A 2.0 kg cart is released from rest at a height of 0.8 m on a frictionless track. Ignore air resistance. What speed does it reach at the bottom?

A 2.0 kg cart is released from rest at a height of 0.8 m on a frictionless track. Ignore air resistance. What speed does it reach at the bottom?

• A. 2.0 m/s
• B. 3.0 m/s
• C. 3.96 m/s
• D. 4.5 m/s

Answer: (C)

Conservation of mechanical energy on a frictionless track means the gravitational potential energy the cart has at the top becomes kinetic energy at the bottom. Start with m g h = 1/2 m v^2, with the cart released from rest so initial kinetic energy is zero. The mass cancels, giving v^2 = 2 g h, so v = sqrt(2 g h). Plugging in g ≈ 9.8 m/s^2 and h = 0.8 m: v ≈ sqrt(2 × 9.8 × 0.8) = sqrt(15.68) ≈ 3.96 m/s. The mass doesn’t affect the speed, since the track is frictionless and energy is conserved. The cart reaches about 3.96 m/s at the bottom.

Conservation of mechanical energy on a frictionless track means the gravitational potential energy the cart has at the top becomes kinetic energy at the bottom. Start with m g h = 1/2 m v^2, with the cart released from rest so initial kinetic energy is zero. The mass cancels, giving v^2 = 2 g h, so v = sqrt(2 g h). Plugging in g ≈ 9.8 m/s^2 and h = 0.8 m: v ≈ sqrt(2 × 9.8 × 0.8) = sqrt(15.68) ≈ 3.96 m/s. The mass doesn’t affect the speed, since the track is frictionless and energy is conserved. The cart reaches about 3.96 m/s at the bottom.