Recent content by Littlemonkey

  1. L

    The elastic energy stored in a wire

    That makes sense thanks Doc.
  2. L

    The elastic energy stored in a wire

    Ok I have the answer: Formula: W = ½ force x extension Answer: W = ½ × 60 N × 2.81×10-3m = 8.43×10-2J or 84 mJ One last question and its why it took me so long to work it out. Why is the strain of 0.0025 mentioned when it does not seem to be relevant when working out this answer.
  3. L

    The elastic energy stored in a wire

    Oh I am a balloon I was conviced that 75*10kN was only 75000N not 750000N. One down one to to go. Thanks
  4. L

    The elastic energy stored in a wire

    A steel wire diameter 0.50mm and length 1.84m is suspended vertically from a fixed point and used to support a 60N weight hung from the lower end of the wire. The Young modulus for steel is 2.0 x 10^11 N m-2 a) The extension of the wire b) The elastic energy stored in a wire given the...
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