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Cut the String - Energy and Friction

  1. Aug 3, 2015 #1
    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
    [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]

    3. 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.
     
  2. jcsd
  3. Aug 3, 2015 #2
    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
  4. Aug 4, 2015 #3

    haruspex

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    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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