Answer to Question #181653 in Calculus for Seagul

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

$F = kx$

$k = \dfrac Fx= \dfrac{10N}{0.2m} = 50\ N/m$

$W = \int^b_a F(x)\ dx$

$W = \int^b_a kx\ dx$

$W = k \int^b_a x\ dx$

$W = k\int_{0.2}^{0.5} x \ dx$

$W = k [\dfrac{x^2}{2}]^{0.5}_{0.2} = 50[\dfrac{0.5²}{2} - \dfrac{0.2²}{2}]$

$W =50[0.105] = 5.25\ J$