Figure 12-57 shows an approximate plot of stress versus strain for a spider-web
ID: 1465238 • Letter: F
Question
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 2.00. The vertical axis scale is set by values a = 0.12 GN/m^2. b = 0.30 GN/m^2, and c = 0.80 GN/m^2. Assume that the thread has an initial length of 0.80 cm. an initial cross-sectional area of 8.0 times 10^-12 m^2, and (during stretching) a constant volume. The strain on the thread is the ratio of the change in the thread's length to that initial length, and the stress on the thread is the ratio of the collision force to that initial cross-sectional area. Assume that the work done on the thread by the collision force is given by the area under the curve on the graph. Assume also that when the single thread snares a flying insect, the insect's kinetic energy is transferred to the stretching of the thread, How much kinetic energy would put the thread on the verge of breaking? What is the kinetic energy of a fruit fly of mass 6.00 mg and speed 1.70 m/s and a bumble bee of mass 0.388 g and speed 0.420 m/s? Would the fruit fly and the bumble bee break the thread?Explanation / Answer
area of the graph = 0.5(1*a)+(0.5*(a+b)*0.4)+(0.5*(b+c)*0.6)=0.06+0.084+0.33=0.474GN/m^2 =0.474*10^9 N/m^2
volume = 0.80*10^-2*8*10^(-12)=6.4*10^(-14) m^3
work done=area of graph*volume=3.0336*10^(-5) J
(a)3.0336*10^(-5) Joule of kinetic energy would put the thread on the verge of breaking
(b) kinetic energy of fruit fly=0.5mv^2=8.67*10^(-6) J
(c)kinetic energy of bumble bee=0.5mv^2=3.422*10^(-5) J >3.336*10^(-5) J
(d) kinetic energy of fruit fly<3.0336*10^(-5) J
Since kinetic energy of fruit fly is less than kinetic energy required for breaking thread so fruit fly will not break thread
(e)kinetic energy of bumble bee >3.0336*10^(-5) J
Since kinetic energy of bumble bee is more than kinetic energy required for breaking thread so bumble bee will break thread
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