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Problem finding Spring Constant

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data
    In a physics lab experiment, a compressed spring launches a 31g metal ball at a 25 degree angle. Compressing the spring 19cm causes the ball to hit the floor 2.0m below the point at which it leaves the spring after traveling 5.2m horizontally.
    What is the spring constant?

    2. Relevant equations
    EPE=1/2*k*x^2
    KE=1/2*m*v^2
    PE=m*g*h
    Kinematics?


    3. The attempt at a solution
    I know that the initial energy of the system is EPE+PE, and the only unknown is k, the spring constant, but I'm not sure where to go from there. I figured I could find the KE when it hits the ground and use the law of conservation of energy to equate this to my initial energy and solve for k but all attempts at finding that velocity failed.
     
  2. jcsd
  3. Nov 26, 2009 #2
    Start by arbitrarily assigning the initial velocity of the metal ball as v upon leaving the spring. Then, you are able to resolve the motion of the ball in the vertical and horizontal directions. Since you know the horizontal and vertical displacements of the ball, and the fact that the time taken in both directions is obviously the same, you can solve them simultaneously for v. (Yup, kinematics) From there, you can get the kinetic energy of the ball upon release and consequently the spring constant k.
     
  4. Nov 26, 2009 #3
    Okay so i set EPE=KE right?
    Then I solve for k.
    k=mv^2/x^2.

    Then if I say Ydistance=Vsin(theta)t+1/2(g)t^2
    and Xdistance=Vcos(theta)t
    and then solve both for t and set them equal.
    Then when I solve for V^2 I get:

    (Xdistance^2*g^2) / (2*Ydistance*g*cos^2(theta)-2*Xdistance*g*sin(theta)*cos(theta))

    then I plug this into my equation for k and i get the value 31N/m, which is not correct.

    What am I doing wrong?

    Thanks
     
  5. Nov 27, 2009 #4
    Well, I need to know how the set-up looks like. "at a 25 degree angle" with respect to?
    I'm surprised that you can even get an answer: the part of the expression (2*Ydistance*g*cos^2(theta)-2*Xdistance*g*sin(theta)*cos(theta))
    = 2*2*9.81*cos^2 (25 deg) - 2*5.2*9.81*sin (25 deg) * cos (25 deg)
    = - 6.85 is negative!
    v^2 surely cannot be negative -> v is imaginary otherwise.
     
  6. Nov 27, 2009 #5
    Assuming the ball is launched 25 degrees upward from horizontal, the equation for y motion will be
    y=vo*t*sin(theta)-1/2gt^2 (y is positive upwards) or withe the signs reversed (if y is positive
    downwards). The two terms do not have the same sign anyway.
     
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