Answer to Question #180941 in Calculus for Larkin

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

Find the spring constant $k$ . To do this, we will use the condition: to compress the spring by $x=0.2\,\,\text{meters}$ , it is necessary to apply a force of $F=10\,\,\text{N}$ .

$$F=kx\longrightarrow 10=k\cdot0.2\longrightarrow\boxed{k=50\,\,\frac{\text{N}}{\text{m}}}$$

Now, let's find a general formula that will describe the amount of work that needs to be done in order to stretch the springs by $x$ meters.

$$W(x)=\int\limits_0^xF(x)dx=\int\limits_0^x kx'dx'=\left.\frac{k\cdot\left(x'\right)^2}{2}\right|_0^x=\frac{kx^2}{2}$$

In our case,

$$k=50\longrightarrow W(x)=\frac{50x^2}{2}\longrightarrow\boxed{W(x)=25x^2}\\[0.3cm] W(0.5)=25\cdot0.5^2=25\cdot0.25=6.25\,\,\text{N}$$

The plot of energy versus elongation is as follows

ANSWER

$$x=0.5\longrightarrow W=6.25\,\,\text{N}$$