A cable is $5.00 \mathrm{~m}$ long and has a diameter of $1.50 \mathrm{~mm}$. The elastic modulus of the cable is $2.50 \times 10^{10}\mathrm{~N} / \mathrm{m}^{2}$. A force is applied to the end of the cable and the cable stretches by $3.00 \mathrm{~mm}$. What is the applied force?
A) $26.5 \mathrm{~N}$
B) $32.6 \mathrm{~N}$
C) $20.5 \mathrm{~N}$
D) $36.6 \mathrm{~N}$
E) $16.5 \mathrm{~N}$

Question

Quizplus · Accepted Answer

The correct answer is A because the applied force is given by the equation F = YA(delta L/L), where Y is the elastic modulus, A is the cross-sectional area of the cable, and delta L/L is the fractional change in length. Plugging in the given values, we get F = (2.50 x 10^10 N/m^2) * (pi * (1.50 x 10^-3 m)^2) * (3.00 x 10^-3 m / 5.00 m) = 26.5 N.

A Cable Is 5.00 m5.00 \mathrm{~m}5.00 m Long and Has a Diameter Of 1.50 mm1.50 \mathrm{~mm}1.50 mm

A Cable Is $5.00 \mathrm{~m}$ Long and Has a Diameter Of $1.50 \mathrm{~mm}$