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Block going up frictionless incline

  1. Feb 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37 degrees

    2. Relevant equations

    3. The attempt at a solution

    Tried a few different things but can't seem to fully grasp the problem, any help would be wonderful and if possible, an explanation of how you got the answer or came to the conclusion.

    (In all honesty, tell me where to look and why and i'll come up with the answer.)
  2. jcsd
  3. Feb 18, 2007 #2
    Well...where is the question? :P Is it asking for how far the block travels up the incline?

    If so, you can use the principle of conservation of energy (U1 + K1 = U2 + K2) to find how far upward the block will travel. Be careful though--make sure to measure the h in U = mgh as the VERTICAL distance from the ground rather than distance traveled along the incline, they are very different things!
    Last edited: Feb 18, 2007
  4. Feb 18, 2007 #3
    HAHAHA, yah....sorry.:rofl:

    The first part of the question asked how fast was the block moving and I got 3.11 m/s. The second is wondering how far it went up the incline.
  5. Feb 18, 2007 #4
    Your answer to the first part is correct. Now you can choose two different points at which to calculate the energy of the block just like you did for the first part of the problem. Hint: with what speed is the block traveling when it reaches its maximum height?
    Last edited: Feb 18, 2007
  6. Feb 18, 2007 #5
    So find the E first and that'll tell me the distance right? Should I just plug in a distance to reference when the block stops?

    I apologize for sounding reallllly stupid or thick-headed, i've been studying allllll day and am kind of having a brain fart.

    Basically, without telling me the answer, can you possibly show me how to do it?
  7. Feb 18, 2007 #6
    Well, when you're using U1 + K1 = U2 + K2, you're saying the energy when the block moves between these two points 1 and 2 remains constant. Personally, I would choose my point 1 to be when the spring is completely compressed at the bottom of the slope (where its potential energy U1 = (kx^2)/2), and choose my point 2 to be when the object stops at its maximum height (where its v = 0). Can you write the energy equations for this choice now?
  8. Feb 18, 2007 #7
    (Kx^2/2) = mg(.13m) ?

    Using wrong reference in my notes lol.
    Last edited: Feb 18, 2007
  9. Feb 18, 2007 #8
    Wait, you are solving for h--why did you include a value for h in your equation? 0.14 would be the h that the block moves in compressing the spring but that is not the maximum height the block reaches when the block is released!

    The form of your equation is right (kx^2 / 2 = mgh), but let's make sure of which points we're using.

    Point 1 is at the bottom of the incline where the spring is compressed 0.22 m (so it has an elastic potential energy of kx^2 / 2). Here the block is not moving, so its kinetic energy is 0. Let's call this height h=0 as our reference point for gravitational potential energy mgh.

    Now we release the block, so the potential energy of the spring is transferred to the kinetic energy of the block, and the block moves up the incline (increasing its h above the starting point).

    Point 2 is where the block stops moving up the incline, so its v = 0, making kinetic energy mv^2 / 2 = 0. Since it is above the starting point by a height h, it now has gravitational potential energy mgh.

    You know k, x, m, and g--so your only unknown is h!
    Last edited: Feb 18, 2007
  10. Feb 18, 2007 #9
    49.39 is what I got using this calculation:

    U1 + K1 = U2 + K2, but since K1 and K2 = 0, U1 = U2, being that [U1] 1/2kx^2 = [U2] mgh, so; 1/2 (400n/m)(.22m)^2 = 2(9.8m/s^2)h

    968. = 19.6h

    h = 49.39 m
  11. Feb 18, 2007 #10
    It looks like you might have missed a decimal point when plugging it in, maybe on the .22, because your answer is 2 orders of magnitude too large. 1/2 (400)(.22) = 9.68. Fix that little error and you have the right answer for h!

    Now it's just a matter of finding the distance travelled on the incline from the value for h.
  12. Feb 18, 2007 #11
    ohh, i got .49m as the distance. But the HW is saying that's wrong.

    *starting to get frustrated lol*
  13. Feb 18, 2007 #12
    You're in my class. Um, professor Vivek Sharma?

    You're plugging in the wrong height, that's why you're getting .49. I think you're solving for d and plugging that in instead of h right? Check that over again.
  14. Feb 18, 2007 #13
    Yes, we're not done yet! 0.49 is the vertical distance the block has traveled...just like if you travel from San Francisco (elevation 60 ft) to New York (1000 ft) your h above sea level will have increased 940 ft, but you have traveled much more than this! So, you know the angle of the incline and how much the height of has changed. How far along the incline has the block moved?
  15. Feb 18, 2007 #14
    i got h= .13 m

    i got .13m from .22sin 37 degrees

    What do i do from that?

    And yes, you know me? lol, what a small world haha.
  16. Feb 18, 2007 #15
    Are you sure? We have just calculated h that the object rises "above sea level." Pretend this is our incline in the diagram. The height the object has traveled would be A:


    and we want to find the distance of C, knowing the angle a (37 degrees).
    Last edited: Feb 18, 2007
  17. Feb 18, 2007 #16
    .8142m is the answer i got for C

    .49/sin37 is how i got that.
    Last edited: Feb 18, 2007
  18. Feb 18, 2007 #17
    Yup! Is that the right answer?
  19. Feb 18, 2007 #18
    well, if i have your blessing i'll input the answer lol.
  20. Feb 18, 2007 #19

    thank you sooooo much gabee. HUGE help. it actually makes sense now too lol.

    thank you again.
  21. Feb 18, 2007 #20
    No problem, glad I could help. :)
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