Suppose an acorn with a mass of 3.17 g falls off a tree. At a particular moment during the fall, the acorn has a kinetic energy of 0.259 J and a potential energy of 0.207 J with respect to the ground. How does the acorn's potential and kinetic energy change as it falls

where [tex]E_{ki}[/tex] is the initial kinetic energy, [tex]E_{pi}[/tex] is the initial potential energy, [tex]E_{kf}[/tex] is the final kinetic energy and [tex]E_{pf}[/tex] is the final potential energy.

so, when the acorn fall the final potential energy is going to decrease because the potential energy is equal to:

[tex]E_p = mgh[/tex]

Where m is the mass, g the gravity and h the height.

Therefore, in order to fullfil with the law of conservation of energy is necesary that the final kinetic energy increase.

Now, we know that the kinetic energy is equal to:

[tex]E_k = \frac{1}{2}mv^2[/tex]

where v is the velocity, so, it means that the velocity of the acorn is going to increase.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-29T14:05:58+02:00

As the acorn falls, the final potential energy is going to decrease and the final kinetic energy increases.

What is the acorn's potential and kinetic energy change as it falls?

Question Parameter(s):

Suppose an acorn with a mass of 3.17 g falls off a tree.

the acorn has a kinetic energy of 0.259 J

and potential energy of 0.207 J

Generally, the equation for the law of conservation of energy is mathematically given as

Eki +Epi = Ekf+Epf

Therefore,the potential energy is

Ep = mgh

m = mass,

g = gravity

h= height.

and

E_k = 0.5mv^2