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Initial Compressed Spring Force

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A spring has a relaxed length of 0.17 m, and its stiffness is 15 N/m. The spring sits vertically on a table. You place a block of mass 0.058 kg on the spring and push down on the block until the spring is only 0.11 m long. You hold the block motionless on the compressed spring, then remove your hand.

    At the instant you release the block, what is the force exerted by the spring on the block? (Consider the initial stretch of the spring.) Use the usual axis system, with the +y axis running vertically up.

    L=0.11m
    L0=0.17m
    Ks=15N/m

    2. Relevant equations

    |s|(absolute value of s)=|L-L0|

    |Fspring|(magnitude of the force) = Ks*|s|

    3. The attempt at a solution

    I am asked to find the force exerted by the spring initially. I plugged in the variables I was given.

    |s| = |0.11m-0.17m|
    |s| = 0.06m

    |Fspring| = 15N/m*0.06m
    |Fspring| = 0.9N

    So my answer for the force acted on the block initially by the spring was 0.9N. However, I am told this is not right. Can anyone explain why not?
     
  2. jcsd
  3. Sep 15, 2012 #2

    lewando

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    Gold Member

    Your method looks right. Are you being penalized for not enough SDs? You can get an extra SD by multiplying before subtracting.

    [great user name, btw :) ]
     
  4. Sep 15, 2012 #3

    SammyS

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    I think not.

    15(0.17-0.11) = 15(0.17) - 15(0.11)

                 = 2.55 - 1.65

    Each of which is accurate to the nearest tenth (according to SigDig). So while each of these has 2 SigDig, when you subtract you're left with 1 SigDig.

    However, if you multiply both lengths, 0.17 & 0.11, by 2.0000, before subtracting, the result will have 2 SigDig. Then if you multiply by 15 followed by division by 2.0000, the result will have 2 SigDig, showing the folly of being too rigid about applying the rules of SigDig.

    Alternatively, if you look at the question, there does appear to be an emphasis on directionality. Should the answer of a magnitude be accompanied by a direction?
     
  5. Sep 15, 2012 #4
    I don't think my professor is that strict with SDs. It is possible though.

    This is possible but isn't it implied to be going up the y-axis since it is a positive number? Maybe I should have wrote it as a vector: <0,0.9,0>N?
     
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