Cut the String - Energy and Friction

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SUMMARY

The discussion focuses on the mechanics of a box-block-spring system involving energy conservation and kinetic friction. The initial mechanical energy, represented as elastic potential energy (U_{elastic} = \frac{1}{2} k x^2), is balanced against the work done by kinetic friction (W^{NC}_{f,i} = \int_{0}^{d} \vec{f^k_{box,block}} \cdot d\vec{r}). The calculation for the coefficient of kinetic friction, \mu_k, yields a negative value, indicating an error in the approach. The correct interpretation emphasizes that friction does negative work, necessitating a reevaluation of the energy balance.

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  • Understanding of mechanical energy concepts, including potential and kinetic energy.
  • Familiarity with Newton's Second Law and normal force calculations.
  • Knowledge of frictional forces and their role in energy dissipation.
  • Proficiency in calculus for evaluating integrals in work calculations.
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  • Study the effects of kinetic friction on motion and energy transfer.
  • Learn about the correct application of Newton's Second Law in dynamic systems.
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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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