A 2.0 kg block is sliding down a frictionless inclined plane that makes a 30° angle with the horizontal. If the block starts from rest at a height of 3.0 m, what is its speed at the bottom of the incline?

Core Concept

This is a conservation of mechanical energy problem. At the top, all energy is gravitational potential energy (mgh). At the bottom, all energy is kinetic energy (½mv²). Setting them equal: mgh = ½mv². The mass cancels out, giving v = √(2gh) = √(2 × 9.8 × 3.0) = √58.8 ≈ 7.7 m/s. Option A is too low, possibly from using h/2 instead of h. Option B might result from using sin(30°) incorrectly. Option D is the speed after falling 5 m, not 3 m.