Answer
given,
k = 250 N/m
q = 900 N/m³
(FSp)s=−kΔs−q(Δs)^3
work done = Force x displacement
[tex]W = \int {F. dx}[/tex]
limits are x = 0 to x = 0.15 m
work done
[tex]W = \int_0^{0.15} (k x + q x^3)\ dx[/tex]
[tex]W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15[/tex]
[tex]W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}[/tex]
W = 3.375 + 0.1139
W = 3.3488 J
b) % cubic term =[tex]\dfrac{0.1139}{3.3488}[/tex]
% cubic term =[tex]3.4\ %[/tex]
The work done by the spring and the cubic term is mathematically given as
Question Parameters:
How much work must you do to compress this spring 15cm?
Consider a spring with k = 250 N/m and q = 900 N/m3.
a)Generally the equation for the Force is mathematically given as
(FSp)s=−kΔs−q(Δs)^3
Where Workdone is
W = \int {F. dx}
Hence
[tex]W = [\frac{kx^2}{2}+\frac{qx^4}{4}+ C]_0^0 *15[/tex]
[tex]W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}[/tex]
W=3.3488J
b)
Where cubic term is mathematically given as
C=\frac{0.1139}{3.3488}
C=%3.4
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