Conserved Energy in a moving frame of reference

[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