FredericChopin
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Homework Statement
http://imgur.com/FZM5gqC,RlLeGmP#1
http://imgur.com/FZM5gqC,RlLeGmP#0
Homework Equations
f^k_{AB} = \mu_k N_{AB}
W_{f,i} = \int_{r_i}^{r_f} \vec{F} \cdot d\vec{r}
E^{mech}_f = E^{mech}_i + W^{NC}_{f,i}
U_{elastic} = \frac{1}{2} k x^2
The Attempt at a Solution
Since the blocks are at rest after the release of the spring, the final mechanical energy, E^{mech}_f, is 0. The initial mechanical energy, E^{mech}_i, is the elastic potential energy of the spring (U_{elastic} = \frac{1}{2} k x^2). 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 (W^{NC}_{f,i} = \int_{r_i}^{r_f} \vec{f^k_{AB}} \cdot d\vec{r}). The force of kinetic friction acts through a displacement d in the same direction as force, so W^{NC}_{f,i} = \int_{0}^{d} \vec{f^k_{box,block}} \cdot d\vec{r} = f^k_{box,block}d.
Let's consider the system of the box, the block, and the spring.
The final mechanical energy of this system will be:
E^{mech}_f = E^{mech}_i + W^{NC}_{f,i}
Substituting in terms:
0 = U_{elastic} + f^k_{box,block}d
, which becomes:
0 = \frac{1}{2} k x^2 + \mu_k N_{box,block}d
The block is not accelerating in the vertical direction, and so due to Newton's Second Law, N_{box,block} must be equal in magnitude to m_{box}g:
0 = \frac{1}{2} k x^2 + \mu_k m_{box}gd
Solving for \mu_k yields:
\mu_k = \frac{-kx^2}{2m_{box}gd}
It's strange that there is a negative sign in the answer as \mu_k should be a positive scalar. It also turns out this answer is incorrect.
What went wrong?
Thank you.