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Work and Energy Conservation, Need Help Please!

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

    The spring in the figure has a spring constant of 1000 N/m. It is compressed 13.0 cm, then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.220. What distance does the block sail through the air?

    knight_Figure_11_72.jpg

    2. Relevant equations
    .5(k)(x^2)
    .5(m)(V^2)
    Wnc=Ef-Ei

    3. The attempt at a solution
    I think I'm pretty close to solving this problem but I'm messing up somewhere...

    I know I need to find the final velocity of the mass right when it gets to the top of the incline so I'm gonna use Wnc=Ef-Ei. (To find Ei I used potential energy of a spring formula .5(k)(x^2).

    So...
    -μk(cosθ)(m)(g)(L)=.5(m)(Vf^2)+(m)(g)(h)-.5(k)(x^2)

    The distance the mass traveled up the incline is L so L=2/sin45

    -.22(cos45)(.2)(9.8)(2/sin45)=.5(.2)(Vf^2)+.2(9.8)(2)-.5(1000)(.13^2)

    I get that Vf=1.38m/s

    To find the distance I simply use R=Vf^2/g and get .19 but it's wrong? Can someone help me out!?!?!
     
  2. jcsd
  3. Jun 8, 2012 #2
    I ended up using all my attempt on mastering physics and the answer ended up being 3.74m but I still dunno how to come up with that value?
     
  4. Jun 8, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I assume you can calculate the initial energy given to the block by the spring. The kinetic energy at the top of the incline will be that minus
    (1) the potential energy at the top of the incline.
    (2) the work done against friction on the incline.
     
  5. Jun 9, 2012 #4
    I found this interesting as I'll be working on this soon in class, so I tried it myself. HallsofIvy's method worked perfectly, thanks! Answer came out to 3.738 m.
     
    Last edited: Jun 9, 2012
  6. Jun 9, 2012 #5
    By the way, it also works to use linear motion formulas, though it takes a tiny bit longer. You can get the initial velocity (v0) at the bottom of the ramp from the initial KE, then (I'm using a coordinate system of +x up the ramp), find the x components of acceleration due to kinetic friction and gravity. Integrate this acceleration and use the v0 in the velocity function, then integrate again leaving s0 = 0. Now you have linear motion functions for the ramp portion. You know the ramp is 2sqrt(2) m long, so set the position function s(t) = 2sqrt(2) and solve for t, then plug that into the velocity function to get the initial velocity at the top of the ramp for your projectile motion problem.

    I screwed this up the first several times I tried to calculate it by trying to take the velocity coming off the spring, and multiplying it by cos45 to get the initial velocity for the ramp's linear motion formulas. That was wrong, the velocity doesn't change magnitude, just direction.
     
  7. Jun 9, 2012 #6
    ]So...
    -μk(cosθ)(m)(g)(L)=.5(m)(Vf^2)+(m)(g)(h)-.5(k)(x^2)

    The distance the mass traveled up the incline is L so L=2/sin45

    -.22(cos45)(.2)(9.8)(2/sin45)=.5(.2)(Vf^2)+.2(9.8)(2)-.5(1000)(.13^2)

    I get that Vf=1.38m/s

    To find the distance I simply use R=Vf^2/g and get .19 but it's wrong? Can someone help me out!?!?!

    ...........................
    Everything ok
    Check your calculation.

    v2f=36.67
    d=36.67/9.8=3.7m
     
    Last edited: Jun 9, 2012
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