Proving that total energy is constant?

Question

A mass $m$ is moving along the $x$-axis under the influence of a force $F(x)$ according to Newton’s Second Law, i.e., $F = ma$ . Note that $F(x)$ depends only on $x$, so that it is a conservative force, i.e. there exists a potential energy function $U(x)$ such that $F(x) = -U'(x)$ . Prove that the total energy $E(t)$ , i.e. the sum of the kinetic energy and the potential energy, is constant.

So far,I've defined $V(x)$ as velocity and said the $V'(x) = a$. From there kinetic energy $K(x) = (1/2)m V(x)^2$. The derivative of $K(x)$ is $mV'(x)$ or $ma$ which is equal to $F(x)$. For $E(t)$ to be constant, its derivative must be 0. I used the equation $E(t) = K(x) + U(x)$ and found the first derivative: $K'(x) - (-U'(x))$ which is equal to $F(x) - F(x) = 0$. Therefore, the total energy is constant.

Is this a valid solution?

Answer 1

Assuming $\dot{m} = 0$ (e.g. no rocket), we get: \begin{align} \frac{dE}{dt} &= \frac{d}{dt} \left( \frac{m}{2}{\dot{x}^2} + U(x) \right) \\ &= m \, \dot{x} \, \ddot{x} + \dot{U} \\ &= F \, \dot{x} + \dot{U}\\ &= - \frac{dU}{dx} \frac{dx}{dt} + \dot{U}\\ &= - \dot{U} + \dot{U} \\ &= 0 \end{align}

Answer 2

The solution you posted is not valid. To show that energy is conserved, you must show that the derivative of the energy function with respect to time is equal to zero. Also, the acceleration is the derivative of the velocity with respect to time.

You know that

\begin{align*} E(t) & = K(t) + U(t)\\ & = \frac{1}{2}mv^2 + U(x(t))\\ \end{align*}

where $m$ is the mass. The velocity, $v$, is the derivative of the position, $x$, with respect to time. The acceleration, $a$, is the derivative of the velocity with respect to time. We write

\begin{align*} v & = \dot{x}\\ a & = \dot{v} = \ddot{x} \end{align*} where the dot signifies that the derivative is taken with respect to time and the double dot represents the second derivative with respect to time.

You also know that the derivative with respect to $x$ of the potential energy $U$ is $-F(x)$, where $F(x) = ma$ is the force. Thus,

$$\frac{dU}{dx} = -F(x) = -ma$$

Since the force depends only on $x$, the mass is not changing. Hence, $\dot{m} = 0$. Differentiating implicitly with respect to time yields

\begin{align*} \dot{E}(t) & = \dot{K}(t) + \dot{U}(t)\\ & = \frac{1}{2}\dot{m}v^2 + 2 \cdot \frac{1}{2}mv\dot{v} + \frac{dU}{dx}\frac{dx}{dt}\\ & = 0 + mva - Fv\\ & = mav - Fv\\ & = Fv - Fv\\ & = 0 \end{align*}

where we apply the Product Rule to differentiate

$$K(t) = \frac{1}{2}mv^2$$

and the Chain Rule to differentiate $U(x(t))$.