91Ó°ÊÓ

k = 400 N/m

Kinetic energy20.0J

Frictional force 10.0 N

Apply Eq. 8-31,$\mathbf{λ}{\mathit{E}}_{\mathbf{t}\mathbf{h}}\mathbf{=}{\mathit{f}}_{\mathbf{k}}\mathit{d}$and relate initial kinetic energy${\mathit{K}}_{\mathbf{i}}$to the potential energy${\mathit{U}}_{\mathbf{r}}$to calculate the distance and relate${\mathit{U}}_{\mathbf{r}}$to the "second" kinetic energy${\mathit{K}}_{\mathit{s}}$it has at the unstretched position to calculate the kinetic energy.

An appropriate picture (once friction is included) for this problem is Figure 8-3 in the textbook. We apply Eq. 8-31,$\mathrm{λ}{E}_{th}=f_{k}d$ and relate initial kinetic energy$K_i$ to the "resting" potential energy$U_r$:

$$\begin{gathered} K_i+U_i=f_{k}d+k_r+U_r \\ 20.0\mathrm{J}+0=f_k\mathrm{d}+0+\frac{1}{2}k{d}^2 \end{gathered}$$

Where $f_k=10.0\mathrm{N}\mathrm{and}K=400\mathrm{N}/\mathrm{m}$. We solve the equation for dusing the quadratic formula or by using the polynomial solver on an appropriate calculator, with $d=0.292\mathrm{m}$being the only positive root.

Hence the solution is$d=0.292\mathrm{m}$

We apply Eq. 8-31 again and relate$U_r$ to the "second" kinetic energy$K_s$ it has at the unstretched position.

$$\begin{gathered} K_r+U_r=f_{k}d+k_s+U_s \\ \frac{1}{2}k{d}^2=f_{k}d+k_s+0 \end{gathered}$$

Using the result from part (a), these yields$K_s=14.2\mathrm{J}$

Hence the kinetic energy is 14.2 J