Cuesta College, San Luis Obispo, CA
Assume that a golf ball can be considered to be a cube with a cross-sectional area of 1.4×10–3 m2. A steam roller then applies 3.9×104 N of downwards force on the golf ball, compressing it down to half of its original vertical height.[*][**] The overall Young's modulus for this golf ball is: (A) 55 N/m2.
(B) 2.0×104 N/m2.
(C) 2.8×107 N/m2.
(D) 5.6×107 N/m2.
Correct answer (highlight to unhide): (D)
Hooke's law for elastic materials is given by:
(F/A) = Y·(∆L/L),
where the cross-sectional area A of the wire is 1.4×10–3 m2, the force F applied is 3.9×104 N, and the ratio ∆L/L that the golf ball is compressed to is 1/2. The (overall) Young's modulus of the golf ball can then be solved for:
Y = (F/A)/(∆L/L),
Y = ((3.9×104 N)/(1.4×10–3 m2))/(0.5),
Y = 5.5714285714×107 N/m2,
or to two significant figures, Y = 5.6×107 N/m2.
(Response (A) is F·A; response (B) is F/2; and response (C) is the applied stress F/A.)
Sections 70854, 70855, 73320
Exam code: quiz06rn3T
(A) : 3 student
(B) : 6 students
(C) : 9 students
(D) : 37 students
Success level: 69%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73