Question

a spring, whose equilibrium length is 3 ft,3 ft, extends to a length of 6 ft6 ft when a force of 13 lb13 lb is applied. if 8 ft-lb8 ft-lb of work is required to extend this spring from its equilibrium position, what is its total length? (use symbolic notation and fractions where needed.) length:length:

Answer 1

The total length of the spring is approximately 4.92 ft.

To remedy this question we need to apply Hooke's law. In physics, Hooke's regulation is an empirical law which states that the force (F) had to enlarge or compress a spring by using a ways (x) scales linearly with admire to that distance.

Mathematically F = kx, where ok is the spring constant and x is spring stretch/compression.

Length of the spring=3ft

Extended length=6ft

Length extended(x)=(6-3)ft=3ft

Force applied(F)=13 lb

Work done to extend the spring(w)=8ft-lb

Now by using Hooke's Law:

F=kx

or,$13=k\times 3$

or,$k=\frac{13}{3}$

Work done by the spring can be defined as

$w=\frac{1}{2} k x^2$

Substituting the values we get:

$or,8=\frac{1}{2}\times \frac{13}{3} \times x^2\\or,x^2=\frac{48}{13}\\or,x=1.921...$

Therefore total length of the spring=3+1.92 ft=4.92 ft.

To learn more about equilibrium length of spring:

https://brainly.in/question/13551116

#SPJ4