Figure 12-57 shows an approximate plot of stress versus strain for a spider-web thread, out to the point of breaking at a strain of 0.200. The vertical axis scale is set by values a=0.12 GN/m², b=0.30 GN/m², and c=0.80 GN/m² . Assume that the thread has an initial length of 0.80 cm, an initial cross-sectional area of 8.0x10^-12 m², and (during stretching) a constant volume. Assume also that when the single thread snares a flying insect, the insect’s kinetic energy is transferred to the stretching of the thread. (a) How much kinetic energy would put the thread on the verge of breaking? What is the kinetic energy of (b) a fruit fly of mass 6.00 mg and speed 1.70 m/s and (c) a bumble bee of mass 0.388 g and speed 0.420 m/s ? Would (d) the fruit fly and (e) the bumble bee break the thread?

The vertical axis scale is set by valuesis as follows:

a=0.12 GN/m²

b=0.30 GN/m²

c= 0.80 GN/m²

Initial length of thread is 0.80 cm and initial cross-sectional area of $8.0\times {10}^{-12}{\mathrm{m}}^2$

From the area of the graph, find the work done. find the kinetic energy of the given insects and objects using the formula in terms of mass and velocity. Comparing these two values, we can decide whether thread would break or not.

Consider the formula for the work done:

$\mathrm{W}=\int \mathrm{Fdx}$ ….. (i)

Here, W is work done, F is force, and dx is change in displacement.

Solve for the kinetic energy as:

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ ….. (ii)

Here, K is kinetic energy, m is mass and v is speed.

Consider the formula for the work done:

$\mathrm{W}=\int \mathrm{Fdx}$

Consider the formula for the force as:

$\mathrm{F}=\mathrm{stress}\times \mathrm{area}$

Consider the case of very small length is:

$\mathrm{dx}=\mathrm{strain}\times \mathrm{length}$

Consider the formula for the work done as:

$$\begin{gathered} \mathrm{W}=\int \mathrm{stress}\times \mathrm{A}\times \mathrm{strain}\times \mathrm{L} \\ =\mathrm{AL}\int \mathrm{stress}\times \mathrm{strain} \\ =\mathrm{cross}\mathrm{sectional}\mathrm{area}\mathrm{of}\mathrm{thread}\times \mathrm{length}\mathrm{of}\mathrm{thread}\times \mathrm{area}\mathrm{under}\mathrm{the}\mathrm{curve}\mathrm{of}\mathrm{length} \end{gathered}$$

The area under the curve is obtained as:

$$\begin{gathered} \mathrm{Area}\mathrm{under}\mathrm{the}\mathrm{curve}\mathrm{of}\mathrm{graph}=\frac{1}{2}\left({\mathrm{as}}_1\right)+\frac{1}{2}\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{s}}_2-{\mathrm{s}}_1\right)+\frac{1}{2}\left(\mathrm{b}+\mathrm{c}\right)\left({\mathrm{s}}_3-{\mathrm{s}}_2\right) \\ =\frac{1}{2}\left[{\mathrm{as}}_2+\mathrm{b}\left({\mathrm{s}}_3-{\mathrm{s}}_1\right)+\mathrm{c}\left({\mathrm{s}}_3-{\mathrm{s}}_2\right)\right] \\ =\frac{1}{2}\left[\left(0.12\times {10}^9\mathrm{N}/{\mathrm{m}}^2\right)\left(1.4\right)+\left(0.30\times {10}^9\mathrm{N}/{\mathrm{m}}^2\right)\left(1.0\right)+\left(0.80\times {10}^9\mathrm{N}/{\mathrm{m}}^2\right)\left(0.60\right)\right] \\ =4.74\times {10}^8\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the kinetic energy can be calculated as,

$\begin{array}{rcl}\mathrm{K}& =& \left(8.0\times {10}^{-12}\mathrm{N}/{\mathrm{m}}^2\right)\left(8.0\times {10}^{-3}\mathrm{m}\right)\left(4.74\times {10}^8\mathrm{N}/{\mathrm{m}}^2\right)\\ & =& 3.03\times {10}^{-5}\mathrm{J}\end{array}$

Kinetic energy that would put the thread on the verge of breaking is $3.03\times {10}^{-5}\mathrm{J}$

Consider the expression for the kinetic energy as:

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}& =& \frac{1}{2}{\mathrm{m}}_{\mathrm{f}}{{\mathrm{v}}_{\mathrm{f}}}^2\\ & =& \frac{1}{2}\times \left(6.00\times {10}^{-6}\mathrm{kg}\right){\left(1.70\mathrm{m}/{\mathrm{s}}^2\right)}^2\\ & =& 8.67\times {10}^{-6}\mathrm{J}\end{array}$

The kinetic energy of fruit fly of mass 6.00 mg and speed 1.70 m/s² is $8.67\times {10}^{-6}\mathrm{J}$

Consider the formula for the kinetic energy as:

${\mathrm{K}}_{\mathrm{b}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{b}}{{\mathrm{v}}_{\mathrm{b}}}^2$

Substitute the values and solve as:

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{b}}& =& \frac{1}{2}\times \left(3.99\times {10}^{-4}\mathrm{kg}\right){\left(0.420\mathrm{m}/{\mathrm{s}}^2\right)}^2\\ & =& 3.42\times {10}^{-5}\mathrm{J}\end{array}$

The kinetic energy of a bumble bee of mass 0.388 g and speed 0.420 m/s² is $3.42\times {10}^{-5}\mathrm{J}$.

From the explanation it is clear that:

Fruit fly will not be able to break the thread, because K_f<W .

From the explanation it is clear that:

The bumble bee will be able to break the thread, because K_b>W .