Question #181728

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 using 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.


1
Expert's answer
2021-05-11T06:05:49-0400

The work done in stretching a spring a distance x from its rest position is:

W=kx2/2W=kx^2/2

k=F/xk=F/x

FF is the force to stretch a spring

Then:

k=10/0.2=50 H/mk=10/0.2=50\ H/m


The work done to stretch the spring 0.5 m from the equilibrium position:


W=500.52/2=6.25 JW=50\cdot0.5^2/2=6.25\ J

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