1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cut the String - Energy and Friction

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data

    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:
    $$\Rightarrow U-\int N\mu_k\cdot dr=0$$
    The reason is obvious that we all know the friction does negative work here.
    Last edited: Aug 3, 2015
  4. Aug 4, 2015 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Which way does the force of friction act? Which way is the displacement vector? What is the sign of their dot product?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted