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Mass falling on a vertical spring

  1. Mar 6, 2009 #1
    A 195 g block is dropped onto a relaxed vertical spring that has a spring constant of 1.6 N/cm. The block becomes attached to the spring and compresses the spring 14 cm before momentarily stopping.

    (a) While the spring is being compressed, what work is done on the block by the gravitational force?

    .26754 joules

    (b) While the spring is being compressed, what work is done on the block by the spring force?

    -1.568 joules

    (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

    3.65212969 m/s

    (d) If the speed at impact is doubled, what is the maximum compression of the spring?

    This is the only part that I can not get
     
  2. jcsd
  3. Mar 6, 2009 #2
    You can count the speed of the block at impact. You double it, which essentially gives a new value for its kinetic energy. This kinetic energy (and some potential energy as the spring is compressed, the block comes closer to the ground, i.e. it loses potential energy) is then transformed into potential energy in the spring. From that you can get the maximal compression.

    Potential energy in spring=0.5kx^2

    where k is the spring constant and x is the compression. Maybe that helps a little bit?
     
  4. Mar 6, 2009 #3

    minger

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    I'm you didn't understand Ofey, its pretty simple really. You know that energy cannot be created or destroyed, only transferred. When a mass is dropped onto a spring, potential energy is converted to kinetic.
    [tex] (mgz)_1 = \left(\frac{1}{2}m v^2\right)_2[/tex]
    At the bottom of the fall, the kinetic energy is converted into potential energy in the spring.
    [tex] \left(\frac{1}{2}m v^2\right)_1 = \left( \frac{1}{2}k z^2\right)_{spring}[/tex]
    Keeping the z notation. The question is if V = 2V, find z.

    *note for simplicity I have neglected potential energy conversion during the spring compression as Ofey mentioned. For complete accuracy, you should take into account (i.e. however this might be an iterative process since you don't know the compression beforehand).
     
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