Question

A nylon string on a tennis racket is under a tension of 550N. If its diameter is

1.58 mm, by how much is it lengthened from its untensioned length of 45.0 cm?

Give your answer in cm. [Young’s Modulus for nylon = 5 GPa]

Answer 1

The elongation is given by the Hook's law:

$$\Delta L = \dfrac{FL}{ES}$$

where $F = 550N, L = 45cm = 0.45m, E = 5\times 10^9Pa$. $S$ is the cross-sectional area, given as follows:

$$S = \dfrac{\pi d^2}{4}$$

where $d= 1.58mm = 1.58\times 10^{-3}m$. Thus, obtain:

$$\Delta L = \dfrac{4FL}{\pi d^2E} \Delta L = \dfrac{4\cdot 550\cdot 0.45}{\pi (1.58\times 10^{-3})^2\cdot 5\times 10^9} \approx 0.03m=3cm$$

Answer. 3cm.