A 2 kg block is released from rest at height 5 m on a frictionless track. What is its speed at the bottom?

A 2 kg block is released from rest at height 5 m on a frictionless track. What is its speed at the bottom?

• A. About 7.0 m/s
• B. About 12 m/s
• C. About 9.9 m/s
• D. About 14 m/s

Answer: (C)

On a frictionless track, mechanical energy is conserved, so the loss in gravitational potential energy equals the gain in kinetic energy. That gives m g h = 1/2 m v^2, and the mass cancels, leaving v = sqrt(2 g h). With h = 5 m and g ≈ 9.8 m/s^2, v ≈ sqrt(2 × 9.8 × 5) = sqrt(98) ≈ 9.9 m/s. The speed at the bottom depends only on the height dropped, not on the mass, so the block reaches about 9.9 m/s.

On a frictionless track, mechanical energy is conserved, so the loss in gravitational potential energy equals the gain in kinetic energy. That gives m g h = 1/2 m v^2, and the mass cancels, leaving v = sqrt(2 g h). With h = 5 m and g ≈ 9.8 m/s^2, v ≈ sqrt(2 × 9.8 × 5) = sqrt(98) ≈ 9.9 m/s. The speed at the bottom depends only on the height dropped, not on the mass, so the block reaches about 9.9 m/s.