If A Force Of 50N Stretches A Wire From 20m To 20.01m; What Is The Amount Of Force Required To Stretch The Same Material From 20m To 20.05m?

A. 250N
B. 200N
C. 100N
D. 50N

Correct Answer:

Option A – 250N

Explanation

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 = $\frac{F_{1}}{e_{1}} = \frac{F_{2}}{e_{2}} ⟹ \frac{50}{0.01} = \frac{F_{2}}{0.05} ⟹ F_{2} = \frac{50 \times 0.05}{0.01} = 250 N$

$\frac{F_{1}}{e_{1}} = \frac{F_{2}}{e_{2}} ⟹ \frac{50}{0.01} = \frac{F_{2}}{0.05} ⟹ F_{2} = \frac{50 \times 0.05}{0.01} = 250 N$

If a force of 50N stretches a wire from 20m to 20.01m; what is the amount of force required to stretch the same material from 20m to 20.05m?