Consider a guanabana (5.8 kg) falling from its branch. How much kinetic energy (in joules) does it gain during the time needed for it to drop 6.0 m?

The Kinetic Energy of the guanabana is, K. E = 341.04 J

Explanation:

Given data,

The mass of the guanabana, m = 5.8 kg

The height of the fruit from the ground, h = 6 m

Let the potential energy at the ground be 0

According to the law of conservation of energy, the total energy of a system is conserved.

So, when the guanabana reaches the ground, the P. E is fully converted into K. E

The P. E at height h is

P. E = mgh

= 5.8 x 9.8 x 6

= 341.04 J

At ground 6 m drop,

P. E = K. E

Therefore, the Kinetic Energy of the guanabana is, K. E = 341.04 J