Interactive Practice

Mass-Spring on a Rough Surface: Friction That Grows With Distance

IKey Formulas

Formula	Name	What you need
$f = \mu N = b x \, m g$	Position-dependent friction	The model $\mu = bx$ on a horizontal table, where $N = mg$
$W_{nc} = \displaystyle\int_{0}^{x_f} -b x m g \, dx = -\tfrac{1}{2} b m g \, x_f^{2}$	Work of a varying force	The whole path, not just the endpoints
$W_{nc} = \Delta E_m = E_f - E_i$	Work–energy theorem with friction	$E_i = \tfrac{1}{2} m v_0^{2}$ and $E_f = \tfrac{1}{2} k x_f^{2}$
$x_f^{2} = \dfrac{m v_0^{2}}{k + b m g}$	Stopping position	One linear equation in $x_f^{2}$ — no square root needed

Key Insight: $\mu$ is dimensionless and $\mu = bx$ , so $[b] = \mathrm{m}^{-1}$ and $[b m g] = \mathrm{N/m}$ — exactly the dimensions of $k$ . Friction drains energy the way a second spring of stiffness $bmg$ would store it, so the initial kinetic energy splits in the ratio $k : bmg$ and the loss is $|\Delta E_m| = \tfrac{1}{2} m v_0^{2}\left(1 + \tfrac{k}{bmg}\right)^{-1}$ .

IIVisualization 1 — Friction's Work Is a Triangle

A friction force that grows with distance accumulates as the area under a rising line, not as force times distance.

Stroke t / t_stop 0.00

💡 On the inward stroke the block is an exact harmonic oscillator of stiffness $k + bmg$ , which is why $v(x) = v_0\sqrt{1 - x^{2}/x_f^{2}}$ . The analogy dies at the turning point: the spring hands its $\tfrac{1}{2}k x_f^{2}$ back, while friction takes another bite on the way out.

IIIVisualization 2 — Where the Energy Budget Runs Out

The block stops exactly where the spring and friction together have spent the whole kinetic energy budget.

Your stopping guess x [m] 0.350

IVVisualization 3 — Who Gets the Kinetic Energy

One ratio, $k/(bmg)$ , decides how the initial kinetic energy splits between stored spring energy and heat.

Spring constant k [N/m] 24.5 Friction ramp b [1/m] 1.25

VQuiz Questions

Problem 1 · The Friction Work Integral

Given: a block of mass $m$ slides from $x = 0$ to $x = x_f$ on a horizontal table whose friction coefficient is $\mu = bx$ — find the work done by friction.

✅ Correct! The integrand grows linearly from

0

bmg\,x_f

, so its integral is the triangle

\tfrac{1}{2}(b m g x_f)(x_f)

❌ That is the rectangle, not the triangle. Multiplying the largest friction force

bmg\,x_f

by the full distance

x_f

counts every earlier, weaker part of the path at its final strength — exactly twice the true magnitude.

❌ Check the integral.

\int_0^{x_f} x\,dx = \tfrac{1}{2}x_f^{2}

, not

\tfrac{1}{2}x_f

— the

x

\mu = bx

and the

dx

each contribute a power.

❌ Check the sign. Friction points opposite the displacement, so every element

-\mu N\,dx

is negative and the total work must be negative.

Show solution

On a horizontal table the normal force is $N = mg$ , so the friction force has magnitude

f = \mu N = (b x)(m g) = b m g \, x

It grows with position, so it cannot be pulled out of the work integral. Over each step $dx$ friction does negative work:

W_{nc} = \int_{0}^{x_f} -\mu N \, dx = -\int_{0}^{x_f} b m g \, x \, dx = -b m g \left[\frac{x^{2}}{2}\right]_{0}^{x_f}

W_{nc} = -\tfrac{1}{2} b m g \, x_f^{2}

Note the shape: $\tfrac{1}{2}(bmg)x_f^{2}$ is the same $\tfrac{1}{2}(\text{stiffness})x^{2}$ form as a spring's stored energy, with $bmg$ playing the role of the stiffness.

The trap: the shortcut $W = -\mu m g d$ is only valid for constant $\mu$ . Using the final coefficient $\mu_f = b x_f$ over the whole distance gives $-b m g x_f^{2}$ — precisely twice too much, because the rectangle of height $bmg x_f$ contains the triangle twice.

Problem 2 · Why $k + bmg$ Is Legal

Given: the stopping position came out as $x_f^{2} = \dfrac{m v_0^{2}}{k + b m g}$ — what must the dimensions of $b$ be for that sum to make sense?

✅ Correct!

bmg

is a stiffness, so

k + bmg

adds like with like and

k/(bmg)

is a pure number.

❌ Then $\mu$ itself would carry dimensions.

\mu = bx

with dimensionless

b

would make

\mu

a length, but a friction coefficient is a pure number — so

b

must cancel the length in

x

❌ Not quite. Start from

[\mu] = 1

\mu = b x

, then feed that through

bmg

and compare with

[k] = \mathrm{N/m}

Show solution

A friction coefficient is a ratio of two forces, so it is dimensionless:

[\mu] = 1, \qquad \mu = b x \;\Longrightarrow\; [b] = \frac{1}{[\,x\,]} = \mathrm{m^{-1}}

Then $mg$ is a force, and

[b m g] = \frac{\mathrm{N}}{\mathrm{m}} = [k]

So $k + bmg$ adds two quantities of the same kind — a legitimate sum, and $bmg$ deserves to be read as a second stiffness. It follows that

\frac{k}{b m g} \ \text{is a pure number,}

which is why the whole answer can depend on that single ratio:

W_{nc} = -\frac{m v_0^{2}}{2}\left(1 + \frac{k}{b m g}\right)^{-1}

Dimensional check of the result itself: $m v_0^{2}$ is an energy and the bracket is dimensionless, so $W_{nc}$ is an energy ✓.

Problem 3 · Stopping Point and the Loss

Given: the block reaches $x = 0$ with speed $v_0$ , the spring has constant $k$ , and $\mu = bx$ beyond $x = 0$ — find where it first comes momentarily to rest, and what fraction of $\tfrac{1}{2}mv_0^{2}$ friction takes.

Stopping position

Fraction lost to friction

✅ Correct! The two stiffnesses share the budget in the ratio

k : bmg

, and the shares

\frac{k}{k+bmg}

and

\frac{bmg}{k+bmg}

add to

1

❌ Check the stopping position. Every term in the balance carries

\tfrac{1}{2}

and every

x_f^{2}

collects on one side:

m v_0^{2} = (k + bmg)x_f^{2}

. Dropping friction leaves only

k

; moving its term to the wrong side flips the sign.

❌ Check the fraction. Substitute

x_f^{2}

into

\tfrac{1}{2}bmg\,x_f^{2}

and divide by

\tfrac{1}{2}mv_0^{2}

— the

\tfrac{1}{2}mv_0^{2}

cancels completely, leaving no stray factor of

\tfrac{1}{2}

and no bare

k

in the denominator.

Show solution

Step 1 — the two energy states. Initially the spring is uncompressed, so there is no potential term; finally the block is at rest, so there is no kinetic term:

E_i = \tfrac{1}{2} m v_0^{2}, \qquad E_f = \tfrac{1}{2} k x_f^{2}

Step 2 — the friction work.

W_{nc} = -\int_{0}^{x_f} b m g \, x \, dx = -\tfrac{1}{2} b m g \, x_f^{2}

Step 3 — the balance $W_{nc} = E_f - E_i$ .

-\tfrac{1}{2} b m g \, x_f^{2} = \tfrac{1}{2} k x_f^{2} - \tfrac{1}{2} m v_0^{2}

Multiply by $2$ and collect the $x_f^{2}$ terms — one linear equation in $x_f^{2}$ , so no square root is needed:

m v_0^{2} = (k + b m g)\, x_f^{2} \qquad\Longrightarrow\qquad x_f^{2} = \frac{m v_0^{2}}{k + b m g}

Step 4 — the loss. Substitute back:

W_{nc} = -\frac{b m g}{2} \cdot \frac{m v_0^{2}}{k + b m g} = -\frac{1}{2} m v_0^{2} \cdot \frac{b m g}{k + b m g}

Dividing by $E_i = \tfrac{1}{2}mv_0^{2}$ leaves the fraction lost:

\frac{|W_{nc}|}{E_i} = \frac{b m g}{k + b m g} = \left(1 + \frac{k}{b m g}\right)^{-1}

The spring keeps the complement $\dfrac{k}{k + bmg}$ , and the two shares sum to $1$ : every joule is accounted for.

Problem 4 · A Steeper Ramp: $\mu = c x^{2}$

Given: the same block, spring and table, but now the coefficient ramps quadratically, $\mu = c x^{2}$ with $c > 0$ constant — find the friction work, and decide whether the stopping position still follows from a stiffness sum.

Friction work from $0$ to $x_f$

Does the stiffness-sum shortcut survive?

✅ Correct! The method is unchanged — integrate, then balance — but only a linear

\mu

produces the

\tfrac{1}{2}(\text{stiffness})x^{2}

shape that can merge with the spring term.

❌ Check the integral.

\int_0^{x_f} x^{2}\,dx = \tfrac{1}{3}x_f^{3}

: the power of

x

\mu

rises by one on integration, and its reciprocal becomes the coefficient.

❌ Look at the powers, and at the units. The theorem still holds — that is how the equation was written at all. But the friction term now carries

x_f^{3}

, and

cmg

has units

\mathrm{N/m^{2}}

, so it can never be added to

k

Show solution

The integral. With $N = mg$ still and $\mu = c x^{2}$ :

W_{nc} = -\int_{0}^{x_f} c m g \, x^{2} \, dx = -c m g \left[\frac{x^{3}}{3}\right]_{0}^{x_f} = -\tfrac{1}{3} c m g \, x_f^{3}

The balance. The work–energy theorem is untouched — $W_{nc} = E_f - E_i$ holds for any force law:

-\tfrac{1}{3} c m g \, x_f^{3} = \tfrac{1}{2} k x_f^{2} - \tfrac{1}{2} m v_0^{2}

\tfrac{1}{2} m v_0^{2} = \tfrac{1}{2} k x_f^{2} + \tfrac{1}{3} c m g \, x_f^{3}

This is a cubic in $x_f$ . The friction term is no longer of the form $\tfrac{1}{2}(\text{stiffness})\,x_f^{2}$ , so it cannot be folded into $k$ — the clean collapse to $x_f^{2} = m v_0^{2}/(k + bmg)$ was a gift of the linear ramp $\mu = bx$ .

Units confirm it. $[\mu] = 1$ and $\mu = cx^{2}$ give $[c] = \mathrm{m^{-2}}$ , so $[c m g] = \mathrm{N/m^{2}}$ — not a stiffness, and not addable to $k$ .

The transferable idea: whenever a force varies along the path, integrate first, then balance the energy books. Only the integral changes; the bookkeeping never does.

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