A ball weighing 300 g freely falls on a horizontal floor surface from a height of 5 m. After hitting the floor

A ball weighing 300 g freely falls on a horizontal floor surface from a height of 5 m. After hitting the floor, the ball bounces to a maximum height of 2 m. How much heat is released when the ball hits the floor?

h1 = 5 m.

h2 = 2 m.

g = 9.8 m / s2.

m = 300 g = 0.3 kg.

Q -?

When the ball falls, the potential energy En1 transforms into kinetic energy Ek1, at the moment of impact, part of the energy goes into internal energy Q, and the lagging part again goes over first into kinetic, and then into potential energy En2.

According to the law of conservation of energy, the difference in the potential energy of the initial Еп1 and the final position of the body Еп2 passed into internal energy in the form of the amount of heat Q: Q = Еп1 – Еп1 = m * g * h1 – m * g * h2 = m * g * (h1 – h2).

Q = 0.3 kg * 9.8 m / s2 * (5 m – 2 m) = 8.82 J.

Answer: upon impact, the amount of heat released Q = 8.82 J.