Question #181223

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


1
Expert's answer
2021-05-14T05:48:09-0400

F(x)=kxF(x) = kx

10=k(0.2)10 = k(0.2)

k=50 N/mk = 50\ N /m


W=abF(x) dxW = ∫ ^b_ a F(x)\ dx


W=abkx dxW = ∫ ^b _a kx\ dx


W=kabx dxW = k∫ ^b_ a x\ dx


W=k[x22]ab=k[x22]00.5W = k [\dfrac{x²}{2} ] ^b_a = k [\dfrac{x²}{2} ] ^{0.5}_0


W=50[0.522022]=50×0.252=6.25 JW = 50[\dfrac{0.5²}2-\dfrac{0²}{2}] = 50× \dfrac{0.25}{2} = 6.25\ J


6.25J\therefore 6.25 J of work should be done to stretch the spring to its maximum extent.


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