1/2 m vi^2 + mgh - mu mg cosθ = 1/2 m vf^2
And you notice all the m's cancel.
a block slides along a path that is without friction until the block reaches the section of length L = 0.75 m,
which begins at height h = 2.0 m on a ramp of angle θ = 30°. In that section, the coefficient of kinetic friction is 0.40. The
block passes through point A with a speed of 8.0 m/s. If the block can reach point B (where the friction ends), what is its
speed there, and if it cannot, what is its greatest height above A?
1 answer