How does my book get ##\frac{1}{2}## by this derivation?

[n3pix]

TL;DR

A question about derivation of the formula of Work.

The integral is called the line integral of $F$ from $A$ to $B$. The work done in the displacement by the force is defined as,

$W(A\rightarrow B)=\int_A^B \vec{F}.dr$

where the limits $A$ and $B$ stand for the positions $r_A$ and $r_B$.

We now return to the free particle subject
to forces. We want to generalize $Eq. (5.6)$, which we here repeat,

$\frac{1}{2}Mv^2-\frac{1}{2}Mv_0^2=\vec{F}.(y-y_0)$

to include applied forces that vary in direction and magnitude but are known as functions of position throughout the region where the motion occurs. By substituting $\vec{F}=M\frac{d\vec{v}}{dt}$ into $Eq. (5.12)$, where $\vec{F}$ is the vector sum of the forces, we find for the work done by these forces,

$W(A\rightarrow B)=M\int_A^B \frac{d\vec{v}}{dt}.d\vec{r}$

Now

$d\vec{r}=\frac{d\vec{r}}{dt}.dt=\vec{v}dt$

So that

$W(A\rightarrow B)=M\int_A^B (\frac{d\vec{v}}{dt}.v)dt$

where the limits $A$ and $B$ now stand for the times $t_A$ and $t_B$ when the particle is at the positions designated by $A$ and $B$. But we can rearrange the integrand,

$\frac{d}{dt}v^2=\frac{d}{dt}(\vec{v}.\vec{v})=2\frac{d\vec{v}}{dt}.\vec{v}$

Question 1: In the above equation, how could we gather $\vec{v}$ because one of them is in the derivation ($\frac{d\vec{v}}{dt}$) and another one is free ($\vec{v}$) and together they do $\frac{d}{dt}(\vec{v}.\vec{v})$. Is it legal to do?

so that

$2\int_A^B(\frac{d\vec{v}}{dt}.\vec{v})dt=\int_A^B(\frac{d}{dt}v^2)dt=\int_A^Bd(v^2)=(v_B^2-v_A^2)$

Question 2: Here, there were $2$ in the first equation but where is it in the second equation? It's confusing...

On substitution in $Eq. (5.14)$ we have an important result:

$W(A\rightarrow B)=\int_A^B \vec{F}.d\vec{r}=\frac{1}{2}Mv_B^2-\frac{1}{2}Mv_A^2$

for the free particle. This is a generalization of $Eq. (5.6)$. We recognize,

$K=\frac{1}{2}Mv^2$

Thanks...

 

[PeroK]

$\frac{d}{dt}v^2=\frac{d}{dt}(\vec{v}.\vec{v})=2\frac{d\vec{v}}{dt}.\vec{v}$

Question 1: In the above equation, how could we gather $\vec{v}$ because one of them is in the derivation ($\frac{d\vec{v}}{dt}$) and another one is free ($\vec{v}$) and together they do $\frac{d}{dt}(\vec{v}.\vec{v})$. Is it legal to do?

In general:

$\frac{d}{dt}(\vec{u}.\vec{v})= \vec u \cdot \frac{d\vec{v}}{dt} + \frac{d\vec{u}}{dt} \cdot \vec{v}$

Hence:

$\frac{d}{dt}(\vec{v}.\vec{v})= 2\vec v \cdot \frac{d\vec{v}}{dt}$

so that

$2\int_A^B(\frac{d\vec{v}}{dt}.\vec{v})dt=\int_A^B(\frac{d}{dt}v^2)dt=\int_A^Bd(v^2)=(v_B^2-v_A^2)$

Question 2: Here, there were $2$ in the first equation but where is it in the second equation? It's confusing...

On substitution in $Eq. (5.14)$ we have an important result:

$W(A\rightarrow B)=\int_A^B \vec{F}.d\vec{r}=\frac{1}{2}Mv_B^2-\frac{1}{2}Mv_A^2$

for the free particle. This is a generalization of $Eq. (5.6)$. We recognize,

$K=\frac{1}{2}Mv^2$

Thanks...

I don't understand your question here. That's a just a straight integration of the above equation.

 

[n3pix]

In general:

$\frac{d}{dt}(\vec{u}.\vec{v})= \vec u \cdot \frac{d\vec{v}}{dt} + \frac{d\vec{u}}{dt} \cdot \vec{v}$

Hence:

$\frac{d}{dt}(\vec{v}.\vec{v})= 2\vec v \cdot \frac{d\vec{v}}{dt}$
I don't understand your question here. That's a just a straight integration of the above equation.

I mean, in this quote;

$W(A\rightarrow B)=M\int_B^A (\frac{d\vec{v}}{dt}.\vec{v})dt$

where the limits $A$ and $B$ now stand for the times $t_A$ $t_A$ when the particle is at the positions designated by $A$ and $B$. But we can rearrange the integrand,

$\frac{d}{dt}{\vec{v}}^2=\frac{d}{dt}(\vec{v}.\vec{v})=2\frac{d\vec{v}}{dt}.\vec{v}$

In the integral ($W(A\rightarrow B)=M\int_B^A (\frac{d\vec{v}}{dt}.\vec{v})dt$) the integrand is $(\frac{d\vec{v}}{dt}.\vec{v})$ and it can not be equal to $(\frac{d}{dt}{\vec{v}}^2)$. But the derivation says that it is equal to this. Check it again please.

In my second question, I mean that

$2\int_A^B(\frac{d\vec{v}}{dt}.\vec{v})dt=\int_A^B(\frac{d}{dt}v^2)dt=\int_A^Bd(v^2)=(v_B^2-v_A^2)$

In this integral, first stage has $2$ ($2\int_A^B(\frac{d\vec{v}}{dt}.\vec{v})dt$) but in the second stage it doesn't have $2$ ($=\int_A^B(\frac{d}{dt}v^2)dt$) Where the $2$ gone?

And I want to add another question.

In the last equation ($W(A\rightarrow B)=\int_A^B \vec{F}.d\vec{r}=\frac{1}{2}Mv_B^2-\frac{1}{2}Mv_A^2$) Where the $\frac{1}{2}$ came from? I can't see any $\frac{1}{2}$ in the all of the derivation.

Thanks again.

 

[PeroK]

In this integral, first stage has $2$ ($2\int_A^B(\frac{d\vec{v}}{dt}.\vec{v})dt$) but in the second stage it doesn't have $2$ ($=\int_A^B(\frac{d}{dt}v^2)dt$) Where the $2$ gone?

Just look at the previous equation:

$\frac{d}{dt}(\vec{v}.\vec{v})= 2\vec v \cdot \frac{d\vec{v}}{dt}$

That's where the $2$ comes from!

The final $\frac 1 2$ arises simply by dividing both sides of the equation by $2$.