Conservation of Mechanical Energy — Question 7

Question 7

Inferring stiffness from two moving states

A 1 kg slider attached to a horizontal ideal spring moves without friction. Its speed is 3 m/s at extension 0.10 m and 1 m/s at extension 0.30 m.

1. Infer the spring stiffness.

2. Find the turning amplitude.

3. Could the same motion have speed 2 m/s at extension 0.20 m?

Question 7 - Solution

Inferring stiffness from two moving states

(a) Eliminate the unknown total energy.

Equate total energies and eliminate the common total: $$\tfrac12m(v_1^2-v_2^2)=\tfrac12k(x_2^2-x_1^2).$$ $$\boxed{k=\frac{1(9-1)}{.30^2-.10^2}=100\ \mathrm{N/m}}.$$

(b) Determine the conserved total.

The common energy is $$E=\tfrac12(1)(3)^2+\tfrac12(100)(.10)^2=5\ \mathrm J.$$ $$\tfrac12kA^2=E\Rightarrow\boxed{A=\sqrt{.10}=.3162\ \mathrm m}.$$

(c) Check the third state against the same E.

At $x=.20$ m, $$U=50(.20)^2=2\ \mathrm J,\quad K=5-2=3\ \mathrm J\Rightarrow\boxed{v=\sqrt6\ \mathrm{m/s}}.$$ The proposed 2 m/s gives $E=\tfrac12(1)(2)^2+2=4$ J, contradicting the conserved 5 J.