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The horizontal distance the train will move before its speed reduces?

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data

    A 60,000 kg train is being pulled up a 1% gradient by an engine exerting 3 kN. The frictional force opposing the motion of the train is 4 kN. The train’s initial speed is 12 m/s. Through what horizontal distance will the train move before it’s speed is reduced to 9 m/s?

    The answer is 275m

    2. Relevant equations

    Fpush/pull=mgsinθ (component along the ramp)+Ff (friction force)

    I know that the equation above is used to solve the problem but i do not understand it or know how to use it

    3. The attempt at a solution

    s=ΔE/F

    Vf=1/2mv^2=2430000
    Vi=1/2mv^2=4320000

    ΔE=-1890000

    then i did not know how to work out the force to put it in the s=ΔE/F formula to work out the horizontal distance.

    But i know my attempt is completely wrong, my major problem with friction problems is identifying which forces are acting on the object and the types of forces and then finding the right formula to work out the answer.

    Any help very much appreciated, thank you.
     
  2. jcsd
  3. Nov 2, 2013 #2

    tiny-tim

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    hi hbk69! :smile:

    you need to find the total work done, including by gravity (per distance)

    1% gradient means tan = .01, but it's so small i think you're entitled to assume sin = .01 also :wink:
     
  4. Nov 2, 2013 #3
    Let d be the distance that the train travels before its speed is reduced to 9 m/s. What is the net force acting on the train in the direction down the grade during the time it is traveling that distance. In terms of d, how much work do these forces do on the train while it is traveling this distance? What is the change in kinetic energy of the train? How is the change in kinetic energy related to the work done by the external forces?
     
  5. Nov 2, 2013 #4
    Hi! thanks for the response. Workdone=F*s, and the s is the horizontal distance which i am looking for? but how can i work out F?
     
  6. Nov 2, 2013 #5
    thanks for the reply, how can i determine the net force down the gradient? which i think would help me work out the work done?

    And this was the change in kinetic energy i calculated:

    Vf=1/2mv^2=2430000
    Vi=1/2mv^2=4320000

    ΔE=-1890000

    i am not sure how it relates to the work done or whether i worked it out correctly, as i don't know how to determine the external forces/net force
     
  7. Nov 2, 2013 #6

    tiny-tim

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    hi hbk69! :smile:
    not really

    work done = F.s

    s is the displacement

    it's a vector, and (in this case) it's along the slope,

    ie it's one unit up for every 100 units horizontally​

    F is the total force (from the engine, the friction, and the weight)
    F is always the total force (the net force).

    along the slope ("horizontally"), it's 3000 N forwards, and 4000 N backwards

    vertically, it's mg
     
  8. Nov 2, 2013 #7
    Do a free body diagram on the engine and draw vectors for all the forces acting on the train, in the directions both parallel and perpendicular to the grade. Tiny Tim pretty much laid it out for you, but forgot to mention the component of the weight 0.01 mg acting parallel to the grade.

    Recall that the work done on the train by the external forces is equal to its change in kinetic energy. This gives you an equation for calculating s.

    Chet
     
  9. Nov 2, 2013 #8
    How can the horizontal distance be found mathematically also how could i find the net force mathematically?

    If 3000 is forward and 4000 is backwards, would the net force be 4000*mg(vertical component) - 3000*mg
     
  10. Nov 2, 2013 #9

    tiny-tim

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    nope, you haven't got this at all

    let's go over it slowly …

    (let's assume for the moment that the train goes a displacement of s = 1 metre forwards along the slope)

    there's three forces, and they're all vectors

    it's convenient to deal with the weight separately from the other two

    the weight of course is Fg = 60,000g downwards

    what is the work done (Fg·s) ?​

    the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

    what is the work done by those forces?​
     
  11. Nov 2, 2013 #10
    Tiny Tim,

    It seems you again forgot to include the component of the train weight parallel to the grade:
    mg(0.01)=6000 N backwards.

    Chet
     
  12. Nov 2, 2013 #11
    Sorry, i have a very very weak understanding of physics

    So in this case would i add the Fnet to the Fg downwards? when s=1m, W= (1000+60,000) *1 to get the answer?
     
  13. Nov 2, 2013 #12

    tiny-tim

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    no, you can't add vectors like that

    you can't add 60,000 in one direction to 1000 in a different direction to get 61,000, can you? :wink:

    you have to add them as vectors, ie either use a vector triangle, or use components (eg x and y), and add the components, separately

    in this case, adding the vectors (as vectors) is a bit awkward (since they're not exactly perpendicular), so i recommended that instead of adding the vectors (as vectors), and then "dotting" with s, you "dot" each vector with s first, and then add the results (which of course are scalars)

    ok, now answer my previous questions (assuming for the moment that s = 1 m) …

    the weight of course is Fg = 60,000g downwards

    what is the work done (Fg·s) ?​

    the other two are vectors along the slope: 3000 N forwards and 4000 N backwards, making a total vector of 1000 N backwards along the slope

    what is the work done by those forces?​
    no, this says it all …
     
  14. Nov 2, 2013 #13
    lol sorry still confused, not sure how i can add the vectors in that case because as they are not perpendicular i can't use pythogras theorem either where i square both of them add the result and take the square root to get Fnet, and what do you mean by "dot each vector with s" don't know how to to it.

    and answering that question, you said s=1m, Fg=60,000 so the Work=Fg*s= 60,000J? it's wrong again isn't it.
     
  15. Nov 2, 2013 #14

    tiny-tim

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    no, is said that the displacement vector s is 1 m along the slope

    since you don't know how to do a dot product, yes it's wrong :redface:
    for dot product, see eg http://en.wikipedia.org/wiki/Dot_product

    i don't see how you can understand work done if you don't understand the dot product

    do you have a ta or a tutor or a study-buddy who can help you with this?

    if not, i think you ought to ask your professor to go over it with you
    you have to add the x and y components of the vectors (separately)

    have you not done addition of vectors?
     
  16. Nov 2, 2013 #15
    I have done it in maths but don't know how to apply it to physics problems takes a bit longer before i understand the maths side of things, don't have a tutor re-taking a physics module without attendance. Thanks for trying to help me
     
  17. Nov 2, 2013 #16

    tiny-tim

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    ok, let's see …

    can you add the vectors (4,6,0) and (1,1,3) ?
     
  18. Nov 2, 2013 #17
    (5,7,3)
     
  19. Nov 2, 2013 #18

    tiny-tim

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    that's right :smile:

    now when you want to add two vectors, you split each vector into its x y and z components, and then you add the components (exactly like that example)

    that's called the component method for adding vectors (you can also use the vector triangle method)

    ok, now let's do dot product

    (dot product is the same as inner product, but most people call it dot product because you write it with a dot, unlike the cross product!)

    can you do the dot product of the vectors (4,5,6) and (2,3,4) ? :wink:
     
  20. Nov 2, 2013 #19
    but sometimes you may have to substract them and involve angles etc and that is where it gets really complicated for me with friction problems involving inclined planes and finding the right formulas to solve the problem when it come to friction

    oohhh is that what the dot product its like one of those where you know someone by face but not their name, i think i can do that but applying it to physics is the problem

    (4)(2),(5)(3),(6)(4)

    (8,15,24)
     
  21. Nov 2, 2013 #20

    haruspex

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    Almost... but the result is not a vector. You add up the products to get a scalar: 8+15+24.
     
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