physics/maths question -srry didn't get any help from physics forum

https://www.khanacademy.org/science/...w-ap-physics-1

So for this question, what I don't understand is why they don't use the equations separately for the 2 parts of the motion
..maybe i am not clear enough so in other words..
we can say there are 2 parts to the motion - the motion with the ramp is different to the motion on the rough horizontal surface.. but they use e.g.
K0​+U0​+WNC​​=K+U to represent the 2 parts as a whole

can someone clear this up for me? Thanks

page not found.

page not found.

So for this question, what I don't understand is why they don't use the equations separately for the 2 parts of the motion
..maybe i am not clear enough so in other words..
we can say there are 2 parts to the motion - the motion with the ramp is different to the motion on the rough horizontal surface.. but they use e.g.
K0​+U0​+WNC​​=K+U to represent the 2 parts as a whole

can someone clear this up for me? Thanks
Here is the question below

I guess energy is conserved everywhere, so the LHS of the equation is at the ramp, which should be the same as the total energy on the flat surface at rest (the RHS)

help.docx
there are 2 pages to it thanks1!!!!!!

Sure you can split it up to find the velocity when it reaches the bottom of the ramp, and then find it's decceleration.
This is just an alternative method. Nothing wrong with that.
It's a common theme in mechanics that problems can be tackled in more than one way.

Hi! that's what I actually tried before but I just couldn't get the velocity when it reaches the bottom of the ramp -that's what limited me

SUVAT?
Just make sure you work out the distance it actually goes down the ramp (and not just vertical distance)
And make sure you get the component of acceleration right.

It's been a while from my M1 exam
ok so do we split up the acceleration into components? and then i dunno what to do
can you help me with the first part of the motion please? am a bit lost lol

Ignore my prev comment, I didn't realise it didn't give you a horizontal length.
You have to do it their way working strictly in terms of energy.
So you have to find the GPE lost = mgh
and find a speed by saying KE= GPE lost
and KE= 1/2 mv^2
Solve for v will give you its velocity at the bottom of the ramp