A ball weighing 200 g falls from a height of 20 m with an initial velocity equal to zero.

A ball weighing 200 g falls from a height of 20 m with an initial velocity equal to zero. What is its kinetic energy at the moment before hitting the ground, if the energy loss due to air resistance is 4 J?

m = 200 g = 0.2 kg.

h = 20 m.

g = 9.8 m / s ^ 2.

Epot = 4 J.

Ek -?

According to the law of conservation of energy, when falling, the ball has only potential energy En, the value of which is determined by the formula: En = m * g * h, where m is the mass of the ball, g is the acceleration of gravity, h is the height of the fall.

During the fall of the ball, part of the potential energy En passes into kinetic energy Ek, and part to overcome the resistance force Epot.

En = Ek + Epot.

Ek = m * g * h – Epot.

Ek = 0.2 kg * 9.8 m / s ^ 2 * 20 m – 4 J = 35.2 J.

Answer: when hitting the ground, the ball will have kinetic energy Ek = 35.2 J.