Answer:
The velocity and translational kinetic energy of the acorn when hitting the ground are approximately 19 meters per second and 3 joules, respectively.
Explanation:
Let suppose that the acorn is a conservative system. By Principle of Energy Conservation, we understand that initial potential gravitational potential energy ([tex]U_{g}[/tex]), in joules, which is related to initial height above the ground, is equal to the final translational kinetic energy ([tex]K[/tex]), in joules, related to the instant just before hitting the ground. Let suppose that ground has a height of zero. That is:
[tex]U_{g} = K[/tex] (1)
[tex]m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1b)
Where:
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Height, in meters.
[tex]v[/tex] - Speed, in meters per second.
If we know that [tex]m = 0.017\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]h = 18.5\,m[/tex], then the velocity and the translational kinetic energy of the acorn just before hitting the ground is:
[tex]m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]v = \sqrt{2\cdot g \cdot h}[/tex]
[tex]v \approx 19.049\,\frac{m}{s}[/tex]
[tex]K = \frac{1}{2}\cdot m\cdot v^{2}[/tex]
[tex]K = 3.084\,J[/tex]
The velocity and translational kinetic energy of the acorn when hitting the ground are approximately 19 meters per second and 3 joules, respectively.
