Question

an elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. what is the original length of the screen?

Answer 1

For Hooke's law $F = Ke$

⟹$\frac Fe=K$

$\frac{f_1}{e_1}=\frac{f_2}{e_2}$

Let the original length=t₀

therefore $e_1=(36−t_0)$ cm; $e_2=(46−t_0)$ cm

if $f_1=40N\ \& \ f_2=60N$

$Then \ \frac{40}{ 36−t_0}=\frac{60}{(45−t_0)}$

⟹$40(45−t_0)=60(36−t_0)$

therefore $1800−40t_0 = 2160−60t_0$

$60t_0−40t_0=2160−1800$

$20t_0 = 360$

$t_0 = 18cm$