Answer to Question #180867 in Calculus for Jasmine

Question

A manufacturer wants to build a spring that takes a force 10 N (in negative direction) to compress it 0.2 m from the equilibrium position. The spring should be able to stretch 0.5 m from the equilibrium position.

As a mechanical engineer, you were asked to present how much work should be done to stretch the spring.The presentation shall include important components of the problem, complete and correct computations and a logical and organized explanation.

Hints:

First, create a force function F(x) by finding the spring constant k in F(x) = kx where F(x) is the force and x is the position from the equilibrium.

Then, calculate work ustng the concept of the definite integral.

If a variable force F(x) moves an object in a positive direction along the x-axis from point a to point b, then the work done on the object is W = ∫ ^b a F(x) dx

Answer 1

The restoration force on the spring was f(x)=kx

Then small work done in streching the spring throghh length dx-

$dw=f(x)dx=kxdx$

Then Total work done in streching the spring from 0.2 to 0.5m-

$W=\int_{0.2}^{0.5}dw=\int_{0.2}^{0.5}kxdx$

$k\dfrac{x^2}{2}|_{0.2}^{0.5}=k(\dfrac{0.25-0.04}{2})=k\dfrac{0.21}{2}~~~~~~-(1)$

The spring constant $k =\dfrac{F}{x_o}=\dfrac{10}{0.2}=50N/m$

Eqn.(1) $\Rightarrow W=\dfrac{0.21}{2}\times 50=5.25J$