Please help me with this physics review question. please explain.?

[Physics Guy]

A block with mass m = 12.3 kg slides down an inclined plane of slope angle 36.4 ° with a constant velocity. It is then projected up the same plane with an initial speed 2.60 m/s. How far up the incline will the block move before coming to rest?

[m_bud]

since the block slide down at a constant speed, all forces are balance.

the force that causes the block to slide down is
sin(36.4)mg, which is equal to the friction force

friction = ? cos(36.4)mg

?cos(36.4)mg = sin(36.4)mg
?cos(36.4) = sin(36.4)
? = 0.737

final energy = initial energy - work done by friction

the final energy of the block is the potential energy
initial Energy is the kinetic energy
work done by friction = ?cos(36.4)mg sin(36.4)/h

mgh = .5mv² - ?cos(36.4)mg sin(36.4)/h
gh = .5v² - 0.737 cos(36.4)g sin(36.4)/h
9.8h = .5(2.6)² - 0.737 cos(36.4)(9.8) h/sin(36.4)
9.8h = 3.38 - 9.79h
19.59h = 3.38
h = 0.172m [this is the altitude]

sin(36.4) = h/L
L = .172 / sin(36.4)
L = 0.29m <== answer

hope it helps :-)

[m_bud]

since the block slide down at a constant speed, all forces are balance.

the force that causes the block to slide down is
sin(36.4)mg, which is equal to the friction force

friction = ? cos(36.4)mg

?cos(36.4)mg = sin(36.4)mg
?cos(36.4) = sin(36.4)
? = 0.737

final energy = initial energy - work done by friction

the final energy of the block is the potential energy
initial Energy is the kinetic energy
work done by friction = ?cos(36.4)mg sin(36.4)/h

mgh = .5mv² - ?cos(36.4)mg sin(36.4)/h
gh = .5v² - 0.737 cos(36.4)g sin(36.4)/h
9.8h = .5(2.6)² - 0.737 cos(36.4)(9.8) h/sin(36.4)
9.8h = 3.38 - 9.79h
19.59h = 3.38
h = 0.172m [this is the altitude]

sin(36.4) = h/L
L = .172 / sin(36.4)
L = 0.29m <== answer

hope it helps :-)