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Potential Spring Energy

  1. Sep 30, 2007 #1
    A 10kg mass, attached by means of two springs to the ceiling, is held against the floor and is then released. How fast will it be traveling when it hits the ceiling? The spring constant of each spring is 80 N/m, and each spring has an unstretched length of 1 m. Assume that springs become loose and floppy once they're at their rest length.

    distance between ceiling and floor: h_o=4m
    horizontal distance from mass to each spring: 2m

    Springs are at an angle from mass.

    I used pythagorean theorem to find stretched length of spring:
    s=sqrt(4^2+2^2)=sqrt(20)

    I found theta to find h after the mass is released:
    tan(theta)=4/2
    theta=63.43494

    I found h using trig:
    h=unstretched length of string*sin(theta)=.8944

    The answer is v=9.92m/s

    I used this equation to solve for v with my datum at the floor:

    PE(sp)=KE+PE(g)
    1/2ks^2=1/2mv^2+mgh
    1/2(80)(sqrt(20))^2=1/2(10)v^2+(10)(9.81)(.89)
    v=11.93m/s
     
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2

    learningphysics

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    Why did you use 0.89 instead of 4m, for mgh?

    Also you need to use 2*((1/2)kx^2) at the bottom instead of just (1/2)kx^2, since there are two springs...

    also, the stretched amount of each spring is sqrt(20) - 1, so

    initial energy =

    2*(1/2)*80*(sqrt(20)-1)^2

    also, for final energy, there's still energy stored in the springs... the horizontal distance is 2m, hence each spring is stretched by 1m.
     
  4. Oct 1, 2007 #3
    Thank you! i got the answer. i just want to know for final energy, the 1 m the springs are still stretched is that from the unstretched length or an additional stretch?
     
  5. Oct 1, 2007 #4

    learningphysics

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    additional stretch... total of 2m... 1m unstretched, 1m stretched.
     
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