A 5 kg block on a frictionless surface experiences a constant acceleration of 2.4 m/s^2 for 4 s. What is its kinetic energy at the end?

A 5 kg block on a frictionless surface experiences a constant acceleration of 2.4 m/s^2 for 4 s. What is its kinetic energy at the end?

• A. 180.0 J
• B. 210.0 J
• C. 225.0 J
• D. 230.4 J

Answer: (D)

Kinetic energy depends on speed, not acceleration. If the block starts from rest and accelerates at 2.4 m/s^2 for 4 s, its final speed is v = at = 2.4 × 4 = 9.6 m/s. Then the kinetic energy is K = 1/2 m v^2 = 1/2 × 5 × (9.6)^2 = 2.5 × 92.16 = 230.4 J. The frictionless surface means all the work done goes into kinetic energy, with no losses. So the kinetic energy at the end is 230.4 joules.

Kinetic energy depends on speed, not acceleration. If the block starts from rest and accelerates at 2.4 m/s^2 for 4 s, its final speed is v = at = 2.4 × 4 = 9.6 m/s. Then the kinetic energy is K = 1/2 m v^2 = 1/2 × 5 × (9.6)^2 = 2.5 × 92.16 = 230.4 J. The frictionless surface means all the work done goes into kinetic energy, with no losses. So the kinetic energy at the end is 230.4 joules.