Question

Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)s=−kΔs−q(Δs)3 , where q q is a constant. Consider a spring with k kk = 350 N/m N/m and q qq = 750 N/m 3 N/m3 .How much work must you do to compress this spring 15cm? Note that, by Newton's third law, the work you do on the spring is the negative of the work done by the spring.By what percent has the cubic term increased the work over what would be needed to compress an ideal spring?

Answer 1

Answer with Explanation:

We are given that

Restoring force,$(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

We have to find the work must you do to compress this spring 15 cm.

$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done=$\int_{0}^{0.15}-Fd(\Delta s)$

W=$-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

Ideal spring work=$0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J$

Percentage increase in work=$\frac{4.033-3.938}{3.928}\times 100=2.4$%