Work Done by a Spring Force — Question 10

Question 10

Work bounds depend on both stiffness and deformation

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. An ideal spring stiffness lies in $[190,210]$ N/m. A compressed endpoint has extension $x_i\in[-0.11,-0.09]$ m and the final extension is exactly $x_f=0.05$ m. The bounds are independent.

1. Is spring work necessarily positive?

2. Find the tight minimum and maximum work.

3. Explain why the most negative initial extension gives the largest squared deformation.

Question 10 - Solution

Work bounds depend on both stiffness and deformation

(a) Compare deformation magnitudes.

The governing relation is $$\boxed{W_s=\tfrac12k(x_i^2-x_f^2),\qquad x_i^2\in[.09^2,.11^2]}$$ Every allowed $|x_i|$ exceeds 0.05 m, so $$W_s=\tfrac12k(x_i^2-.05^2)>0.$$

(b) Choose compatible extreme inputs.

$$W_{\min}=\tfrac12(190)(.09^2-.05^2)=0.532\ \mathrm J.$$

$$W_{\max}=\tfrac12(210)(.11^2-.05^2)=1.008\ \mathrm J.$$ Independence makes both endpoint combinations allowable.

(c) Square only after interpreting the sign.

Negative x labels compression, not negative stored capacity for spring work. On this negative interval, moving x toward $-0.11$ increases $x^2$; using the numerically largest x for the largest work reverses the bound.