Question: A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
Options:
Correct Answer: 10 m/s
Solution:
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.