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Hooke's Law problem

  1. Jan 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A mass of 1.0kg is attached to a spring obeying Hooke's Law F = k.s, where F is the force applied and s the spring extension. The spring constant, k is 50 N/m. The spring and the object lie on a surface tilted 45 degrees with respect to the vertical Neglect friction and answer the following questions:

    a. What is the extension of the spring?

    b. What is the work done by gravity in extending the spring by the above extension?

    2. Relevant equations

    F = k.s

    W = f.d

    3. The attempt at a solution


    1kg = 9.8N (force)

    Then resolve force into vert and horiz components= Tan45 x 9.8 = 9.8 (as tan45 is 1).

    Rearrange F = k.s into F/k = s

    9.8/50 = 0.196m = s


    Work done by gravity:

    9.8N x 0.196 = 1.92 Joules

    Does anyone know if this is right?

  2. jcsd
  3. Jan 5, 2010 #2
    This is wrong. Draw the right triangle representing the force and its components; the hypotenuse is the weight force (directed straight downward), so the components must be less than that.
  4. Jan 5, 2010 #3


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    Homework Helper

    Where did you get tan45 from? The parallel component of the weight is 9.8*sin(45).
    No, because W=F*d only works if both F and d remain constant. F certainly does, but not d. Use the formula for potential energy instead: U=(1/2)kx^2. That's equal to the work done by gravity.
  5. Jan 5, 2010 #4
    Ahh yes that makes a lot more sense! Thanks for the help :)

    So I now make that:

    a) 0.139m

    b) 0.483 J
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