Cut the String - Energy and Friction

[FredericChopin]

Homework Statement

http://imgur.com/FZM5gqC,RlLeGmP#1
http://imgur.com/FZM5gqC,RlLeGmP#0

Homework Equations

[itex]f^k_{AB} = \mu_k N_{AB}[/itex]

[itex]W_{f,i} = \int_{r_i}^{r_f} \vec{F} \cdot d\vec{r}[/itex]

[itex]E^{mech}_f = E^{mech}_i + W^{NC}_{f,i}[/itex]

[itex]U_{elastic} = \frac{1}{2} k x^2[/itex]

The Attempt at a Solution

Since the blocks are at rest after the release of the spring, the final mechanical energy, [itex]E^{mech}_f[/itex], is 0. The initial mechanical energy, [itex]E^{mech}_i[/itex], is the elastic potential energy of the spring ([itex]U_{elastic} = \frac{1}{2} k x^2[/itex]). There are no external non-conservative forces acting on the box-block-spring system, but there is the internal non-conservative force of kinetic friction acting on the box and block ([itex]W^{NC}_{f,i} = \int_{r_i}^{r_f} \vec{f^k_{AB}} \cdot d\vec{r}[/itex]). The force of kinetic friction acts through a displacement [itex]d[/itex] in the same direction as force, so [itex]W^{NC}_{f,i} = \int_{0}^{d} \vec{f^k_{box,block}} \cdot d\vec{r} = f^k_{box,block}d[/itex].

Let's consider the system of the box, the block, and the spring.

The final mechanical energy of this system will be:

[itex]E^{mech}_f = E^{mech}_i + W^{NC}_{f,i}[/itex]

Substituting in terms:

[itex]0 = U_{elastic} + f^k_{box,block}d[/itex]

, which becomes:

[itex]0 = \frac{1}{2} k x^2 + \mu_k N_{box,block}d[/itex]

The block is not accelerating in the vertical direction, and so due to Newton's Second Law, [itex]N_{box,block}[/itex] must be equal in magnitude to [itex]m_{box}g[/itex]:

[itex]0 = \frac{1}{2} k x^2 + \mu_k m_{box}gd[/itex]

Solving for [itex]\mu_k[/itex] yields:

[itex]\mu_k = \frac{-kx^2}{2m_{box}gd}[/itex]

It's strange that there is a negative sign in the answer as [itex]\mu_k[/itex] should be a positive scalar. It also turns out this answer is incorrect.

What went wrong?

Thank you.

 

[tommyxu3]

My opinion:
Your calculation is like that the L-shape block is fixed on the floor.
Besides, the sign of ##\mu_k## is usually positive for its just a ratio between ##f_k## and ##N.## That is, your relation may turn to:
$$U+W_f=0$$
$$\Rightarrow U-\int N\mu_k\cdot dr=0$$
The reason is obvious that we all know the friction does negative work here.

 

[haruspex]

Which way does the force of friction act? Which way is the displacement vector? What is the sign of their dot product?