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Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is 7.50 104 N. The minimum effective cross-sectional area of the bone is 3.90 10-4 m2, its length is 0.59 m, and Young's modulus is Y = 9.4 109 N/m. The mass of the man is 66 kg. He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.h = _________.

Respuesta :

Answer:

[tex]H = 0.7 m[/tex]

Explanation:

As we know by the formula of elasticity we will have

[tex]F = \frac{YA}{L} x[/tex]

if we compare this equation by the spring force we will have

[tex]F = kx[/tex]

[tex]k = \frac{YA}{L}[/tex]

so we will have

[tex]k = \frac{9.4 \times 10^9)(3.90 \times 10^{-4})}{0.59}[/tex]

[tex]k = 6.21 \times 10^6 N/m[/tex]

now we know that maximum compression in the bone will be given as

[tex]F = kx[/tex]

[tex]7.50 \times 10^4 = (6.21 \times 10^6) x[/tex]

[tex] x= 0.012 m[/tex]

now the energy stored in the bone is given as

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]U = \frac{1}{2}(6.21 \times 10^6)(0.012)^2[/tex]

[tex]U = 452.4 J[/tex]

Now we can say that maximum initial energy must be equal to this energy

[tex]mgH = 452.4[/tex]

[tex]66(9.81)H = 452.4[/tex]

[tex]H = 0.7 m[/tex]

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