Recent content by sanctifyd83

  1. S

    Spring constant based on change in potential energy

    Thanks so much everyone. It helped out a lot.
  2. S

    Spring constant based on change in potential energy

    That absolutely makes sense, I couldn't quite grasp that. It seemed like I was just arbitrarily inputting units. Thanks so much.
  3. S

    Spring constant based on change in potential energy

    Forgive me as I'm trying to understand the material but I think that x1[/SUB's units would be in cm? If this is the case -5.5 cm turns into -.055 m which when squared becomes .003 m2. 276J/.003 m2 = 92000N/m. How do you know what the variable's initial units are? I assumed we were working with...
  4. S

    Spring constant based on change in potential energy

    Thank you for that. Yes, the value ought to be negative. This gives the value for x1 slightly different negative values which, when squared, become positive. The result is still an extremely large value for k. x1 would equal -5.5 or -.7 Using -5.5 k would be around 90000N/m and -.7 would...
  5. S

    Spring constant based on change in potential energy

    Thank you, rude man. Yes I input the middle term incorrectly. It should be 2.5 m/x1. I did it correctly when I worked it out though, I don't believe it affects the rest of my "solution."
  6. S

    Spring constant based on change in potential energy

    Homework Statement A compressed spring has 138 J of energy stored in it. When it is decompressed by 125 cm, it now stores 82.6 J of energy. What is the spring constant Homework Equations General: ΔU = 1/2kxf2 - 1/2kxi2 U(x) = 1/2kx2 U1 = 1/2kx12 U2 = 1/2k(x1 + 1.25 m)2 The...
Back
Top