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Mass on spring

  1. Jul 22, 2005 #1
    I'm half way through this problem, but I'm stuck at the end.

    A block of mass m = 4.5 kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = 5 m. The spring has spring constant k = 11 N/m and is relaxed when hanging in the vertical position. The block is pulled d = 4 m to one side. In this problem, the block is always constrained to move on the floor (i.e. it never leaves the floor).

    I found the following:

    change in L: 1.4 m (length the spring is extended)
    Potential energy stored in spring: 10.78 J
    velocity max it attains: 2.19 m/s

    Now suppose this: When the spring is vertical (hence, unstretched), the block is given an initial speed equal to 1.3 times the max speed it attains. thus, initial speed is 1.3 x 2.19 = 2.85 m/s

    Find: How far from the initial (unstretched) point does the block go along the floor before stopping?

    Attached Files:

  2. jcsd
  3. Jul 22, 2005 #2


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    Use the same conservation of energy principles you have been using to work you way back from the maximum velocity. Find the initial kinetic energy, which gets converted to potential energy of the spring, to find the maximum spring extension. That will give you the hypotenuse of the triangle. You know the triangle height. Find the base leg.
  4. Jul 22, 2005 #3
    Can you please tell if the block was allowed to let go from left-side position or Was the block at rest at the centre when it was given the velocity of 1.3 times the max?

    Though the application process remains the same, whatever velocity the black has/gets when it is the the centre position (where the spring is unstretched) , this K.E the block gets converted into the potential energy of the spring.So there comes a point when the block stops at the extreme right. From energy-conservation you can get the length of the spring when it is at the extremem right , when the block stops for an instant before going back. Knowing the length of spring in extreme right stretched position , and you do know the relaxed length of the spring at the centre, Now Pytha.Theorem is all you need!

    Last edited: Jul 22, 2005
  5. Jul 22, 2005 #4
    ok, i got that part down, its distance max is 4.64 m,

    but here's another part of the problem:
    what is the magnitude of the acceleration of the block at this point? (when spring is stretched farthest?)

    what I did was I tried to identify the force exerted by the spring, find the x comp of that force, then apply F=ma and solve for a. however, I seem to be doing something wrong when I tried to find the force of the spring. any hints?
  6. Jul 22, 2005 #5
    I think you managed correctly. The force of the spring F=-k*dL, where dL is change in L.
  7. Jul 22, 2005 #6
    You seem to be doing right. You just need to get the extension in spring when at extreme position , let that extension be x.

    x= Length of the spring in extended position - Length of the spring when unstretched.

    F=kx is the force on the block along the spring , so you basically need to take the component of this force F in negative x-direction , then equate it with ma, 'a' is the required acceleration at that particular instant.

    I think you are doing mistake in calculation of 'x'.Check it agains as I told above.


    For more practise , try to find acceleration of the block as a function of time as the spring goes from the extreme right stretched position to central unstretched position.

  8. Jul 22, 2005 #7
    ok, i got it. i found F, and i realized that i have to take the angle at which the spring was stretched into account. Thanks for the help!
    Last edited: Jul 22, 2005
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