5.Y There are types of springs that don\'t behave exactly like an ideal Hooke\'s

Question

5.Y There are types of springs that don't behave exactly like an ideal Hooke's law spring. Consider a

spring where the force of the spring is given by F(x)=kx – Ax2 + Bx3 where k=200 N/m and A=600

N/m2 and B=9000 N/m3.

A) Make a graph of force versus distance for this spring for distances between x=-3 m and x=3 m

using matlab and turn in graphs and a copy of your matlab code.

B) How much force is required to hold the spring at x=+1.5 m?

C) At x=-1.5 m?

D) Now make a graph of energy stored in the spring as a function of distance between x=-3 m and

x=+3 m. How much work is required to stretch the spring to x=-1.5 m? To x=1.5m? ( This involves

computing the integral of F(x). If you have not seen integrals in calculus yet, come see me to learn

how to do this. )

Plz detailed explanation.

Explanation / Answer

F(+1.5) = 200*1.5 - 600*(1.5^2) + 9000*(1.5^3) =300-1350+30375= 29325 N F(-1.5)= -300 -1350 - 30375 = -32025 N Work Done = Integral F.dx = kx2/2 - Ax3/3 + Bx4/4 Work Done ( 0 to 1.5) = (200*1.5^2 )/ 2 - (600*1.5^3 )/3 + (9000*1.5^4 )/4 = 225 - 675 + 11390.625 = 10940.625 J Work Done ( 0 to -1.5) = (200*1.5^2 )/ 2 + (600*1.5^3 )/3 + (9000*1.5^4 )/4 = 225 + 675 + 11390.625 = 12290.625 J Total Work Done = 10940.625 + 12290.625 = 23231.25 J

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.