The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same?

Question

please see the link for question

http://answerboard.cramster.com/physics-topic-5-94227-0.aspx

Answer 1

A) All potential energy since it is not moving
B) Some of each since it is moving horizontally
C) H, since all of the kinetic energy gets converted to potential energy on the longer track
You can also derive this
H/Sin(q)=v0*t-.5*sin(q)*g*t^2
while
H1/Sin(q)=v0*t1-.5*sin(q)*g*t1^2
and
H2=H1+sin(v0-sin(q)*t1)*t2-.5*g*t^2

You can show that for any t>t1 H>(H1+H2)

D) for the longer track v(t)=v0*t-sin(q)*g*t
and the H(t) is given above
solve for H(t) when v(t)=0

E) using the equation above for H2, solve for t1 and compute v1(t1) on the shorter track as
v1(t1)=v0*t1-sin(q)*g*t1
then compute H2 when the vertical velocity is zero.

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