Conserved Energy in a moving frame of reference

[Doc Al]

mg in the -y direction
that is why the sin is there

The weight acts downward, so is already totally in the y direction. No need for any sinθ.

mg sin theta cos theta in the y
mg cos ^2 theta in the x

I think you have these reversed.

 

[Tinhorn]

so mg for weight

and mg sin theta cos theta in the x
mg cos ^2 theta in the y

how does that make fr = mgsinθcosθ x+ mgsin^2y

 

[Tinhorn]

Can i say mgsinθ is the force moving the box down the slope

since the x and y components of the force A is
$A_{x}$= A*cosθ
$A_{y}$= A*sinθ

I can say mg sin θ = mg sinθcosθ + $mgsin^{2}$

and since weight or gravity is also working on the box
which is -mg on the y component.

dot product would be

$mgsinθcosθ _{x}$- $mgsin^{2} _{y}$
which is the resultant (i copied the sign wrong last time)

 

[Doc Al]

Can i say mgsinθ is the force moving the box down the slope

since the x and y components of the force A is
$A_{x}$= A*cosθ
$A_{y}$= A*sinθ

I can say mg sin θ = mg sinθcosθ + $mgsin^{2}$

and since weight or gravity is also working on the box
which is -mg on the y component.

dot product would be

$mgsinθcosθ _{x}$- $mgsin^{2} _{y}$
which is the resultant (i copied the sign wrong last time)

That's the correct resultant, but I don't follow your reasoning. Here's how you get it by adding the force components you found earlier:

Add the x components (there's only one): mgcosθsinθ
Add the y components: mgcos²θ - mg = -mgsin²θ

 

[Tinhorn]

I get it.
but i had always thought the x and y components of the force A is
Ax= A*cosθ
Ay= A*sinθ

why does normal force have sin in its component
and cos in its component

 

[Doc Al]

I get it.
but i had always thought the x and y components of the force A is
Ax= A*cosθ
Ay= A*sinθ

Right, when θ is given with respect to the horizontal. (It's reversed when θ is with respect to the vertical.)

why does normal force have sin in its component
and cos in its component

Using your notation, A is the normal force, thus A = mgcosθ. When you find its components, you'll get the additional sine and cosine factors.

 

[Tinhorn]

One last question

if ΔKE = $\frac{1}{2}$$mv^{2}_{x}$ + $\frac{1}{2}$$mv^{2}_{y}$

why is ΔKE also -$\frac{1}{2}$$mv^{2}_{x}$ + $\frac{1}{2}$$mv^{2}_{y}$

should'nt it be -ΔKE

 

[Doc Al]

One last question

if ΔKE = $\frac{1}{2}$$mv^{2}_{x}$ + $\frac{1}{2}$$mv^{2}_{y}$

That's an expression for KE, not ΔKE, right?

why is ΔKE also -$\frac{1}{2}$$mv^{2}_{x}$ + $\frac{1}{2}$$mv^{2}_{y}$

should'nt it be -ΔKE

I don't know what you are asking.

 

[Tinhorn]

got it
wasnt clear about the question but i got it