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Block on Inclination

  1. Feb 1, 2008 #1
    [SOLVED] Block on Inclination

    1. The problem statement, all variables and given/known data
    Suppose the incline is frictionless. The angle of inclination is 22 degree, the spring constant is 275N/m and the mass of the block is 5.4 kg. The block is released from rest with the spring initially unstretched.
    a. How far x does it move down the incline before coming to rest?
    b. Chooing down the incline as the positive direction, waht is its acceleration at its lowerst point?
    [​IMG]
    This is the picture of it.

    I've tried:
    a. I find the weight of the block by mg=5.4*9.8. Then I find the force of spring by (mg)sin22 degree. and find x from there. But i don't think it was right though.
    b. I have no idea! Give me hint please!!
    Thanks you yall!!!
     
  2. jcsd
  3. Feb 2, 2008 #2

    malawi_glenn

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    I dont know if it is wrong with me, the forums description of this subforum, or the posters. But many many problems posted here are introductory mechanics problems (like this one).

    "Upper level undergraduate physics" it says, how shall one interprent that?

    Since so many of the posters dont know at what level they are stuying, maybe we should just have one HW forum for physics?
     
    Last edited: Feb 2, 2008
  4. Feb 2, 2008 #3
    If it is. Can you please move it to introductary mechanic subforum then?
     
  5. Feb 2, 2008 #4
    Guyss...help me w/ this one, please!
     
  6. Feb 2, 2008 #5

    malawi_glenn

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    Hints: draw force arrows.

    and: Use [tex] \sum \vec{F} = ma [/tex]

    Now find all forces, as a function of x

    How would the forces on the block be if it didnt was attached to the spring?

    How does the force from the spring act?

    Then add the forces, for what x is the sum of the forces zero?

    and for Q# b) you need the height, to find out the lowet point I guess.
     
  7. Feb 2, 2008 #6

    Dick

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    You are probably better off posting in the Introductory Physics section, you'd get more patient replies. Compare this to your bedspring problem. It's almost exactly the same. When the spring is at it's maximum extension the change in the gravitational potential energy of the block is equal to the change of the potential energy of the spring. Now that you have that position look at the forces acting on the block. There is a gravitational force down the incline and a tension force exerted by the spring pulling up. They aren't equal. Use the difference to compute the acceleration.
     
  8. Feb 2, 2008 #7
    but to set up the equation mgh=(1/2)kx^2, i need to know h, it ain't give h in the question.
     
  9. Feb 2, 2008 #8

    malawi_glenn

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    how is h related to x ? use basic trigonometry.

    you can always set mgh = 0 at the top, and then when x is increasing the block gets lower pot E by putting a minus sign in front:

    -mgh
     
    Last edited: Feb 2, 2008
  10. Feb 2, 2008 #9
    is h=xsin22 ?
     
  11. Feb 2, 2008 #10

    malawi_glenn

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    that is correct.
     
  12. Feb 2, 2008 #11
    YESS a is solved. So on b, use Fnet=Fspring - mgsin22 = ma. Find acceleration from that? But since they ask for acceleration at its lowest point. So i don't think it right though
     
    Last edited: Feb 2, 2008
  13. Feb 2, 2008 #12

    Dick

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    If the spring is at maximum extension then the block is at it's lowest point. That's the same point where the block is stationary and there is no kinetic energy. I.e. the value of x you should have just computed.
     
  14. Feb 2, 2008 #13
    I found fs=-39.6
    so Fnet=mgsin22-(-39.6)=59.424. Find acceleration from ma equation. But it doesnt make sence, cuz the block is moving down, so the spring force must be smaller than the gravitational force acting on it.
     
  15. Feb 2, 2008 #14
    What is I.e. ?
     
  16. Feb 2, 2008 #15

    malawi_glenn

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  17. Feb 2, 2008 #16

    malawi_glenn

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    The force of gravity and the spring force is acting in opposite directions.

    Fnet=mgsin22-(-39.6)=59.424

    Seems strange.
     
  18. Feb 2, 2008 #17
    spring forces keep it up and gravity force move it down. So it is in opposite directition. Isn't it?
     
  19. Feb 2, 2008 #18

    malawi_glenn

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    Then why did you put a minus sign between?

    Let positive x direction be to the left.

    Then F_spring = -kx[itex] \hat{x} [/tex]
    F_grav = mgsin22 [itex] \hat{x} [/tex]

    F_net = (mg sin22 + (-kx))[itex] \hat{x} [/tex]

    so at the lowest point , x as you found in a), find the acceleration of the block just before it stops.
     
  20. Feb 2, 2008 #19
    [itex] \hat{x} [/itex]
    what is that mean
     
  21. Feb 2, 2008 #20

    malawi_glenn

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    nope, it is x-hat: i.e the basis vector for the x-direction. Forces are vectors.
     
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