91Ó°ÊÓ

Consider the formula for Newton’s law of motion:

${\mathbf{v}}^{\mathbf{2}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{gH}$ (1)

Here, vis final velocity,${\mathbf{v}}_{\mathbf{0}}$ is initial velocity, g is gravitational acceleration, H is height.

Here we have, Young’s modulus of bone is$\mathrm{Y}=1.4×{10}^{10}\mathrm{Pa}$

Cross-sectional area is $\mathrm{A}=3×{10}^{-4}{\mathrm{m}}^2$

Change in length is 1.0%.

(a)

Derive the formula for the force as:

$\begin{array}{r}Y=\frac{F_{âŠ\yen }{I}_0}{A\mathrm{Δ}I}\\ F_{âŠ\yen }=YA\frac{\mathrm{Δ}I}{I_0}\end{array}$ (2)

Consider the percentage by which the length is changed is 1.0% .

So, $\frac{∆l}{l_0}=1.0\%$

Substitute the values in equation (2) and solve as:

$\begin{array}{r}{F}_{âŠ\yen }=\left(1.4×{10}^{10}\mathrm{Pa}\right)\left(3×{10}^{−4}{\mathrm{m}}^2\right)\left(1.0\mathrm{\%}\right)\\ =\left(1.4×{10}^{10}\mathrm{Pa}\right)\left(3×{10}^{−4}{\mathrm{m}}^2\right)\left(0.01\right)\\ =42×{10}^3\mathrm{N}\end{array}$

Hence, the maximum force that can be applied to a bone whose minimum cross-sectional area is$3.00{\mathrm{cm}}^2\mathrm{is}42×{10}^3\mathrm{N}$

(b)

Simply the equation (1) as follows:

$\begin{array}{r}{v}^2=v_0^2+2gH\\ v^2=2gH(âˆ\mu v_0=0)\\ H=\frac{v^2}{2g}\left(3\right)\end{array}$

Determine the expression for velocity from Newton’s second Law:

$\begin{array}{r}2F_{max}=\frac{\mathrm{Δ}p}{\mathrm{Δ}t}\\ 2F_{max}=\frac{mv}{\mathrm{Δ}t}\\ V=\frac{2F_{max}\mathrm{Δ}t}{m}\end{array}$

Substitute the values and solve as:

$\begin{array}{r}{\mathrm{v}}_{max}=\frac{2\left(42×{10}^3\mathrm{N}\right)(0.030\mathrm{s})}{\left(70\mathrm{kg}\right)}\\ =36\frac{\mathrm{m}}{\mathrm{s}}\\ \mathrm{Now},\mathrm{to}\mathrm{find}\mathrm{maximum}\mathrm{height}\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{v}}_{\max }\mathrm{in}\mathrm{equation}\left(3\right).\\ {\mathrm{H}}_{max}=\frac{{\mathrm{v}}_{max}^2}{2\mathrm{g}}\\ =\frac{{\left(36\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{2\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\\ =65\mathrm{m}\end{array}$

Hence, the maximum height from which a 70 kg man could jump and not fracture his tibia is 65 m .