A body weighing 2 kg freely falls to the surface of the earth for 6 seconds. What is the kinetic energy

A body weighing 2 kg freely falls to the surface of the earth for 6 seconds. What is the kinetic energy of the body at the moment of falling?

To find out the required value of the kinetic energy of the presented body at the moment of falling, consider the formula: Ek = 0.5 * mt * V ^ 2 = 0.5 * mt * (g * t) ^ 2, where mt is the mass of the falling body (mt = 2 kg); g – acceleration due to gravity (g ≈ 9.8 m / s2; const); t is the duration of the fall (t = 6 s).

Let’s perform the calculation: Ek = 0.5 * mt * (g * t) ^ 2 = 0.5 * 2 * (9.8 * 6) ^ 2 = 3457.44 J.

Answer: At the moment of the fall, the presented body should have had a kinetic energy of 3457.44 J.