Posted by Becky on .
A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track.It leaves the track horizontally, flies through the air, and subsequently strikes the ground. gravity= 9.81 m/s^2.
height1=4.1m is initially dropped from here
height2=2.4m part where curve flattens out
mass of block=0.546kg
What is the speed v of the block when it
leaves the track (in m/s)?
What is the horizontal distance x the block travels in the air (in m)?
What is the speed of the block when it hits the ground (in m/s)?



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Physics (Algebrabased) 
Elena,
PE → KE1
KE1 → Work(fr) +KE2 =>
PE → Work(fr) +KE2
If the initial height is ‘h₁’,
the height of the horizontal section is ‘h₂’,
the length of the rough section is ‘s’, then
mg (h₁h₂)=μmgs +mv²/2,
v=sqrt{ 2g[(h₁h₂)μs]}.

h₂=gt²/2,
t=sqrt(2h₂/g).
x=vt,
v(y)=gt.
V(near the ground)=sqrt{v²+v(y)²} 
Physics (Algebrabased) 
Becky,
Thank you!