Question

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 400 N/m; the other end of the spring is fixed in place. The cookie has a kinetic energy of 20.0 Jas it passes through the spring’s equilibrium position. As the cookie slides, a frictional force of magnitude 10.0 Nacts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?

Short answer

$$\begin{gathered} \mathrm{a})d=0.292\mathrm{m} \\ \mathrm{b})K_s=14.2\mathrm{J} \end{gathered}$$

Step-by-step solution

Given Data

k = 400 N/m

Kinetic energy20.0J

Frictional force 10.0 N

Understanding the concept

Apply Eq. 8-31,$\mathbf{\lambda }{\mathit{E}}_{\mathbf{t}\mathbf{h}}\mathbf{=}{\mathit{f}}_{\mathbf{k}}\mathit{d}$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.

Step 3(a): Calculate how far the cookie will slide from the equilibrium position before coming momentarily to rest 

An appropriate picture (once friction is included) for this problem is Figure 8-3 in the textbook. We apply Eq. 8-31,$\lambda E_{th}=f_{k}d$ 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$d=0.292\mathrm{m}$

Step 4(b): Calculate the kinetic energy of the cookie as it slides back through the equilibrium position 

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