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A spring is 16.2 cm long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to 28.0 N , causing the spring to stretch to a length of 19.0 cm .

Part A
What is the force constant of this spring?
Part B-How much work was required to stretch the spring from 16.2 cm to 19.0 cm ?
Part C- How long will the spring be if the 28.0 N force is replaced by a 56.0 N force?

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Jess
2 answers

  1. Nvm guys, I got it !
    A- k=F(delta x)=28N/(0.190-0.162)=1000N/m
    B- w=1/2mv^2= 1/2(1000)(0.190-0.162)^2=0.392J
    C- delta x = F/k=56N/1000=0.056 + 0.162= 0.218m= 21.8cm

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    Jess

  2. B. Work = F*d = 28 * (0.19-0.162) = 32.8 Joules.

    C. Length = 1m/1000N * 56N = 0.056 m.
    or 56N/(1000N/m) = 0.056 m.

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    Henry

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