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Help on momentum and energy

  1. Jun 1, 2006 #1
    Hi guys, in need again for some help...thanks to everyone who helped before on my previous thread on potential energy i understand it a little better now. again im stuck on three problems if you guys could provide me some insight i'd greatly appreciate it.

    1) Sliding in socks. Suppose that the coefficient of friction between your feet and the floor, while wearing socks, is 0.250. Knowing this, you decide to get a running start and then slide across the floor. If your speed is 3.00m/s when you start to slide, what distance will you slide before stopping?

    Ok i have no idea where to even begin with this one but i know it has to be simple but im no seeing it? where can i begin?

    2) The Great Sandini is a circus performer with mass 60.0kg who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1400N/m that he will compress with a force of 4900N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 32.0N during the distance of 4.30m that he moves in the barrel.

    Ok on this one i began by using hooke's law and getting the distance the spring would be compressed so 4900N/(1400N/n)=3.50m
    Now i jumped into Ki+Ui+Wother=Kf+Uf
    (1/2)60(0)^2+(1/2)1400(3.50)^2+32*4.30=(1/2)*60*vf^2+60(9.8)(2.9)
    the vf i got was 15.3 but its comming out wrong. Anyone see where im goin' wrong here?

    3) A projectile of mass 20.5kg is fired at an angle of 58.0 degrees above the horizontal and with a speed of 85.0m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. How far from the point of firing does the other fragment strike if the terrain is level?

    Ok so first i broke down the velocity into x and y components:
    Vx=85cos(58)=45.0m/s and Vy=85sin58=72.1m/s
    then i calculated the time it took to reach the highest point using Vf=Vi+at where i used 0=-9.80t+72.1, t=7.36s. I calculated the highest point to be at 265m and the x distance to be 331m using x=45(7.36)
    Now i went to momentum. I found Vx to be 85m/s and using that i did the following: 85*7.36+331= 957m. I entered this answer and it says im close but not there yet...what am i doing wrong?

    thanks for any help in advance
     
  2. jcsd
  3. Jun 1, 2006 #2
    k i figured out the 3rd one...any word on the first two?
     
  4. Jun 2, 2006 #3

    Hootenanny

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    For question one, think conservation of energy. Work done by the frictional force must equal the intial kinetic energy.

    These look to me like homework questions.

    ~H
     
  5. Jun 2, 2006 #4

    vanesch

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    (moved to homework).


    I'm not sure that an energetic approach is the best technique when dealing with non-conservative forces (friction).

    For 1) I'd say that probably the best thing to do is just to draw a force diagram (weight of the person + reaction of the floor in balance, and the friction force).

    Given that the friction coefficient is given, this gives a relationship between the weight and the friction force. Now, a simple application of Newton's second law should do (take the mass as an unknown and see if you really need it in the end).

    Again, for 2) I'd not go through an energetic approach, because there is this small friction force. I think with all the elements given and Hooke's law, that you can determine the force at every point along the trajectory. Application of Newton's second law should then do.
    You have several pieces: first, the piece where the spring acts (over 3.5 meter if your calc is right). I guess (it is not clear from the problem) that during these 3.5 meters, the friction force is also at work.
    Next, there is the motion only in the barrel, but without the spring. It is not 100% clear to me whether this is still 4.3 meters, or whether we should consider that the 3.5 meters of the spring are contained in it.

    cheers,
    Patrick.
     
  6. Jun 2, 2006 #5
    How would i apply F=ma to this? I drew the free body diagram but i dont see it yet....how will Newtons second law help me get distance?
     
  7. Jun 2, 2006 #6

    Hootenanny

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    In my personal opinion consiering energy would be the way to go here.

    ~H
     
  8. Jun 2, 2006 #7
    ok i tried the energy aproach where like you said the work done by frictional force equal the intial kinetic energy so what i did looked soemthing liek this:
    uk*n*d=(1/2)mv^2
    d=v^2/(uk*g)
    d= 3.00^2/(0.250*9.80)=3.67m

    Its still not the right answer...did i use your approach wrong?

    nvm i missed the 2 at the denominator i got the answer thanks .... i knew it was something simple...its the simple ones i always miss tho grrrr
     
    Last edited: Jun 2, 2006
  9. Jun 2, 2006 #8

    Hootenanny

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    Its surprising how often one does something stupid like that:rolleyes:

    ~H
     
  10. Jun 2, 2006 #9
    lol yeaaaaaah....ok the same question has a second part to it and im trying to apply energy to it also. it goes:
    Now, suppose that your young cousin sees you sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of friction as yours). Instead of getting a running start, she asks you to give her a push. So, you push her with a force of 125N over a distance of 1.00m. If her mass is 20.0kg, what distance d does she slide (i.e., how far does she move after the push ends)? Remember that the friction force is acting anytime that she is moving.

    what i did was: Wtot=(1/2)mvf^2-(1/2)mvi^2
    and i substituted with: 125d-49d=-(1/2)mv^2
    mad-ukmgd=-(1/2)mv^2
    md(a-ukg)=-(1/2)mv^2
    d=-v^2/(2m(a-ukg))

    Im stuck though because i dont have v .... am i approaching this all wrong?
     
  11. Jun 2, 2006 #10

    Hootenanny

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    Work done = force [itex]\times[/itex] distance moved.

    ~H
     
  12. Jun 2, 2006 #11
    ok W=F*d, but what force would go where and on the left woulndt i need d there too, but if its on both sides then d cancels...im goin in circles....
     
  13. Jun 2, 2006 #12

    Hootenanny

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    Something like this perhaps;

    Work Done by you - Work done by you against friction = Work done by friction against your friend.

    [tex]125-\mu mg = \mu mg d[/tex]

    Alternitively one could use newton's second law and kinematic equations to determine the distance travelled. However, considering the first part of the question used conservation of energy, then I would think it is more in the spirit of the question to use energetic means.

    ~H
     
    Last edited: Jun 2, 2006
  14. Jun 2, 2006 #13
    oh that makes sense :rolleyes: why didnt i think fo that....as for the great sandini question i was jsut off by a sig fig...it shoulda been 4 and i had 3......stupid mistakes are killin me in physics... thanks for all your help...understanding in progress...
     
  15. Jun 2, 2006 #14

    Hootenanny

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    No problem, and remember its not about getting the right answer; it's knowing how to get there. :smile:

    ~H
     
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