Conservation of Mechanical Energy — Question 6

Question 6

A nonlinear potential sets unequal-looking algebraic tasks

A 2 kg particle moves on a line with $U(x)=x^4+2x^2$ J when x is in meters. At x=0 its speed is $\sqrt3$ m/s, directed right. This is the only interaction doing work.

1. Find total mechanical energy.

2. Find both turning coordinates.

3. Find speed at x=0.50 m and explain why the negative algebraic value of $x^2$ must be rejected.

Question 6 - Solution

A nonlinear potential sets unequal-looking algebraic tasks

(a) Evaluate the initial state.

At the initial point, $U(0)=0$: $$\boxed{E=\tfrac12(2)(\sqrt3)^2+0=3\ \mathrm J}.$$

(b) Reduce the turning equation.

At a turning point, $K=0$ and $U=E$. With $z=x^2$, $$x^4+2x^2=3\Rightarrow z^2+2z-3=(z-1)(z+3)=0.$$ The physical root is $z=1$, giving $\boxed{x=\pm1\ \mathrm m}$.

(c) Recover K at the intermediate position.

At $x=.50$ m, $$U=\frac1{16}+\frac12=\frac9{16}\ \mathrm J,\qquad K=3-\frac9{16}=\frac{39}{16}\ \mathrm J.$$ $$\boxed{v=\sqrt{2K/m}=\frac{\sqrt{39}}4=1.561\ \mathrm{m/s}}.$$ The other algebraic root $x^2=-3$ is impossible for a real coordinate.