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Average Force of meteor

  1. Jul 16, 2014 #1
    1. The problem statement, all variables and given/known data
    A meteor with a mass of 1 kg moving at 20 km/s collides with Jupiter's atmosphere. The meteor penetrates 100 km into the atmosphere and disintegrates. What is the average force on the meteor once it enters Jupiter's atmosphere? (Note: ignore gravity).


    2. Relevant equations

    K = 1/2 mv^2
    W = F x d


    3. The attempt at a solution

    I initially started this problem with F = ma in mind. I then solved for time using d = vt and got 200 seconds. Then I found acceleration using V = Vi + at and got 100 m/s^2. By plugging in 100 m/x^2 into F=ma, I get (1 kg)(100 m/s^2) = 100 N.

    The correct answer is 2 x 10^5. What am I doing wrong? Can't this problem be solved without using kinetics and work?

    Thanks.
     
  2. jcsd
  3. Jul 16, 2014 #2

    Nathanael

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    d = 100km, right?

    What did you use for "v" and why?
     
  4. Jul 16, 2014 #3
    Yes, d = 100 km = 1,000,000 m is what I used. I used 20,000 m/s for the initial velocity because the problem states that the meteor disintegrates in the atmosphere = 0 m/s.
     
  5. Jul 16, 2014 #4
    What is an alternate definition for work?
     
  6. Jul 16, 2014 #5
    Energy via a force--kinetic energy. But I am trying to solve this problem using kinematics. Can this be done?
     
  7. Jul 16, 2014 #6

    Nathanael

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    100 km is only 100,000 m

    Also you can't use v = 20,000 m/s in that equation (d=vt) because the velocity is changing.




    The correct answer is supposed to be 200,000 N ?
    With every method I use, I keep getting an answer of 2,000 N. I probably shouldn't be helping you if I can't even solve the problem correctly..
     
  8. Jul 16, 2014 #7
    If you use a work energy approach you can use:

    1/2m v^2 = f x

    And solve for x. I got 2*10^5

    And the velocity is in km/s so 200,000,000m/2
     
  9. Jul 16, 2014 #8
    Sorry thats m/s at the end, not m/2
     
  10. Jul 16, 2014 #9
    Wait. km/s should actually be 20,000 which gives me an answer of 2*10^3
     
  11. Jul 16, 2014 #10
    I may be wrong here, so beware. But I think the reason you cannot use a simple constant acceleration formula like v^2 = u^2 + 2ax is beacuse there is not constant acceleration inside jupiter.

    So in my mind (my mind mind you) the 'x' is given beacuse there is no constant acceleration and it asks you to find the average.
     
  12. Jul 16, 2014 #11
    I am so sorry, yes, the correct answer is 2 s 10^3. Sorry to confuse everyone. Okay, so to clarify, I cannot use d =vt if it is assumed an object does not have constant velocity. Also, when acceleration is not constant, kinematic equations cannot be used.

    Would it be safe to say that in situations like the problem, to use energy and work formulas?

    Also, 100km = 100 km x 1000 m/1 km = 100,000 m. I stand corrected. I think I did this conversion correct and perhaps should keep it in scientific notation for a standardized test.
     
    Last edited: Jul 16, 2014
  13. Jul 16, 2014 #12
    100,000m

    And don't take too much of what I said to heart beacuse i am not too sure myself.

    However, yes. It is definately easier to use a work-energy approach for these types of questions
     
  14. Jul 16, 2014 #13

    Nathanael

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    Solve for x?
    You mean solve for f using the fact that it travelled 100km? (x=100,000)
    (P.S. What are your units? m/s? Isn't the answer supposed to be a force?)

    You sure can use a constant acceleration formula and get the same answer (2,000).

    The possible problem you're thinking of (that acceleration is not constant,) is also apparent in your above method (the "energy method").

    The "f" in your equation would not actually be a simple constant, it would be a function of some sort. But to solve for the "average force" is to solve it as if "f" were a constant.

    So in the "acceleration method" you can use constant acceleration equations because you're solving the problem as if the acceleration (force) were constant (and average).
     
  15. Jul 16, 2014 #14
    So just to clarify; in free fall motion, kinematic equations are used because the acceleration of gravity is constant but velocity constantly changes. In other words, to use kinematics, either velocity or acceleration must be constant. Similarly, the equation d =vt can be used for problems where there is no change in velocity. Am I on the right track?
     
  16. Jul 16, 2014 #15
    Note that the m/s was not my answer. And yes, you are correct. I should have said solve for f
     
  17. Jul 16, 2014 #16

    Nathanael

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    Yes, because the acceleration of gravity is approximately constant near the surface of the Earth.
    (It is also approximately constant in other similar situations where the distances involved are small in comparison with the distance of the gravitation).

    [edit: Also free fall questions that use those equations equation ignore air resistance (they call it negligable). In this situation, though, the forces of the atmosphere are not negligable, on the contrary, they are the basis of this problem.]

    Yes, to use those specific equations requires constant acceleration, because those equations were derived with the assumption that acceleration is constant.

    Similarly, d=vt applies to constant velocities, because that equation was "derived" with the assumption of constant velocity. (I don't know if that equation is technically "derived" because it's just a rearrangement of the definition of velocity)



    BUT, those equations can be used when dealing with averages. This is because the definition of "average" is essentially "the constant that has the same result" (if that doesn't make sense don't worry about it, it's just the way I think of averages).

    For example:
    "Average speed" is the speed that has the same result (distance travelled) as the "actual speed" (which is non-constant)
    In this example, it only makes sense to speak of average speed over a certain time period

    You can also look at averages in terms of totals.
    For example, average speed is the Total Distance over the Total Time


    But at any rate, an average is essentially a constant, and so you can use such equations.


    Side note:
    If something is constant, then the average is the same thing as the "actual"
     
    Last edited: Jul 16, 2014
  18. Jul 16, 2014 #17

    Nathanael

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    Just saying it can tell you what it is. (For example, 2 times 4 would be "Two fours")

    100 times 1,000 is "One hundred thousands"

    But you can also use scientific notation:
    [itex]10^2*10^3=10^5=100,000[/itex]



    You can also use your original method (what I call the "acceleration method"). See my above post for details.
    In fact, I would personally suggest only using the "energy method" if you already understand the "acceleration method" because the "energy method" is essentially just a shortcut that is based off the "acceleration method" (No use using a shortcut if you don't understand how it works.)
     
  19. Jul 16, 2014 #18
    Note: went back and tried:

    V^2 = u^2 + 2ax

    Worked completely fine.
    See I knew something was wrong all along. ;p
     
  20. Jul 16, 2014 #19

    Nathanael

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    Ok let's get back to the problem, starting with your first step.

    You actually can use d=vt but instead of using the initial velocity, you would need to use the average velocity

    So what would be the correct value for the time it takes?
     
  21. Jul 16, 2014 #20
    Well, the problem says to "ignore gravity" which in my interpretation, means that acceleration is constant. I will go back and review the mistakes I made. It looks like I totally messed up an easy conversion from km->m. This thread clarified so much! Thank you everyone!
     
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