Deriving formula for kinetic energy

[billard]

TL;DR

Issue deriving 1/2 *mv^2 from some pre-assumed equations

Hello! I am new to the differential version of classical physics, and I am trying to work how to derive kinetic energy from some pre-assumed equations:

Assume that we know: $\ddot{z} = 0$ and $m\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = 0$This results in $$\frac{1}{2}m\dot{r}^2 = W = const.$$

How is the kinetic energy given here with our pre-assumptions? I am sure this is very simple, forgive me, I am a beginner.

 

[anuttarasammyak]

Work done per time is innerproduct of Force F and velocity v. F =ma. When no work done, you can deduce that KE is conserved.

 

[topsquark]

TL;DR Summary: Issue deriving 1/2 *mv^2 from some pre-assumed equations

Hello! I am new to the differential version of classical physics, and I am trying to work how to derive kinetic energy from some pre-assumed equations:

Assume that we know: $\ddot{z} = 0$ and $m\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = 0$This results in $$\frac{1}{2}m\dot{r}^2 = W = const.$$

How is the kinetic energy given here with our pre-assumptions? I am sure this is very simple, forgive me, I am a beginner.

If the applied force is constant, then $\ddot{\textbf{r}}$ is constant, and in particular, so is the direction of $\dot{\textbf{r}}$, ala Newton's 2nd.

So
$\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = \ddot{r} \dot{r} \, cos( \theta )$
where $\theta$ is the angle between the acceleration and the velocity is constant.

So now you have
$m\ddot{r} \dot{r} cos( \theta ) = ( m \, cos( \theta ) ) \ddot{r} \dot{r} = 0$

Now note that
$\dfrac{d}{dt} ( \dot{r} )^2 = 2 \ddot{r} \dot{r}$

Can you finish?

-Dan

 

[kuruman]

Start with $K=\frac{1}{2}m(\mathbf{\dot r}\cdot \mathbf{\dot r}).$
Can you show that $\dfrac{dK}{dt}=0~$ if $~\mathbf{\dot r}\cdot \mathbf{\ddot r}=0~?$

Here assume that $\mathbf{r}=x~\mathbf{\hat x}+y~\mathbf{\hat y}+z~\mathbf{\hat z}.$