Vertical cross secion?

Question

a vertical cross section LMN is fixed on a smooth surface. M is the lowest point of the cross-section.L is 2.45m above the level of M, and N is 1.2m above the level of M. A particle of mass 0.5kg is released from the test at L and moves on the surface until it leaves it at N. Find

(1) the kinetic energy of the particle at N

(2) Find the least value of v for which the particle will reach L.

Answer 1

(1) 0.5(9.81)(2.45 - 1.2) = 6.13125 J

(2) Going the other way, the least value of V at N which will get the particle to L is
v = √(2*6.13125/0.5) = 4.9523 m/s