A load weighing 0.4 kg was tied to a thread 1.2 m long. The thread with a load was taken away from the vertical

A load weighing 0.4 kg was tied to a thread 1.2 m long. The thread with a load was taken away from the vertical at an angle of 60. What is the kinetic energy of the load when it passes the equilibrium position?

m = 0.4 kilograms is the mass of the cargo;

L = 1.2 meters – thread length;

y = 60 degrees – the angle by which the load was deflected from the vertical;

g = 10 m / s2 – acceleration of gravity.

It is required to determine Ekin (Joule) – the kinetic energy of the load when passing the equilibrium position.

Let’s find the height to which the load was deflected from the equilibrium position:

h = L – L * cos (y) = 1.2 – 1.2 * cos (60) = 1.2 – 1.2 * 0.5 = 1.2 – 0.6 = 0.6 meters.

Then the potential energy of the load at this height will be equal to:

Epot = m * g * h = 0.4 * 10 * 0.6 = 4 * 0.6 = 2.4 Joules.

According to the law of conservation of energy, the kinetic energy of the load at the moment of passing the equilibrium position will be equal to its potential energy at height h, that is:

Ekin = Epot = 2.4 Joules.

Answer: The kinetic energy will be 2.4 Joules.