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Potential Energy/Work done question AS Level Physics - Mechanics

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    The question posed - Calculate the work done in a steel wire (Esteel = 200GPa) 2.0m long and 0.010cm² in cross-sectional area extends by 2.5mm when loaded.

    so

    Length (L) = 2m
    Area (A) = 1(x10 to the -6)m
    Extension (x) = 0.0025m
    Ep? = 2(x10 to the 11) Nm????


    2. Relevant equations
    I know work done = Force x distance and EP = 1/2F[itex]\chi[/itex]


    3. The attempt at a solution
    my attemp at solving this as follows:

    1/2 2(x10 to the eleven)Nm x 0.0025m
    = 1(x10 to the eleven)Nm x 0.0025m
    = 2.5(x10 to the seven) Nm²

    this just does not seem right to me... I haven't used half of the information give in the problem and I'm pretty sure potential energy is not neccissarly the same as work done. Also if the was the case wouldn't the spring constant reach it's ultimate tensile stress or fracture point.

    Please help I was away when we covered this in college and the notes provided do not seem to help at all.
     
  2. jcsd
  3. Oct 29, 2011 #2

    PhanthomJay

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    yes
    The modulus of elasticity (Esteel) is an inherent property of the material (in this case, steel) and is a measure of its rigidity, with units of N/m^2
    only if the force is constant and the displacement is in the direction of the force
    yes, OK
    And what you have used is not right, ..you have incorrectly equated F with E
    It's the negative of the change in PE that is the same as work done by a conservative force (like a spring, which is the case here)
    Not if you use the proper values and the steel stays within its elastic limit
    What you have mostly missed is the equation for the extension of the steel under an applied axial load. If [itex]\chi[/itex] is the extension, then it can be shown that


    [itex]\chi[/itex] = FL/AE
    or
    F= (AE/L)[itex]\chi[/itex]

    Since the steel member obeys Hooke's law, F=k[itex]\chi[/itex], then

    k=AE/L

    And you can solve for the work done using the same formula you use when assuming the steel rod is a spring (or use your equation W = F/2([itex]\chi[/itex])).
     
  4. Oct 29, 2011 #3
    thank you for those equations they are not in my notes anywhere... and thank you for taking the time to help me with my problem

    so...

    F =
    1(x10 to the -6)m² x 2(x10 to the 11)Nm / 2m = 100000Nm
    100000Nm x 0.0025m
    = 250 Nm²

    Work Done =
    250Nm² x 0.0025m
    = 0.625 J

    still this somehow does not seem right? ie. the units do not seem to match up for me or am I missing something?

    you mentioned that work done = force x distance only if force is a constant what would happen if this was not the case?
     
  5. Oct 29, 2011 #4

    PhanthomJay

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    You're welcome.
    Watch your units.......E has units of N/m^2, so the force is 250 N.
    No, the force is not constant...it varies from 0 as the wire just starts to stretch, to 250 N when it is fully stretched...the work done is the average force times the extension...W = (250/2)(.0025) = .3125 J

    As I noted, this is not the case...when the force is not constant, and the force is in the direction of the displacement, then
    [itex]W = \int F.dx[/itex]
    I don't know if you have taken calculus, but since F =kx, per Hookes Law, then [itex]W =\int kx.dx = 1/2kx^2[/itex]. Or if you don't know calculus, you can use, in this case, since the force varies linearly with the distance, the other formula, [itex] W = (F/2)(x) = (kx/2)(x) = 1/2kx^2[/itex], the same result.

    Note that the work done by the force is the same value as the potential energy change of the steel wire (a spring), but I don't know if you have studied energy methods yet.
     
  6. Oct 30, 2011 #5
    thanks I see now that you had already explained the force was not a constant and can now see where the equation in my notes 1/2kx² comes into play. =) I shall return to you for pointers in the future you've been a great help
     
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