A stone with a mass of m = 0.10 kg falls from a state of rest from a height of h = 8.0 m. Determine how much the internal

A stone with a mass of m = 0.10 kg falls from a state of rest from a height of h = 8.0 m. Determine how much the internal energy of the stone will increase when falling, if the speed of its movement at the surface of the Earth is v = 12 m / s. The coefficient g is assumed to be 10 H / kg.

Given:

H = 8 meters – the height from which the stone falls;

m = 0.1 kg is the mass of the stone;

g = 10 N / kg – acceleration of gravity;

V = 12 m / s – the speed of the stone at the surface of the earth.

It is required to determine the increase in the internal energy of the stone Eвн (Joule) when falling.

Since the stone falls from a state of rest, at the maximum point it has only potential energy Ep:

Ep = m * g * H = 0.1 * 10 * 8 = 8 Joules.

At the surface of the earth, its potential energy is zero, and its kinetic energy:

Eк = m * V ^ 2/2 = 0.1 * 12 * 12/2 = 7.2 Joules.

The energy conservation law for this case looks like this:

Eп = Eк + Eвн, hence:

Eun = En – Ek = 8 – 7.2 = 0.8 Joules.

Answer: The internal energy of the stone will increase by 0.8 Joules.