# The elastic energy stored in a wire

Littlemonkey
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 elastic limit for steel is at a strain of 0.0025

Formula:
Δl=(lo F)/EA

a)Δl=(1.84m×60N)/((2.0×10^11 Nm-2) )×(Π×0.00025^2 ) ) = 2.81×10-3m or 2.81mm

I know the answer to a) is correct I found the book that the original question was taken from I also know the answer to b) is 84mJ but I have no clue where to even start any help would be greatly appreciated.

Also could someone check my calculations on the problem below as again I have the answer from the book but I cannot get the equation to work:

A crane used to lift iron girders on a building site has a steel lifting cable diameter 40 mm. When the crane lifts an iron girder of weight 10 kN, the cable length with the girder just off the ground is 75.0 m. Calculate
a) The extension of the cable for this length and load

Formula:
Δl=(lo F)/EA

Δl=(75m×10kN)/((2.0×10^11 Nm-2) )×(Π×0.02^2 ) ) = 2.98×10-4 m or .298 mm

The answer in the book is 2.98×10-3 m or 2.98 mm I just cant work out where I am going wrong.

## Answers and Replies

Mentor
I also know the answer to b) is 84mJ but I have no clue where to even start any help would be greatly appreciated.
How do you calculate elastic potential energy? (Compare to a stretched spring.)

Δl=(75m×10kN)/((2.0×10^11 Nm-2) )×(Π×0.02^2 ) ) = 2.98×10-4 m or .298 mm

The answer in the book is 2.98×10-3 m or 2.98 mm I just cant work out where I am going wrong.
Double check your arithmetic.

Littlemonkey
Oh im a balloon I was conviced that 75*10kN was only 75000N not 750000N. One down one to to go.

Thanks

Littlemonkey
Ok I have the answer:

Formula:
W = ½ force x extension
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.

Mentor
Why is the strain of 0.0025 mentioned when it does not seem to be relevant when working out this answer.
To calculate the elastic potential energy you need to be assured that the elastic limit is not exceeded. It doesn't factor into the calculation.

Littlemonkey
That makes sense thanks Doc.