Question

A 2-kg particle moves along the x axis under the influence of a conservative force. The potential energy function of the particle is given by the formula U(x) = 10 + 5x2, where x is the coordinate (in m) as U is measured in Joules. What is the farthest point in the positive x direction that the particle can reach (in m) if at x = 2 m its kinetic energy is 50 J?

Answer 1

Answer:

$2\sqrt{5} \approx \rm 4.5\;m$.

Explanation:

Let the x-position of the farthest point be $x_\text{max}$. If there's no energy loss, then the mechanical energy of the point should be the same no matter where it is.

That is:

Mechanical energy at $\rm 2\; m$ = Mechanical energy at $x_\text{max}$.

Note that

Mechanical energy = kinetic energy + potential energy.

The question stated that kinetic energy at $\rm 2\; m$ = $\rm 50\; J$.

Additionally, the potential energy at $\rm 2\; m$ is equal to $U(2) = 10 + 5 x^2 = 10 + 2\times 5^2 = 60$.

Hence, the mechanical energy at $\rm 2\; m$ would be equal to $\rm 50 + 60 = 110\; J$.

If there's no energy loss, the mechanical energy of the particle should stay the same no matter where it is. In other words, the mechanical energy of the particle at $x_\text{max}$ would be the same as the mechanical energy at $2\; \rm m$.

Since the displacement is at its maximum value, the potential energy should also be at its maximum. All of the particle's mechanical energy would now be in the form of potential energy. That is:

$\begin{aligned}U(x_\text{max})=&\;\text{Potential Energy at} ~ x_\text{Max}\\ =&\;\text{Mechanical Energy at} ~ x_\text{Max} \\ =& \; \text{Mechanical Energy at}~ 2\; \text{m}\\ =& \; \rm 110\; J\end{aligned}$.

Since $U(x_\text{max})= \rm 110\; J$,

$10 + 5\left({x_\text{max}}\right)^{2} = 110$.

$\left(x_\text{max}\right)^2 = 20$.

$x_\text{max} = \sqrt{20} = 2\sqrt{5} \approx \rm 4.5\; m$ given that the question is asking for the displacement in the positive x-direction.