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Ideal Spring compression

  1. Dec 10, 2004 #1
    An ideal spring is used to fire a 15.0-g block horizontally across a frictionless table top. The spring has a spring constant of 20 N/m and is initially compressed by 7.0 cm. The speed of the block as it leaves the spring is:


    i do the following:

    energy stored by spring: (kv^2)/2= 0.049

    when the spring is at equilibrium position all the energy wwould be in kinetic thus i solve for the velocity,

    but i get the wrong answer...what am i doing wrong?
     
  2. jcsd
  3. Dec 10, 2004 #2

    ZapperZ

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    Er... if you meant by "v" above as "velocity", then you need to double check the expression for the potential energy of a compressed spring. Hint: it involves the compressed LENGTH, and not dependent on any velocity.

    Zz.
     
  4. Dec 10, 2004 #3

    HallsofIvy

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    It's hard to tell what you are doing wrong since you don't show us HOW you solve for velocity!

    The potential energy stored in the spring, with spring constant 20 N/m, by compressing it 0.07 meters is (1/2)(20)(0.07)2= 0.049 Joules.

    Once the spring has "uncompressed", all of its potential energy goes into the kinetic energy of the block so (1/2)(0.015kg)v2= 0.049 and v= 2.56 m/s approximately.
     
  5. Dec 10, 2004 #4
    aah...my problem was that i forgot to convert 15 grams into kilograms...ahh!!
     
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