Question

stuck with the following equation:

Answer 1

\[mdv . v = d \left( \frac{ mv^2 }{ 2 } \right)\]
the left side: the v is a velocity vector.. I don't understand where the 2 at the bottom comes from @ivancsc1996

Answer 2

So the left side is a mass times (infinitesimal of a v vector dot product v is vector)= the infinitesimal of kinetic energy?

Answer 3

Where did you get the equation and what do you want with it?

Answer 4

ur equation should be an integral.. ^_^

Answer 5

\[\int\limits_{}^{}m vdv \]

Answer 6

oh hoki oki.. i get it.. !!

Answer 7

see. if you integrate the right side.. u end up with v^2/2 right?

Answer 8

we are trying to make the connection between work and velocity first so this is what is written in the book:
\[\delta W= \left( \sum_{i=1}^{n} F _{i}\right) . dR = ma . dR = m \frac{ dV }{ dt } . dR= mdV .\frac{ dR }{ dt } = mdV . V = d(\frac{ mv^2 }{ 2 })\]

Answer 9

\[\int\limits_{}^{}m.vdv=mv^2/2\]

there fore.. if you now differentiate on both sides.. left hand side the integral disappears and right side the differential appears

Answer 10

but why should I integrate all of a sudden?

Answer 11

its a way of writing.. they have not INTEGRATED as such.

for example..

integral of x = x^/2

so i can also write

x = differential of (x^2/2).. thats what that they have done!!

Answer 12

@bronzegoddess what do you wnat to understand of the equation.

Answer 13

what is the scalar product of mdV. V, I dont understand the logic behind the last two

Answer 14

because the V is a vector and all other capital letters are vectors

Answer 15

@Mashy i understand your integration however, the d is still there unless they made a typo

Answer 16

wait, i think i understand what you mean: vdv is the same as (v^2)/2 if we use an integration property

Answer 17

am confused :( the whole point though is to find the connection between work and change in velocity.

Answer 18

sorry i was playing.. OK..
fine.. lets see it this way


work done = force times displacement..

force = mass times acceleration = mass times (rate of change of velocity)

thus

work = Mas X rate of change of velocity X displacement..


hence

\[dw = m (dv/dt) dx = m dv (dx/dt) = m dv v= mvdv\]

Answer 19

yes i understand all of that, my problem is the last part \[d \left( \frac{ mv^2 }{ 2 } \right) \]

Answer 20

@Mashy, sorry my dad explained what you were just explain, thank you for the patience and help!