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Block down ramp colliding with spring

  1. Oct 11, 2004 #1
    A 3.2-kg block is released 34m from a massless spring with force constant 178N/m that is fixed along a frictionless plane inclined at an angle of 27° above the horizontal. Find the maximum compression of the spring.

    This is what I have tried so far.
    equation (1) (1/2 mv^2) = (1/2 kx^2)

    equation (2) (1/2 mv^2)=d*mg(sin theta)

    i used equation (2) to solve for v.
    then using that value of v i solved (1)

    my answer for x=2.333 meters, however that is wrong
     
  2. jcsd
  3. Oct 11, 2004 #2

    Doc Al

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    Staff: Mentor

    In calculating the gravitational PE, don't forget to include the distance the block falls in compressing the spring a distance X.
     
  4. Oct 11, 2004 #3
    ok, i think i understand the concept of that, but I'm not sure how I would set that up in equation form. Because gravity works in the y-component, would i have to use 2 equations. 1 to include for the y-distance the spring compresses and another to solve for the total compression in the sin (theta) component of the spring? If that was the case then I dont know how to set those up because i am trying to solve for maximum compression and in the first equation i woul dneed to know that.
     
  5. Oct 11, 2004 #4

    Doc Al

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    Staff: Mentor

    Think of it this way: How far does the block slide? d = 34m + x. (x = how far the spring compresses.) Use energy conservation: Intial energy (gravitational PE) = final energy (spring PE). You'll get a quadratic equation.

    Hint: measure gravitational PE from the lowest point.
     
  6. Oct 11, 2004 #5
    i set up the problem like this
    U=mg(h+x sin (theta))
    Uspring=1/2 k x^2

    so mg(h+x sin (theta)) = 1/2 kx^2?

    solving for x, x1=1.1331, x2=-1.0965
    those are both wrong.
     
  7. Oct 11, 2004 #6

    Doc Al

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    Staff: Mentor

    Ugrav = mgd sin(theta), where d = 34m +x
     
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