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Homework Help: Spring and wood block problem

  1. Feb 25, 2005 #1
    A 171 g wood block is firmly attached to a very light horizontal spring, as shown in the figure below.
    The block can slide along a table where the coefficient of friction is 0.306. A force of 20.9 N compresses the spring 17.2 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing?

    i know the equation is suppose to be 1/2k(x^2-D^2)=(Uk)mg(x+D)
    where D is the distance the block will travel....
    and then set the equation equals to D = x - (2Ukmg)/D
    i plug in all the numbers and i still get the wrong answer...
    i think my K constant is wrong i thought it was just k = 20.9 N/.172 m = 121.51

    btw my wrong answer is 8.7596 m

    can someone help me?
     

    Attached Files:

    Last edited: Feb 25, 2005
  2. jcsd
  3. Feb 25, 2005 #2
    Frictional force is Umg. Not Ukmg.

    Also Check your calculations. There can not be a D in the denominator.


    I am getting,

    D = x - (2Umg)/k
     
  4. Feb 25, 2005 #3
    I think he meams that Uk is the kinetic coeffiecient of friction (where the k is meant to be a subscript, not the spring constant).
    but other than that, I got the same equation as gamma.

    like gamma said, check your calculations, and also check your units: g -> kg, etc...
     
  5. Feb 25, 2005 #4
    yea mathstudent is right i was talkin about kinetic coefficient and yea i accidently typed the wrong thing for the equation... mathstudent was right i forgot about the units... i used g instead of kg... thanks a lot guys!
     
    Last edited: Feb 25, 2005
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