# Homework Help: Cut the String - Energy and Friction

1. Aug 3, 2015

### FredericChopin

1. The problem statement, all variables and given/known data
http://imgur.com/FZM5gqC,RlLeGmP#1
http://imgur.com/FZM5gqC,RlLeGmP#0

2. Relevant equations
$f^k_{AB} = \mu_k N_{AB}$

$W_{f,i} = \int_{r_i}^{r_f} \vec{F} \cdot d\vec{r}$

$E^{mech}_f = E^{mech}_i + W^{NC}_{f,i}$

$U_{elastic} = \frac{1}{2} k x^2$

3. 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.

2. Aug 3, 2015

### 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.

Last edited: Aug 3, 2015
3. Aug 4, 2015

### haruspex

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