What is the kinetic energy of the 2 kg block at the bottom after sliding down 5 m on a frictionless track?

What is the kinetic energy of the 2 kg block at the bottom after sliding down 5 m on a frictionless track?

• A. 49 J
• B. 19.6 J
• C. 98 J
• D. 196 J

Answer: (C)

When a block slides along a frictionless track, gravitational potential energy lost equals kinetic energy gained. Since it starts from rest, all the drop in potential energy becomes kinetic energy at the bottom: K = m g h. Compute: m = 2 kg, g ≈ 9.8 m/s^2, h = 5 m, so K = 2 × 9.8 × 5 = 98 joules. That’s the energy the block has at the bottom. The other values would come from different heights or masses (for example, 49 J would correspond to a smaller height or mass, 196 J to a larger height or mass). If there were friction, some energy would be lost and the kinetic energy would be less than m g h.

When a block slides along a frictionless track, gravitational potential energy lost equals kinetic energy gained. Since it starts from rest, all the drop in potential energy becomes kinetic energy at the bottom: K = m g h.

Compute: m = 2 kg, g ≈ 9.8 m/s^2, h = 5 m, so K = 2 × 9.8 × 5 = 98 joules. That’s the energy the block has at the bottom.

The other values would come from different heights or masses (for example, 49 J would correspond to a smaller height or mass, 196 J to a larger height or mass). If there were friction, some energy would be lost and the kinetic energy would be less than m g h.