Worked exercise — block on an incline with a spring and friction

Problem

A block of $4\;\text{kg}$ starts from rest at the top of a $30^\circ$ incline, $10\;\text{m}$ long. At the bottom there is a spring ($K = 400\;\text{N/m}$, $\ell_0 = 3\;\text{m}$). Friction is $F_a = 10\;\text{N}$. The block slides down and compresses the spring. Find the maximum compression.

Solution

Step 1 — Interaction diagram. The block interacts with the Earth (weight), with the spring (elastic force), and with the plane (constraint reaction + friction). Friction admits no potential energy: in the balance it appears as an arrow towards $E_{\text{term}}$.

Step 2 — Energy table. Gravitational reference at the base of the incline. Let $l_B$ be the length of the spring in state B (compressed from $\ell_0 = 3\;\text{m}$) and $\Delta s_B$ the distance travelled along the incline.

Starting height: $h_A = 10\cdot\sin 30^\circ = 5\;\text{m}$, so $E_{\text{pot,g}}^A = 4\cdot 9{,}81\cdot 5 = 196{,}2\;\text{J}$.

State	$E_{\text{pot,g}}$	$E_{\text{pot,m}}$	$E_{\text{cin}}$	$E_{\text{term}}$	$E_{\text{tot}}$
A	$196{,}2$	$0$	$0$	$0$	$196{,}2$
B	$4\cdot 9{,}81\cdot h_B$	$200\,(l_B-3)^2$	$0$	$10\cdot\Delta s_B$	$196{,}2$

Step 3 — Geometric conditions. The block starts at the top and compresses the spring, so

$\Delta s_B + l_B = 10\;\text{m}, \qquad h_B = l_B\sin 30^\circ = \tfrac{1}{2}\,l_B$

Step 4 — Conservation equation. $E_{\text{tot}}^B = 196{,}2$:

$4\cdot 9{,}81\cdot h_B + 200\,(l_B-3)^2 + 10\,(10-l_B) = 196{,}2$

Substituting $h_B = l_B/2$:

$19{,}62\,l_B + 200\,(l_B-3)^2 + 100 - 10\,l_B = 196{,}2$

$200\,l_B^2 - 1190{,}38\,l_B + 1703{,}8 = 0$

Solving.

$l_B = \frac{1190{,}38 \pm \sqrt{1190{,}38^2 - 4\cdot 200\cdot 1703{,}8}}{2\cdot 200} = \frac{1190{,}38 \pm \sqrt{53964{,}5}}{400} = \frac{1190{,}38 \pm 232{,}3}{400}$

The two roots are $l_B = 3{,}56\;\text{m}$ and $l_B = 2{,}40\;\text{m}$. Since the spring is compressed, we must have $l_B < \ell_0 = 3\;\text{m}$: we choose $l_B \approx 2{,}40\;\text{m}$.

The maximum compression is therefore

$\ev{\ell_0 - l_B = 3 - 2{,}40 \approx 0{,}60\;\text{m}}$

Balance check ($l_B = 2{,}40$): $E_{\text{pot,g}} = 4\cdot 9{,}81\cdot 1{,}20 = 47{,}0$; $E_{\text{pot,m}} = 200\cdot 0{,}60^2 = 72{,}0$; $E_{\text{term}} = 10\cdot 7{,}60 = 76{,}0$. Sum $\approx 195$ J $\approx E_{\text{tot}}$. ✓

Links

Topics: Work and energy Concepts: Gravitational potential energy · Elastic potential energy · Thermal energy from friction · Kinetic friction · Conservation of mechanical energy Skills: Using the energy table · Force resolution on an incline Methods: State-by-state energy table Objects: Inclined plane · Spring · Block