- #1
n3pix
- 15
- 1
- TL;DR Summary
- 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...
##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...