How to derive the kinetic formula

[physphys]

Homework Statement

There are other ways to derive the kinetic motion equation. show an alternative way using the equations of motion

Homework Equations

Ek= 1/2mv^2

The Attempt at a Solution

W= FD
W= MAD
= m( v2-v1/t)(V1+V2/2) T
= M/2 ( V2^2-V1^2)
= EK= MV^2/2

they need another one with the formulas of motion but i can't solve it. Thanks for the help

 

[Curious3141]

Homework Statement

There are other ways to derive the kinetic motion equation. show an alternative way using the equations of motion

Homework Equations

Ek= 1/2mv^2

The Attempt at a Solution

W= FD
W= MAD
= m( v2-v1/t)(V1+V2/2) T
= M/2 ( V2^2-V1^2)
= EK= MV^2/2

they need another one with the formulas of motion but i can't solve it. Thanks for the help

What formulae of motion have you been taught? List them out (there are three (sometimes four) "standard" ones), and you can see which is the relevant one to use.

Note that those all assume uniform acceleration (as did your solution). The derivation of the equation for kinetic energy can be made without this assumption, but it involves basic calculus. I don't think your question is asking you for this method, so we'll stick with the more basic one.

 

[physphys]

yes it's grade 11 physics. These are the ones they are talking about
v2= v1+ at
d= (v1+v2/2)t
d= v1t+ 1/2a(t)^2
d= v2t+1/2a(t)^2
v2^2=v1^2+2ad

 

[Levis2]

Maybe something along this, since it is a no-calc approach;

W=F*s
W=m*a*s with angle of attack =0, as with kinetic energy
we have from equation of motion that s=1/2*a*t^2
and also v=a*t
t=v/a
inserting in motion equation;
s=1/2*a*(v/a)^2
inserting this in work formula;
W=m*a*1/2*a*v^2/a^2
W=m*1/2*a^2*v^2/a^2
W=1/2*m*v^2

Just my initial guess, never seen it done :)

 

[physphys]

but isn't w= fd? where did fs come from?

 

[Levis2]

but isn't w= fd? where did fs come from?

I'm sorry, here in denmark we use F for force, and s for distance. I take it you use f for force and d for distance of course .. in my previous calculations, s=d :)

 

[Curious3141]

yes it's grade 11 physics. These are the ones they are talking about
v2= v1+ at
d= (v1+v2/2)t
d= v1t+ 1/2a(t)^2
d= v2t+1/2a(t)^2
v2^2=v1^2+2ad

The last equation is what you want. Set v1 (initial velocity) = 0 (since the particle starts from rest, when KE is zero). Set v2 to be equal to V, the final velocity. Manipulate a little till you have a*d on the right hand side in terms of V^2 on the left hand side.

From W (work which is the same as energy) = Fd and F = ma, you can immediately see W = m*a*d.

Now put that expression for a*d from the previous equation into this one and you'll see that W equals the expression you want. Simply conclude by stating that the work done on the object to bring it from rest to speed V is equal to the KE gained, so KE = (1/2)mV^2.

 

[physphys]

Thanks guys it worked out!

 

[Curious3141]

Thanks guys it worked out!

Of course, if you can use calculus, this is the best way to do it:

dW = Fdx = ma.dx = m\frac{dv}{dt}.dx = m\frac{dx}{dt}.dv = mv.dv

The second last step involved a change of variable of integration.

Now integrate:

\int_0^{E_k} dW = \int_0^V mvdv \Rightarrow E_k = \frac{1}{2}mV^2.

And it's actually a shorter proof despite it being a more general result (acceleration does not have to be constant here).