Question

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?

Answer 1

Answer:

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