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Work Done By Stretching a Steel Wire

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    An angler hangs a 5.50 kg fish from a vertical steel wire 1.30 m long and 4.32×10^-3 cm in cross-sectional area. The upper end of the wire is securely fastened to a support.

    Calculate the amount the wire is stretched by the hanging fish. (I found this to be 8.11x10-4 m)

    The angler now applies a force F to the fish, pulling it very slowly downward by 0.569mm from its equilibrium position. For this downward motion, calculate the work done by gravity. (I found this to be 3.07x10^-2 J)

    For this downward motion, calculate the work done by the force F.

    2. Relevant equations

    E=FL_0/A_0ΔL
    W=Fd
    My prof reccomended using: k=EA_0/L_0 and W=0.5kx^2

    3. The attempt at a solution

    Using Young's Modulus, F=EA_0ΔL/L_0
    F=(2x10^11 N/m)(4.32x10^-7 m ^2)/(1.300811 m)
    Solving for F=37.8 N

    W=Fd
    W=(37.8 N)(0.000569 m)
    W=2.2x10^-2 J

    Then I emailed my prof and he said to do it like this:

    k=EA_0/L_0
    k=(2x10^11 N/m)(4.32x10^-7 m^2)/(1.300811 m)
    k=66420 N/m

    W=0.5kx^2
    W=(0.5)(66420 N/m)(0.000569 m)^2
    W=1.08x10^-2 J

    According to Mastering Physics, both of these answers are wrong. Please Help!
     
  2. jcsd
  3. Nov 30, 2011 #2
    The stretch done by the freely hanging fish comes out to .00162 m. Your answer is off by a factor of two, and I'm wondering if you forgot to divide by .5 somewhere.

    You correctly calculated the work done by gravity (during angler stretching).

    Your prof then says to do this:
    k=(2x10^11 N/m)(4.32x10^-7 m^2)/(1.300811 m)

    At this point I must question if he/she is really a prof and really not just a T.A. What he or she has done is to ask you to calculate a SECOND value for k - for k, the spring CONSTANT! The same spring should not have two different values for k. When doing this problem, the distance used in calculating the potential energy should always be measured as the displacement from the wire's prestressed length of 1.30 m. Then you're going to need to subtract from this the work done by gravity calculated using the total displacement.

    Give it another try and give up on the idea that the "prof" is always right.
     
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