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Work and Kinetic Energy Problem involving friction hw problem, DUE SOON

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Part 1.


    A(n) 2300 g block is pushed by an external
    force against a spring (with a 22 N/cm spring
    constant) until the spring is compressed by
    15 cm from its uncompressed length. The
    compressed spring and block rests at the bottom of an incline of 32

    .
    The acceleration of gravity is 9.8 m/s
    2
    .
    Note: The spring lies along the surface of
    the ramp (see figure).
    Assume: The ramp is frictionless.
    Now, the external force is rapidly removed
    so that the compressed spring can push up
    the mass.


    How far up the ramp (i.e., ℓ the length
    along the ramp) will the block move (measured from the end of the spring when the
    spring is uncompressed) before reversing direction and sliding back?
    Remember, the block is not attached to the
    spring.
    Answer in units of cm.

    Part 2.


    Repeat the first part of this question, but
    now assume that the ramp has a coefficient of
    friction of  = 0.423 .
    Keep all other assumptions the same.
    Answer in units of cm

    2. Relevant equations

    Without friction = Ui + Ki = Uf + Kf, where in this problem Ui = 1/2kx^2, Ki = 0, Uf = mgh and Kf = 0. So 1/2 kx^2 = mgh

    With friction = Ui + Ki + (-Fk*l) = Uf + Kf

    So 1/2kx^2 - mu*(mgcosθ)*l = mgh



    3. The attempt at a solution

    I figured out Part 1.

    From the equation above

    h = kx^2/2mg

    l = h/sin32

    Then I subtracted .15 m and converted to cm. I got the correct answer. When I tried to solve for the length with the added friction in, I first forgot to account for the length in (mu*(mgcosθ*l).

    I remembered to add in l, but I think I'm using the wrong l. The first l I used was .45 because that is the length of the ramp. I then tried .30 ( because .45-.15)

    Here is my current equation and answer but I am not sure if I chose the correct l and therefore I don't know if it is correct. Please advise.

    2200 N/m *.15^2m - .423(2.3 kg * 9.8m/s^2*cos32°*.30/2*2.3kg*9.8 m/2^2

    h = 1.044239263 m

    l = h/sinθ

    so: 1.044239263 m/sin32 = 1.97056294 m

    1.97056294 m - .15 m = 1.821 m

    1.821 m * 100 = 182.1 cm

    Sorry for not having a picture.
     
  2. jcsd
  3. Oct 24, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    The two bits I highlighted in red look like you have left out the half in your calculation. Perhaps you had multiplied through the calculation by two before substituting???

    I must say I find this "technique" of including the units with every number while you substitute most distracting [though I have seen other people doing the same], making it all but impossible to follow the calculation to see if you are doing the right thing. It is also a habit that is foreign to anything I have seen in a school or university here where I live.
     
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