Question

A 0.100-kilogram apple falls from a height of 1.50 meters to 1.00 meters. Ignoring frictional effects, what is the kinetic energy of the apple at this height?

Answer 1

before the fall, at 1.50 m, all the energy is potential and none is kinetic.

As it's falling the apple has both kinetic and potential energies. The decrease in potential energy is equal to the increase in kinetic energy

KE = -ΔPE
= -mgΔh
= -(0.100 kg)(9.81 m/s²)(1.00 m - 1.50 m)
= 0.491 J

Answer 2

The kinetic energy of the apple at this height is 0.49 J.

Step-by-step explanation:

Mass of apple, m = 0.1 kg

It falls form a height of 1.5 meters to 1 meters. We need to find the kinetic energy of the apple at this height. As the apple is falling, the decrease in potential energy is equal to the increase in kinetic energy. Using the conservation of energy as :

`E_k=-\Delta E_p`

`E_k=-mg\Delta h`

`E_k=-0.1* 9.8* (1-1.5)`

`E_k=0.49\ J`

So, the kinetic energy of the apple at this height is 0.49 J. Hence, this is the required solution.