A stone is thrown horizontally from a tower 25m high at a speed of 15m / s. Find the kinetic and potential energy

A stone is thrown horizontally from a tower 25m high at a speed of 15m / s. Find the kinetic and potential energy of the stone a second after the start of the movement. Stone weight 0.2kg. Neglect air resistance.

Given:

H = 25 meters – the height of the tower from which the stone was thrown down;

m = 0.2 kg is the mass of the stone;

g = 10 m / s2 – free fall acceleration;

v0 = 15 m / s – initial speed of the stone;

t = 1 second – time interval.

It is required to find E (kin) and E (sweat) – the kinetic and potential energy of the stone after a time interval t after the start of movement.

The speed of the body after a time interval t will be equal to:

v = v0 + g * t = 15 + 1 * 10 = 15 + 10 = 25 m / s.

Then its kinetic energy will be equal to:

E (kin) = m * v ^ 2/2 = 0.2 * 25 ^ 2/2 = 62.5 Joules.

The body height after a time interval t will be equal to:

h = H – v0 * t – g * t ^ 2/2 = 25 – 15 * 1 – 10 * 1 ^ 2/2 =

= 25 – 15 – 5 = 5 meters.

Then its potential energy will be equal to:

E (sweat) = m * g * h = 0.2 * 10 * 5 = 10 Joules.

Answer: kinetic energy will be equal to 62.5 Joules, potential – 10 Joules.