A roller-coaster car of mass 1000 kg starts at height 20 m and uses energy conservation with negligible friction. What speed would it have at the bottom if it started from rest at the top and descended to bottom height 0 m? (v = sqrt(2 g h))

A roller-coaster car of mass 1000 kg starts at height 20 m and uses energy conservation with negligible friction. What speed would it have at the bottom if it started from rest at the top and descended to bottom height 0 m? (v = sqrt(2 g h))

• A. 18.5 m/s
• B. 20.0 m/s
• C. 19.8 m/s
• D. 22.1 m/s

Answer: (C)

Energy conservation in a frictionless roller coaster means the gravitational potential energy at the top converts fully into kinetic energy at the bottom. With no friction, m g h = 1/2 m v^2, so the mass cancels and v = sqrt(2 g h). Using h = 20 m and g ≈ 9.8 m/s^2 gives v = sqrt(2 × 9.8 × 20) = sqrt(392) ≈ 19.8 m/s. So the speed at the bottom is about 19.8 m/s. The mass doesn’t affect the outcome here; only the height difference and gravity matter in this idealized, frictionless scenario.

Energy conservation in a frictionless roller coaster means the gravitational potential energy at the top converts fully into kinetic energy at the bottom. With no friction, m g h = 1/2 m v^2, so the mass cancels and v = sqrt(2 g h). Using h = 20 m and g ≈ 9.8 m/s^2 gives v = sqrt(2 × 9.8 × 20) = sqrt(392) ≈ 19.8 m/s. So the speed at the bottom is about 19.8 m/s. The mass doesn’t affect the outcome here; only the height difference and gravity matter in this idealized, frictionless scenario.