Answer to Question #179089 in Statistics and Probability for tifanny pulido

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 presentationshall include important components of the problems, complete and correct computations and a logical and organizedexplanation.

Answer 1

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

$W=kx^2/2$

$k=F/x$

$F$ is the force to stretch a spring

Then:

$k=10/0.2=50\ H/m$

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

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