Question

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground. Calculate the velocity of the acorn just before it reaches the ground (rounding your answer to the integer) and its kinetic energy when hitting the ground (rounding your answer to the nearest tenth).

Answer 1

The velocity and translational kinetic energy of the acorn when hitting the ground are approximately 19 meters per second and 3 joules, respectively.

Step-by-step explanation:

Let suppose that the acorn is a conservative system. By Principle of Energy Conservation, we understand that initial potential gravitational potential energy (
`U_(g)`), in joules, which is related to initial height above the ground, is equal to the final translational kinetic energy (
`K`), in joules, related to the instant just before hitting the ground. Let suppose that ground has a height of zero. That is:

`U_(g) = K` (1)

`m\cdot g \cdot h = (1)/(2)\cdot m \cdot v^(2)` (1b)

Where:

`m` - Mass, in kilograms.

`g` - Gravitational acceleration, in meters per square second.

`h` - Height, in meters.

`v` - Speed, in meters per second.

If we know that
`m = 0.017\,kg`,
`g = 9.807\,(m)/(s^(2))` and
`h = 18.5\,m`, then the velocity and the translational kinetic energy of the acorn just before hitting the ground is:

`m\cdot g \cdot h = (1)/(2)\cdot m \cdot v^(2)`

`v = √(2\cdot g \cdot h)`

`v \approx 19.049\,(m)/(s)`

`K = (1)/(2)\cdot m\cdot v^(2)`

`K = 3.084\,J`

The velocity and translational kinetic energy of the acorn when hitting the ground are approximately 19 meters per second and 3 joules, respectively.