Verifying Results w/ Newton's & Energy Laws: Justifying Hypotenuse=D

[ChaoticNeutralStuden]

Homework Statement

A block of mass m is released from rest at the top of a ramp with a slope angle θ and height h. The friction between the block and the ramp is small and can be disregarded. After descending the ramp, the block travels a horizontal distance D (in this section the friction is not negligible) until it comes to rest. Determine, using the information given above, the coefficient of kinetic friction between the block and the horizontal plane using:
A) Newton's laws.
B) The energy equations.

Relevant Equations

Newton's laws equations & energy equations

Using the Newton's laws my result was h/hypotenuse and using the energy equations my result was h/D. My results are right ? And if so, how i justify the hypotenuse = D . Sorry for the bad english, the homework was poorly translated.

 

[haruspex]

Using the Newton's laws my result was h/hypotenuse and using the energy equations my result was h/D. My results are right ? And if so, how i justify the hypotenuse = D . Sorry for the bad english, the homework was poorly translated.

Your answer is right (probably... see below), but your explanation of how you got to it makes no sense. How did you get h/hypotenuse from Newton's laws without even knowing what distance 'hypotenuse' refers to in this case?
Please show your working.

There is one awkward thing about the question: it shows the ramp turning to horizontal abruptly. In practice, that would mean an impact; only the horizontal component of velocity would be conserved. But you are probably expected to treat it as a smooth transition.

 

[ChaoticNeutralStuden]

Using Newton's Law :
At the first moment(ramp):
ΣFx : m*g(sin θ )=m*a
a= g*(sin θ )
ΣFy = N-mg(cos θ )= 0
N=mg(cos θ )
When the block starts to travel D
ΣFy = N-mg=0
N=mg
Friction force = N*uc
Friction force = mg*uc

ΣFx : m*g(sin θ )- mg*uc =0
m*g(sin θ )= mg*uc
uc=sin θ
sin θ = h/ hypotenuse
uc= h/ hypotenuse

Using energy equations
At the first moment:
W = m*g(sin θ ) * h/sin θ = mgh
mgh = mv^2/2
v = (2gh)^(1/2)
When the block starts to travel D:
W= -(m*g*uc*D) = - (mv^2)/2

mgucD = m(2gh)/2
uc=h/D

I tried to type my working because I'm without my cellphone to take pictures of my notebook at the moment, I don't know if it's understandable, I considered the hypotenuse by getting the result of the coefficient of kinetic friction = sin θ, and by that considering sin θ = h/hypotenuse. And yeah, we should consider as a smoth transition.

 

[haruspex]

ΣFx : m*g(sin θ )- mg*uc =0

Those two forces act at different times and in different directions . It makes no sense to add or subtract them.

I tried to type my working because I'm without my cellphone to take pictures of my notebook

Good! Images are supposed to be for textbook extracts and diagrams only.

 

[ChaoticNeutralStuden]

Those two forces act at different times and in different directions . It makes no sense to add or subtract them.

Good! Images are supposed to be for textbook extracts and diagrams only.

So... I'm kinda lost now, how do I find the coefficient of friction using the Newton's law in this situation? I considered that a force of the same magnitude of mgsin θ was acting when the block reachs the beginning of the horizontal plane. I know i can reach the result by using kinemactic equations but I don't know if this is what the professor wants...
Edit :
If I considered :
ΣFx : - mg*uc = m *a
and a = ( g(sin θ ) ) , what is wrong with this?

 

[haruspex]

Think about energy.