Kinetic/Thermal Energy Problem

  1. In a volcanic eruption, a 2.0 kg piece of porous volcanic rock is thrown vertically upward with an initial speed of 45 m/s. It travels upward a distance of 45 m before it begins to fall back to the Earth.
    (a) What is the initial kinetic energy of the rock?

    (b) What is the increase in thermal energy due to air friction during ascent?

    (c) If the increase in thermal energy due to air friction on the way down is 70% of that on the way up, what is the speed of the rock when it returns to its initial position?

    Relevant Equations:
    K = 1/2 M * V^2
    External Work = Change in Mechanical Energy + Change in Thermal Energy
    Change in Thermal Energy = Friction Force * Displacement


    Is the answer for part A just
    1/2 M * V^2?
    It seems a little bit too easy.

    For Part B-
    Is to find the change in thermal energy during its ascent
    Change in Thermal Energy = Friction Force * Displacement
    Friction Force = Uk * Normal Force
    How would I find Uk or Normal Force?

    Any help would be grateful.
     
  2. jcsd
  3. Doc Al

    Staff: Mentor

    They can't all be hard.

    Kinetic friction is a model that applies to dry solids rubbing against each other; it doesn't apply here. Luckily, you don't need it. Consider the total mechanical energy the rock had to start with (part A) and compare it to what's left when it reaches its highest point.
     
  4. If I let y=0 when the rock is launched

    Ki + Ui = Kf + Uf?

    The initial Kinetic energy is 1/2 * M * Vi^2
    The initial Potential Energy is 0

    Is the Final Kinetic Energy where it reaches its highest point 0 since the final velocity will be 0 at its highest point
    The final potential energy is MGH where H is 45m

    So I have
    1/2 M * Vi^2 = MGH

    What am I suppose to solve/find out?
     
  5. Nothing, you have everything. Did you check if it's true? I mean, are the two sides equal???
    I don't think so. Why?
     
  6. Ahh, I got it, the difference is 1142.1 J.
    Thanks!

    For Part C:
    (c) If the increase in thermal energy due to air friction on the way down is 70% of that on the way up, what is the speed of the rock when it returns to its initial position?

    70% of 1142.1J = 799.47J

    Would it be
    Ki + Ui = Kf + Uf From the highest point to the initial position?
    0 + MGH = 1/2 M * Vf^2 + 0
    MGH = 1/2 M * Vf^2

    MGH + 799.47 = 1/2 M * Vf^2

    and solve for Vf?
     
  7. Doc Al

    Staff: Mentor

    Almost, but you're getting a bit mixed up. Think of the energy this way:

    What you start with (PE at the top) = KE at the bottom (1/2MVf^2) + Thermal energy (which you've just calculated)

    Rearrange that and solve for Vf.
     
  8. Thanks Doc!
     
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