A box weighing 20 kg, sliding off an inclined plane at an angle of 30 degrees to the horizon

A box weighing 20 kg, sliding off an inclined plane at an angle of 30 degrees to the horizon, moved at a constant speed. Find the change in the potential and kinetic energy of the box along this 1m path section. What is the change in the mechanical energy of the box in this area?

Given:

m = 20 kilograms – box weight;

l = 1 meter – the length of the inclined plane;

a = 30 degrees – the angle of inclination of the plane to the horizon;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine the change in the potential and kinetic energy of the box.

Since the problem statement is not specified, we will not take into account the energy losses for work against the friction force.

Then:

At the top point of movement, the potential energy of the box will be equal to:

Epot = m * g * h, where h is the height of the inclined plane.

Epot = m * g * l * sin (a) = 20 * 10 * 0.5 = 200 * 0.5 = 100 Joules.

At the lowest point, the height of the box will be 0, so the potential energy will also be 0.

Since, according to the condition of the problem, the box was moving at a constant speed, its kinetic energy did not change.

Answer: the change in potential energy is equal to 100 Joules, the kinetic energy of the box did not change.