Determining Potential Energy Values — Question 4

Question 4

Potential energy of a stiffening elastic element

An elastic element exerts $F_x=-kx-\beta x^3$, where $k=40$ N/m and $\beta=200$ N/m$^3$. Set U(0)=0.

1. Determine U(x), including both terms.

2. Find U at x=0.20 m and x=0.40 m.

3. Does doubling extension multiply U by exactly four? Explain.

Question 4 - Solution

Potential energy of a stiffening elastic element

(a) Integrate the negative force.

Integrate both restoring-force terms: $$U(x)=-\int_0^x(-ks-\beta s^3)ds=\boxed{\tfrac12kx^2+\tfrac14\beta x^4=20x^2+50x^4}.$$

(b) Evaluate both deformations.

The two energies are $$\boxed{U(.20)=20(.20)^2+50(.20)^4=.80+.08=.88\ \mathrm J},$$ $$\boxed{U(.40)=3.20+1.28=4.48\ \mathrm J}.$$

(c) Keep the separate scaling powers.

The separate powers scale differently: $$U(2x)=4\left(\tfrac12kx^2\right)+16\left(\tfrac14\beta x^4\right),$$ $$\boxed{U(.40)/U(.20)=4.48/.88=56/11=5.09}.$$ The quartic contribution prevents the Hooke-law factor of four.