A 6 N box is projected with a velocity of 2 m/s at the bottom of the plane inclined 30 degrees above the horizontal. If friction is neglected, how high up along the plane will it travel before coming to rest?

Answer 1

Since friction is negligible, use conservation of potential and kinetic energies.

KE (at bottom) => PE (at top)

mv²/2 = mg*Lsin(θ)
v=2,g=9.81,sin(θ)=1/2
Solve for L. I get 0.82m