Conservation of mechanical energy on a frictionless incline

Steps:

• The block loses gravitational potential energy mgh, where h = x sin θ is the vertical height drop.
• This energy converts fully to kinetic energy (1/2)mv² since friction is negligible.
• Set mgh = (1/2)mv², so gh = (1/2)v².
• Substitute h = x sin θ to get v = √(2gx sin θ).

Why B is correct:

• It derives directly from energy conservation, equating potential energy loss to kinetic energy gain over height x sin θ.

Why the others are wrong:

• A: Underestimates speed by dividing the correct v² by 4, ignoring full energy conversion.
• C: Assumes vertical drop of x, neglecting the incline angle θ.
• D: Gives v² instead of v, omitting the square root from the kinetic energy formula.

Final answer: B