Question: A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
Options:
Correct Answer: 10 m/s
Solution:
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy at the bottom (1/2 mv²). Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 5) ≈ 10 m/s.