If a force of 50N stretches a wire from 20m to ... - JAMB Physics 2004 Question

Original length = 20m

first new length = 20.01m

first increase in length = 20.01m - 20m = 0.01m

Second new length = 20.05m

second increase in length = 20.05 - 20m = 0.05m

From hooke's law = ( \frac{F_1}{e_1} = \frac{F_2}{e_2} \implies \frac{50}{0.01} = \frac{F_2}{0.05} \

\implies F_2 = \frac{50 \times 0.05}{0.01} = 250N )