# A problem on mechanical energy and friction

I'm studying for a mid-term, and I'm trying to solve some mock-papers. The following are two problems which I have solved, but I'm missing a part:

$(1)$ A body of mass $m_1= 8 \rm kg$ passes through $A$ with a speed of $v_A= 10 \rm \frac m s$, and a meight of $h_a = 5 \rm m$ relative to the floor level. The friction is neglibible all through the path, except from in $\overline{CD}=10 \rm m$ where the dynamic friction coefficient is $\mu_d=0.5$. At the end of the path there is an ideal spring of constant $K= 1000 \rm \frac N m$. Find.

$(a)$ The maximum compression of the spring when it is hit by the object. $(b)$ The distance (measured from $C$) the block travels before coming to a still while returning due to the decompression of the spring.

I have solved $(a)$ using the theorem of the work of non-conervative forces. Namely, I have that, being $D'$ the instant of maximum compression, that the work $\mathcal W$ of the non conservatives forces is equal to the change $\Delta$ of the mechanical energy:

$$\mathop \Delta \limits_{{\text{A}} \to {\text{D'}}} {{\text{E}}_{{\text{mec}}}} = \mathop {\mathcal W}\limits_{{\text{A}} \to {\text{D'}}} {{\bf{F}}_{{\text{nc}}}}$$

This translates to

$${\text{E}}_{{\text{mec}}}^{{\text{D'}}} - {\text{E}}_{{\text{mec}}}^{\text{A}} = \mathop {\mathcal W}\limits_{{\text{A}} \to {\text{D'}}} {\bf{N}} + \mathop {\mathcal W}\limits_{{\text{A}} \to {\text{D'}}} {\bf{f}}$$

Where $\rm N$ is the normal and $\rm f$ is the friction. This gives

\eqalign{ & \left( {\frac{1}{2}k \cdot \Delta \ell _{\max }^2 + 0 + 0} \right) - \left( {0 + \frac{1}{2}m \cdot v_A^2 + m \cdot {h_A} \cdot g} \right) = 0 + \overline {{\text{CD}}} \cdot \left\| {\text{f}} \right\| \cdot \cos 180 \cr & \left( {500\frac{{\text{N}}}{{\text{m}}} \cdot \Delta \ell _{\max }^2} \right) - \left( {0 + 4{\text{kg}} \cdot 100\frac{{{{\text{m}}^{\text{2}}}}}{{{{\text{s}}^{\text{2}}}}} + 8{\text{kg}} \cdot 5{\text{m}} \cdot 10\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}} \right) = - 10{\text{m}} \cdot \left\| N \right\|{\mu _d} \cr & \left( {500\frac{{\text{N}}}{{\text{m}}} \cdot \Delta \ell _{\max }^2} \right) - \left( {400{\text{J}} + 400{\text{J}}} \right) = - 10{\text{m}} \cdot m \cdot g \cdot \frac{1}{2} \cr & 500\frac{{\text{N}}}{{\text{m}}} \cdot \Delta \ell _{\max }^2 - 800{\text{J}} = - 10{\text{m}} \cdot 8{\text{kg}} \cdot 10\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}} \cdot \frac{1}{2} \cr & 500\frac{{\text{N}}}{{\text{m}}} \cdot \Delta \ell _{\max }^2 - 800{\text{J}} = - 400{\text{J}} \cr & 500\frac{{\text{N}}}{{\text{m}}} \cdot \Delta \ell _{\max }^2 = 400{\text{J}} \cr & \Delta \ell _{\max }^2 = \frac{4}{5}{{\text{m}}^2} \cr & \Delta {\ell _{\max }} = 2\frac{{\sqrt 5 }}{5}{\text{m}} \cong 0.89{\text{m}} \cr}

I don't know, however, how to solve $(b)$. I suppose I need to consider the energy the body has at $C$ and see what distance it should travel so that it is exhausted by the friction.

ADD: I think one can solve it this way. ¿Are the reasonings correct? (I'm sure the calculations are)

All the elastic energy has to turn into work by the friction, which means

\eqalign{ & \frac{1}{2}k\cdot\Delta \ell _{\max }^2 = \overline {QC} \cdot m \cdot g\cdot{\mu _d} \cr & 500\frac{{\text{N}}}{{\text{m}}}\frac{4}{5}{{\text{m}}^{\text{2}}} = \frac{1}{2}\overline {QC} \cdot 8kg \cdot 10\frac{m}{{{s^2}}} \cr & 400{\text{N}} \cdot {\text{m}} = 40{\text{N}} \cdot \overline {QC} \cr & 10{\text{m}} = \overline {QC} \cr}

where $Q$ is the halting point.

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 This should help you. – Vijay Murthy Jul 2 '12 at 20:48