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Physics problem about springs PLEASE HELP!

  1. Jan 29, 2007 #1
    A 2 kg block hangs from a spring. A 253g body hung below the block stretches it a further 2.5cm. What's the spring constant? Use g=9.8N/kg. Calculate the value to 1 decimal place and use N/m as your unit.

    A 1.8kg block hangs from a spring. A 221g body hung below the block stretches it a further 3.8cm. If the 221g body is removed and the block set into oscillation, what is the period of the motion? Use g=9.8N/kg. Calculate the value to 2 decimal place and use second as your unit.
  2. jcsd
  3. Jan 29, 2007 #2
    In order to get help, you should post an attempted solution.

    Here is a hint: Write the force balance equation for a) just the first block and b) both blocks together. How can you then use these two equations to obtain an expression for k in terms of the known values?

    for the second part, if I were to do it, I would set up a force balance equation (left side in terms of displacement x from the rest postion, right side in terms of acceleration = d^2x/dt^2) with initial conditons dx/dt(0)=0 and x(0)=-x1. This could then be solved, and the period of the solution can be obtained.
  4. Jan 29, 2007 #3
    What a know/think!

    number 1??
    I THINK.. it is
    we know F=KX & W=MG
    since F=W, KX=MG
    so K*d=2kg*G also K*(d+2.5cm)=2.253kg*G
    so K=(2kg*G)/d also K=(2.253kg*G)/(d+2.5cm)
    then (2kg*G)/d = (2.253kg*G)/(d+2.5cm)
    we solve for d: (d+2.5cm)/d= (2.253kg*G)/(2kg*G)
    G cancels out: (d+2.5cm)/d= 2.253kg/2kg
    so 1+(2.5cm/d)= 2.253/2
    2.5cm/d= 2.253/2-1
    d= 2.5cm/(2.253/2-1)=
    now that you have d and d+2.5cm
    you can plug into equation F=KX or K=F/X which is K=MG/X
    use K=MG/X
    K=2kgG/d you know G and d so you have K which is spring constant.

    Number 2..
    NOT A CLUE..
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