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Drag Force and Kinetic Energy

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A stone with mass m = 1.8 kg is thrown vertically upward into the air with an initial kinetic energy of 310 J. The drag force acting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 2.5 N. (a) What is the maximum height reached by the stone? (b) What is the speed of the stone upon impact with the ground?


    2. Relevant equations



    3. The attempt at a solution

    This one completely confused me and I am not sure how to solve it. Any help would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Oct 7, 2009 #2
    In ideal (non-drag) projectile motion, the only force in the y-direction is gravity. The only difference here is you have an additional force in the y-direction always opposite the direction of motion. So if you split up your motion into up and down components, on the way up you have gravity pulling down and drag pulling down. On the way down, you have gravity pulling down and drag pulling up.
     
  4. Oct 8, 2009 #3
    Ok, so tell me if I am understanding this correctly.

    The formula that can be used to solve this problem is the following:

    KE1 + PE1 = KE2 + PE2

    KE1 = Kinetic Energy at the start of the rock being thrown
    PE1 = Potential Energy at the start of the rock being thrown

    KE2 = Kinetic Energy at the highest point reached by the rock
    PE2 = Potential Energy at the highest point reach by the rock

    So, PE1 will equal 0 and KE2 will equal 0.
    We are given KE1 which is 310 J.

    So I get the following:
    KE1 = PE2

    (1/2)mv^2 = mgh

    Fill in the variables:

    310 J = (1.8)(9.8)h

    All I need to do now is solve for (h) but...
    I am not sure how to add in the drag force. Any suggestions?
     
  5. Oct 8, 2009 #4
    you equated the 2. This is not correct because drag exists. your difference in energies will be equal to the energy lost through mechanical heat (aka drag in this case). You know the value of this because it's a force acting on the system (i.e. changing the total energy).

    In an isolated system, your equations are correct, so just imagine that someone is pulling down on the rock as it travels with a force of 2.5N. What is the potential energy at maximum height?
     
  6. Oct 8, 2009 #5
    I actually have a similar problem that I am having trouble with. I am not fully understanding the concepts you presented in the last post. What do mean you by saying that CaptFormal "equated the 2"?
     
  7. Oct 8, 2009 #6

    jambaugh

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    Gold Member

    The total energy, kinetic plus potential will not be conserved because of the drag force.
    You must include the (negative) work done by drag on the stone. For constant force recall that work is force times distance or more precisely Force dot Displacement which comes out negative here since the force is opposite the motion.

    I suggest you let the height it reaches be some variable. Express work and potential energy in terms of it and see what you can work out.
     
  8. Oct 8, 2009 #7
    when i say equated the two, captFormal assumed that the system is isolated and that the total energy is conserved from point to point. I.e. he said PE2 = KE1. This is not true since the drag force does work on the system.
     
  9. Oct 8, 2009 #8
    So... if the system is not isolated, then how to I solve for this? Sorry, I am sure it is quite simple, I am just not seeing how to solve for it.
     
  10. Oct 9, 2009 #9
    Oh, I got it.

    So the formula that I should use would be the following:

    KE = PE + Fd

    Now there is not an isolated system condition and the additional force is being placed into effect.
     
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