Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e Conceptual Question 11.4, Problems 11.1, 11.3
Trilene® XL Super Strong fishing line (Young's modulus 2.0×109 N/m2) and Eagle Claw® Sportfisher fishing line (Young's modulus 3.1×109 N/m2) have the same 10.0 m length [*]. The Trilene® fishing line has a cross-sectional area 1.8 times that of the Eagle Claw®. Both fishing lines are stretched with a tension force of 98 N. The __________ fishing line will stretch more.
(B) Eagle Claw®.
(C) (There is a tie.)
(D) (Not enough information is given.)
[*] S. Ottolini, G. Halpin, P. LaBruzzo, "Tensile Strength of Fishing Line," http://www.santarosa.edu/~yataiiya/E45/PROJECTS/Tensile%20Strength%20of%20Fishing%20Line%20Power%20Point.ppt.
Correct answer (highlight to unhide): (B)
Hooke's law for the Trilene® and Eagle Claw® fishing lines are given by:
(F/ATri) = YTri·(∆LTri/L),
(F/AEagle) = YEagle·(∆LEagle/L),
where tension F and the original, unstretched length L are the same for both fishing lines. The Trilene® fishing line has a cross-sectional area 1.8× that of the Eagle Claw® fishing line:
ATri = 1.8·AEagle.
The amount that the Trilene® fishing line will be stretched is given by:
∆LTri = (F·L)/(ATri·YTri),
∆LTri = ((98 N)·(10.0 m))/((1.8·AEagle)·(2.0×109 N/m2)),
∆LTri = (2.722222222×10–7 m3)/AEagle.
Similarly, the amount that the Eagle Claw® fishing line will be stretched is given by:
∆LEagle = (F·L)/(AEagle·YEagle),
∆LEagle = ((98 N)·(10.0 m))/((AEagle)·(3.1×109 N/m2)),
∆LEagle = (3.161290323×10–7 m3)/AEagle.
Thus this sample of Eagle Claw® fishing line will stretch more than the Trilene® fishing line sample.
Sections 70854, 70855, 73320
Exam code: quiz06eAg7
(A) : 13 students
(B) : 49 students
(C) : 2 students
(D) : 0 students
Success level: 77%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46