# The elastic energy stored in a wire

• Littlemonkey
In summary, we discussed the use of a steel wire with a diameter of 0.50mm and length of 1.84m to support a 60N weight and calculate its extension and elastic energy stored based on the Young modulus for steel. We also clarified the correct answer for the question regarding the extension of a steel lifting cable with a 40mm diameter and the calculation of elastic potential energy. The strain of 0.0025 mentioned in the question is not relevant for the calculation of elastic potential energy.
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 can't work out where I am going wrong.

Littlemonkey said:
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 can't work out where I am going wrong.

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

Thanks

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.

Littlemonkey said:
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.

That makes sense thanks Doc.

## 1. What is elastic energy?

Elastic energy is the potential energy stored in an object when it is stretched or compressed. It is a form of mechanical energy that is stored within the bonds of the object's molecules.

## 2. How is elastic energy related to a wire?

A wire can store elastic energy when it is stretched or compressed. When a force is applied to a wire, it experiences a deformation, and the energy used to deform the wire is stored as elastic energy. This energy is released when the wire returns to its original shape.

## 3. What factors affect the amount of elastic energy stored in a wire?

The amount of elastic energy stored in a wire depends on factors such as the material and thickness of the wire, the amount of deformation, and the force applied. A stiffer wire will store more elastic energy than a more flexible wire when stretched or compressed by the same amount.

## 4. How is elastic energy measured in a wire?

Elastic energy is measured in joules (J) in the International System of Units (SI). It can be calculated using the formula E = 0.5kx², where k is the spring constant of the wire and x is the amount of deformation.

## 5. Can the amount of elastic energy stored in a wire be changed?

Yes, the amount of elastic energy stored in a wire can be changed by altering the factors that affect it, such as the material, thickness, or force applied. Additionally, the wire's shape can also affect the amount of elastic energy stored in it. For example, a coiled wire can store more elastic energy than a straight wire when the same force is applied.

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