Math Problem Statement

A 42.0 cm tibia bone with a cross-sectional area of 160 mm2 is compressed to its ultimate strain point of 2.6%. Assume the elastic modulus is 22 x 109 Pa. How much energy is stored in the bone at its ultimate strain point? (Provide answer in Joules)

Solution

To find the energy stored in the tibia bone at its ultimate strain point, we can treat the bone as an elastic material under compression. The energy stored per unit volume in an elastic material under strain is given by the formula:

Energy density=12Eϵ2\text{Energy density} = \frac{1}{2} E \epsilon^2

Where:

  • EE is the elastic modulus (Pa),
  • ϵ\epsilon is the strain (dimensionless).

Step 1: Convert given values to appropriate units

  • The length of the tibia bone is 42.0 cm, which is 0.42 meters.
  • The cross-sectional area is 160 mm², which is 160×106m2160 \times 10^{-6} \, \text{m}^2.
  • The ultimate strain ϵ\epsilon is 2.6%, which is 0.0260.026 in decimal form.
  • The elastic modulus EE is given as 22×109Pa22 \times 10^9 \, \text{Pa}.

Step 2: Calculate the energy density

Using the formula for energy density:

Energy density=12×22×109×(0.026)2\text{Energy density} = \frac{1}{2} \times 22 \times 10^9 \times (0.026)^2

Now, let's calculate the energy density.

Energy density=12×22×109×0.000676\text{Energy density} = \frac{1}{2} \times 22 \times 10^9 \times 0.000676

Energy density=12×14.872×106\text{Energy density} = \frac{1}{2} \times 14.872 \times 10^6

Energy density=7.436×106J/m3\text{Energy density} = 7.436 \times 10^6 \, \text{J/m}^3

Step 3: Calculate the volume of the bone

The volume VV of the bone is:

V=cross-sectional area×lengthV = \text{cross-sectional area} \times \text{length}

Substitute the values:

V=160×106m2×0.42m=67.2×106m3V = 160 \times 10^{-6} \, \text{m}^2 \times 0.42 \, \text{m} = 67.2 \times 10^{-6} \, \text{m}^3

Step 4: Calculate the total energy stored in the bone

The total energy stored UU is:

U=energy density×volumeU = \text{energy density} \times \text{volume}

U=7.436×106J/m3×67.2×106m3U = 7.436 \times 10^6 \, \text{J/m}^3 \times 67.2 \times 10^{-6} \, \text{m}^3

U=0.499JU = 0.499 \, \text{J}

Final Answer:

The energy stored in the bone at its ultimate strain point is approximately 0.499 Joules.

Would you like more details on any of these steps or have other questions?

Here are 5 related questions:

  1. How does elastic modulus affect the energy stored in materials?
  2. What happens to the bone beyond its ultimate strain point?
  3. How is strain measured in real-world applications?
  4. Can you calculate the stored energy for different materials using the same approach?
  5. How does the cross-sectional area affect the energy stored in a compressed object?

Tip: Strain is a dimensionless quantity, often expressed as a percentage, and it represents deformation per unit length.

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Math Problem Analysis

Mathematical Concepts

Physics
Elasticity
Energy Density
Stress and Strain

Formulas

Energy density = (1/2) E * ε^2
Volume = Cross-sectional area * Length
Total Energy = Energy density * Volume

Theorems

Hooke's Law (applied in the context of elasticity)
Elastic potential energy theorem

Suitable Grade Level

Undergraduate (Engineering, Physics)