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Hookes Law, Finding A Springs Constant

  1. Jan 5, 2014 #1
    A spring extends by 10cm when a mass of 100g is attached to it. What is the spring constant? (calculate your answer in N/m)

    I have done this so far but I don't feel that it is right as the Force (F) is in grams and not Newtons:

    F = K x E
    100 = K x 0.10
    100 ÷ 0.10 = 100
    K = 100 N/m

    Is this right or have I gone wrong somewhere?
    Thanks
     
    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 5, 2014 #2

    Doc Al

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    Staff: Mentor

    What's the weight (in Newtons) of a 100 g mass? (Always convert everything to standard units.)
     
  4. Jan 5, 2014 #3
    [itex] F- kx [/itex]
    100g = 100*10 = 1000 Newtons of force.

    [itex] 1000 = -k (-0.1) \implies k = \boxed{10 000}[/itex]
     
  5. Jan 5, 2014 #4

    Radarithm

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    Gold Member

    You'll need to convert the weight from g to kg; right now you have the units in (g*m/s^2)/m
    N/m is (kg*m/s^2))/m where the unit of Newtons is kg*m/s^2
     
  6. Jan 5, 2014 #5

    Doc Al

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    Staff: Mentor

    This is incorrect.

    Please allow the OP to solve the problem for themselves.
     
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