Cut the String - Energy and Friction

AI Thread Summary
The discussion focuses on the mechanics of a system involving a spring, box, and block, analyzing the energy transformations and the role of kinetic friction. The initial mechanical energy is derived from the elastic potential energy of the spring, while the final mechanical energy is zero when the system comes to rest. The work done by kinetic friction is calculated, leading to an equation that incorrectly suggests a negative coefficient of friction. The error is attributed to misunderstanding the direction of frictional force and displacement, which affects the sign in the energy balance equation. Properly accounting for these factors clarifies that the coefficient of kinetic friction should be a positive value.
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.
 
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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:
Which way does the force of friction act? Which way is the displacement vector? What is the sign of their dot product?
 
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