Problem — Two springs and a box

Problem

A box ($m = 5\;\mathrm{kg}$, $v_A = 2\;\mathrm{m/s}$) moves horizontally between two springs: $K_1 = 1000\;\mathrm{N/m}$ ($\ell_{01} = 2\;\mathrm{m}$) and $K_2 = 500\;\mathrm{N/m}$ ($\ell_{02} = 2{,}5\;\mathrm{m}$). The distance between the walls is $4\;\mathrm{m}$. Where does the box come to rest? Set up the energy table with the constraint $l_1 + l_2 = 4$ and solve.

Solution

Key observation. The two springs, attached to the walls and to the box, have natural lengths summing to $\ell_{01} + \ell_{02} = 4{,}5\;\text{m} > 4\;\text{m}$: both are therefore always compressed and in contact with the box. The system has a single degree of freedom, tied by the constraint

$l_1 + l_2 = 4 \quad\Rightarrow\quad l_2 = 4 - l_1$

Potential energy as a function of $l_1$. The motion is horizontal, so $E_{\text{pot,g}}$ is constant and is ignored. Only the elastic energy remains:

$U(l_1) = \tfrac{1}{2}K_1(\ell_{01}-l_1)^2 + \tfrac{1}{2}K_2(\ell_{02}-l_2)^2 = 500\,(2-l_1)^2 + 250\,(l_1-1{,}5)^2$

Expanding:

$U(l_1) = 750\,l_1^2 - 2750\,l_1 + 2562{,}5$

Equilibrium point (minimum of $U$, zero net force):

$l_1^{\text{eq}} = \frac{2750}{2\cdot 750} = 1{,}83\;\text{m}, \qquad l_2^{\text{eq}} = 2{,}17\;\text{m}$

$U_{\min} = 2562{,}5 - \frac{2750^2}{4\cdot 750} \approx 41{,}7\;\text{J}$

Energy table. Assume the box passes through equilibrium with speed $v_A = 2\;\text{m/s}$ (state A) and look for the turning points (state B, $v = 0$).

State	$E_{\text{cin}}$	$E_{\text{pot,el}} = U(l_1)$	$E_{\text{tot}}$
A	$\tfrac{1}{2}\cdot 5\cdot 2^2 = 10$	$U_{\min} = 41{,}7$	$51{,}7$
B	$0$	$U(l_1)$	$51{,}7$

Turning points — equation $U(l_1) = E_{\text{tot}}$, i.e. $750\,(l_1 - 1{,}83)^2 + 41{,}7 = 51{,}7$:

$750\,(l_1 - 1{,}83)^2 = 10 \quad\Rightarrow\quad (l_1 - 1{,}83)^2 = 0{,}0133 \quad\Rightarrow\quad l_1 = 1{,}83 \pm 0{,}12$

$\ev{l_1 \approx 1{,}72\;\text{m}\ \text{and}\ l_1 \approx 1{,}95\;\text{m}}$

Correspondingly $l_2 \approx 2{,}28\;\text{m}$ and $l_2 \approx 2{,}05\;\text{m}$. Without friction the box does not really stop: it oscillates between these two turning points (where momentarily $v = 0$), symmetric about the equilibrium position $l_1^{\text{eq}} = 1{,}83\;\text{m}$.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Energia potenziale elastica · Energia cinetica · Conservazione dell’energia meccanica · Forze conservative Competenze: Uso della tabella energetica · Impostazione simbolica Metodi: Tabella energetica a stati Oggetti: Molla · Blocco