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Block sliding to a spring on an incline

  1. Nov 28, 2004 #1
    A 26.0 kg box slides 4.0 m down the frictionless ramp shown in the figure, then collides with a spring whose spring constant is 170 N/m.

    1-- What is the maximum compression of the spring?

    2-- At what compression of the spring does the box have its maximum velocity?


    this is killing me...

    so acceleration is [tex]g sin (theta)[/tex] which is [tex]- 4.9 \frac {m}{s^2}[/tex] and after using kinematics I figure out the velocity that the block is hitting the spring with which is [tex]6.26 \frac {m}{s}[/tex] but then what? :eek:

    If I had a gun to my head and had to write something it'd be this:

    [tex]\sum F = - m g sin (theta) + k x [/tex]

    x here is delta x.
     

    Attached Files:

  2. jcsd
  3. Nov 28, 2004 #2
    Just a guess for number 2

    I would guess that the box wouldn't start to slow down until the spring exerted more force on the box than gravity. So I would think that would begin to occur when,...

    [tex]g \sin (\theta) = kx[/tex]

    Keep in mind that this is just a guess from someone with bloodshot eyes who is on his way to bed. :biggrin:
     
  4. Nov 28, 2004 #3
    The work done by the block on the spring is [tex]\frac{-kd^2}{2}[/tex] where d is the distance that the spring is compressed. Then we have that [tex]K_f - K_i = \frac{-m{v_i}^2}{2}[/tex] where K_f is the final kinetic energy of the block (=0 when the block is stopped) and v_i is the initial velocity of the block (the velocity when the block has travelled the 4 meters and touches the spring.) Via the work-kinetic energy-theorem we know that these quantities need to be equal so you can calculate the distance d because you know m, v_i and k.

    marlon
     
  5. Nov 28, 2004 #4
    Not too sure about your second question though... the block slows down once it has contact with the spring. So basically the max velocity is the initial velocity when the compression of the spring is 0

    marlon
     
  6. Nov 29, 2004 #5
    That was my first instinct too, but then I figured that as long as the force of gravity acting on the box is greater than the force of the spring acting on the box, then the box would still be accelerating down the incline. (i.e. Increasing in velocity)

    In other words, the acceleration of the box would begin to decrease when the box contacts the spring but the velocity would not begin to decrease. In fact, the velocity would continue to increase, It would just increase at a slower rate as the spring compresses until,...

    [tex]g \sin \theta = kx[/tex]

    At that moment the total forces on the box are zero and so the box no longer accelerates at that precise point. That should be the point where the box has maximum velocity shouldn't it?

    Then as the spring continues to be compressed the force of the spring becomes the dominant force to slow the box down. But the box shouldn't actually being slowing down until the force of the spring is greater than the force due to gravity. Prior to that point the box should continue to accelerate down the ramp even after it has made contact with the spring because the downward force is still greater than the upward force.

    In other words, as long as the force on the box due to gravity is greater than the force on the box due to the spring, then the net force on the box is still in the downward direction and therefore the box continues to accelerate downward (albeit at a smaller rate) even after it has made contact with the spring. Therefore, it's still accelerating downward after contacting the spring and its speed is still increasing for a while until the spring can muster the force to offset the force of gravity.

    That's the way I see it.
     
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