Question

An isolated system contains a ball at rest 5 meters (m) above the ground. The ball has an initial potential energy of 250 joules (J). If the ball was dropped from its initial position, what would be the total mechanical energy of the ball just before impact with the ground?

Answer 1

Answer:

Total mechanical energy is the sum of potential energy plus kinetic energy. The kinetic energy will be 250 [J] and the potential energy is zero, therefore Total mechanical energy will be 250 + 0 =250[J]

Explanation:

This is a problem that applies the principle of energy conservation, i.e. mechanical energy that will be transformed into kinetic energy. We need to identify what kind of energy we have depending on the position of the ball with respect to the reference axis we take.

The reference axis or reference point is the point at which the potential energy is equal to zero, for this case we will take the ground as our reference point.

We know that the potential energy is defined by:

$E_{p}=m*g*h\\ where:\\m=mass[kg]\\g=gravity[m/s^2]\\h=elevation[m]$

We can clear the mass from this equation:

$m=\frac{E_{p} }{(g*h)} \\m=\frac{250 }{(9.81*5)} \\\\m=5.09[kg]$

When this body falls its potential energy will decrease but its kinetic energy will increase and reach its maximum value when the ball reaches the ground.

In such a way that its potential energy would be transformed into kinetic energy.

$E_{k} = E_{p} \\E_{k} =kinetic energy [J]$

Since the potential energy has been transformed all into kinetic energy the amount of energy is conserved, therefore the total mechanical energy will remain the same.