Classical-Mechanics · Unit 9 · Video 1 · Interactive Practice

The Staircase Problem: Working Backward From Where It Landed

IKey Formulas

Formula	Stage	What it gives
$a_x = -\mu_k g$	Friction slide	Deceleration on the floor
$s = \dfrac{v_0^2 - v_{x,1}^2}{2\mu_k g}$	Friction slide	Slide distance from launch speed
$v_{x,1}^2 = \dfrac{3 g d^2}{2h}$	Free fall	Launch speed from the landing
$s = \dfrac{\,v_0^2 - \dfrac{3 g d^2}{2h}\,}{2\mu_k g}$	Combined	Everything in givens only

Key Insight: The two stages share exactly one unknown — the launch speed. The landing geometry pins it down, and it unlocks the slide.

IIFree Fall Off the Staircase

A horizontal launch keeps $v_x$ constant while gravity grows $v_y$ , curving the path onto the third step.

Time 0.00

IIIHow Far Does It Slide?

The slide distance grows with the launch speed $v_0$ and shrinks as friction $\mu_k$ increases.

Start speed v₀ 7.50 Friction μ_k 0.35

💡 Mass appears nowhere in $s$ : a heavy block and a light block slide exactly the same distance on the same floor.

IVTwo Stages, One Unknown

The launch speed is the single unknown the two stages share.

Step 1

Stage 1 — friction slide

a_x = -\mu_k g \;\Rightarrow\; s = \frac{v_0^2 - v_{x,1}^2}{2\mu_k g}

launch speed

v_{x,1}

still unknown

Stage 2 — free fall

3d = v_{x,1}\,t_2, \qquad 3h = \tfrac{1}{2} g\,t_2^2

\Rightarrow\; v_{x,1}^2 = \frac{3 g d^2}{2h}

Substitute and combine

s = \frac{\,v_0^2 - \dfrac{3 g d^2}{2h}\,}{2\mu_k g}

Does it make sense?

mass cancels

\;\cdot\;

s

grows with

v_0^2

\;\cdot\;

s

shrinks with

\mu_k

VQuiz Questions

Problem 1 · Acceleration on the Floor

Given: a block slides across a horizontal floor with kinetic friction coefficient $\mu_k = 0.5$ and $g = 9.8\ \text{m/s}^2$ — find its horizontal acceleration $a_x$ .

✅ Correct!

a_x = -\mu_k g = -(0.5)(9.8) = -4.9\ \text{m/s}^2

, and the mass cancels.

❌ Close. That is

-g

alone. Friction is

f_k = \mu_k N = \mu_k m g

, so

a_x = -\mu_k g

, not

-g

❌ Not quite. Newton's second law in

x

-\mu_k m g = m a_x \Rightarrow a_x = -\mu_k g

Show solution

The block stays on the floor, so vertically $N - mg = 0 \Rightarrow N = mg$ . Kinetic friction is $f_k = \mu_k N = \mu_k m g$ , pointing backward.

-f_k = m a_x \;\Rightarrow\; -\mu_k m g = m a_x \;\Rightarrow\; a_x = -\mu_k g = -(0.5)(9.8) = -4.9\ \text{m/s}^2

Negative because friction opposes the motion; the mass cancels out.

Problem 2 · Launch Speed from the Landing

Given: a block launches horizontally off the top of a three-step staircase (each step rise $h = 0.5\ \text{m}$ , run $d = 1.0\ \text{m}$ ) and lands at the far corner of the third step, $(3d,\,-3h)$ ; $g = 9.8\ \text{m/s}^2$ . Find the launch speed $v_{x,1}$ .

✅ Correct!

v_{x,1}^2 = \dfrac{3gd^2}{2h} = \dfrac{3(9.8)(1.0)^2}{2(0.5)} = 29.4

, so

v_{x,1} = 5.42\ \text{m/s}

❌ That is one step. You used

v^2 = \dfrac{gd^2}{2h}

. The block clears three steps, so both offsets are

3d

and

3h

, giving

\dfrac{3gd^2}{2h}

❌ Check the fall. The drop is

3h = \tfrac{1}{2} g t_2^2

— keep the

\tfrac{1}{2}

. Dropping it gives

\dfrac{3gd^2}{h}

, which is too large.

❌ Not quite. Eliminate

t_2

between

3d = v_{x,1}t_2

and

3h = \tfrac{1}{2} g t_2^2

Show solution

Horizontal: $3d = v_{x,1} t_2 \Rightarrow t_2 = \dfrac{3d}{v_{x,1}}$ . Vertical: $3h = \tfrac{1}{2} g t_2^2$ .

3h = \tfrac{1}{2} g\left(\frac{3d}{v_{x,1}}\right)^2 = \frac{9 g d^2}{2 v_{x,1}^2} \;\Rightarrow\; v_{x,1}^2 = \frac{9 g d^2}{6 h} = \frac{3 g d^2}{2h}

v_{x,1}^2 = \frac{3(9.8)(1.0)^2}{2(0.5)} = 29.4 \;\Rightarrow\; v_{x,1} = 5.42\ \text{m/s}

Problem 3 · Put the Two Stages Together

Given: $v_0 = 8\ \text{m/s}$ , $\mu_k = 0.3$ , $h = 0.5\ \text{m}$ , $d = 1.0\ \text{m}$ , $g = 9.8\ \text{m/s}^2$ — find the launch speed squared $v_{x,1}^2$ and the slide distance $s$ .

What is $v_{x,1}^2$ ?

What is $s$ ?

✅ Correct!

v_{x,1}^2 = 29.4

and

s = \dfrac{64 - 29.4}{2(0.3)(9.8)} = \dfrac{34.6}{5.88} = 5.88\ \text{m}

❌ Check $v_{x,1}^2$ .

v_{x,1}^2 = \dfrac{3gd^2}{2h} = \dfrac{3(9.8)(1.0)^2}{2(0.5)} = 29.4

❌ Check $s$ . Subtract the launch term first, then divide by

2\mu_k g

s = \dfrac{v_0^2 - v_{x,1}^2}{2\mu_k g}

Show solution

Launch speed: $v_{x,1}^2 = \dfrac{3gd^2}{2h} = \dfrac{3(9.8)(1.0)^2}{2(0.5)} = 29.4\ \text{m}^2/\text{s}^2$ .

Slide distance:

s = \frac{v_0^2 - v_{x,1}^2}{2\mu_k g} = \frac{8^2 - 29.4}{2(0.3)(9.8)} = \frac{64 - 29.4}{5.88} = \frac{34.6}{5.88} = 5.88\ \text{m}

Common slips: forgetting to subtract $v_{x,1}^2$ gives $64/5.88 = 10.9$ ; dropping the factor $2$ gives $34.6/2.94 = 11.8$ ; doubling $\mu_k$ gives $34.6/11.76 = 2.94$ .

Problem 4 · Heavy Block vs Light Block

Given: two blocks start with the same $v_0$ on the same floor (same $\mu_k$ ) and launch off the same staircase; block B has twice the mass of block A. Which slides farther before the edge?

✅ Correct!

s = \dfrac{v_0^2 - 3gd^2/(2h)}{2\mu_k g}

contains no mass — it cancels, so both slide equally.

❌ Mass cancels. Friction scales with mass (

f_k = \mu_k m g

), but so does inertia (

m a_x

), leaving

a_x = -\mu_k g

mass-free — and so is

s

❌ Not quite. Look for

m

s = \dfrac{v_0^2 - 3gd^2/(2h)}{2\mu_k g}

— there is none.

Show solution

Once friction enters, every step is mass-free:

-\mu_k m g = m a_x \;\Rightarrow\; a_x = -\mu_k g

The heavier block feels more friction, but it also carries more inertia; the two effects cancel exactly. Since

s = \frac{\,v_0^2 - \dfrac{3gd^2}{2h}\,}{2\mu_k g}

has no $m$ in it, both blocks slide the same distance.

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