Question:

A nylon string on a tennis racket is under a tension of 275 N. If its diameter is 1.00 mm, by how much is it lengthened from its untensioned length of 30.0 cm?

Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.

The question will be visible after logging in, as required by Pearson Education Inc.

Quick Answer:

$3.5 \textrm{ cm}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. The change in length of the nylon string will be 1 over the elastic modulus, also called the Young's Modulus, times the force applied on the string divided by its cross-sectional area times its original length. So cross-sectional area is*πr squared*,

*r*is the radius of the nylon string, and so we are given 1 millimeter which is 1.00 times 10 to the minus 3 meters for the diameter so we have to divide that by 2 and then square that and times by π to get the cross-sectional area of the nylon string. And Young's Modulus for nylon is 3 times 10 to the 9 newtons per meter squared times by 275 newtons—force that is stretching the nylon string— times its original length of 30.0 centimeters and because I have put in centimeters here, our answer's gonna be in centimeters as well and it should be about 3.5 centimeters of stretching.