I Deriving formula for kinetic energy

AI Thread Summary
The discussion focuses on deriving the formula for kinetic energy, specifically the expression \( \frac{1}{2}mv^2 \), from given assumptions in classical physics. The user starts with the equations \( \ddot{z} = 0 \) and \( m\ddot{\textbf{r}} \cdot \dot{\textbf{r}} = 0 \), leading to the conclusion that kinetic energy is constant when no work is done. A constant applied force implies that acceleration and velocity maintain a consistent direction, allowing for the manipulation of the equations. The conversation emphasizes the relationship between the time derivative of kinetic energy and the conditions under which it remains unchanged. Ultimately, the thread seeks clarification on completing the derivation under these assumptions.
billard
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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.
 
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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.
 
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billard said:
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
 
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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}.##
 
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