Determine the mechanical energy of a bird of mass 200g flying at 12m/s at a height of 50m above the ground.
We can use the conservation of mechanical energy to determine the mechanical energy of the bird. At the initial point when the bird is at rest on the ground, its mechanical energy is zero. At the final point when the bird is flying at a height of 50m above the ground, its mechanical energy is:
$E_f = K + U$
where $K$ is the kinetic energy and $U$ is the potential energy.
We can first find the potential energy of the bird:
$U = mgh$
where $m$ is the mass of the bird, $g$ is the acceleration due to gravity (9.8 m/s^2), and $h$ is the height of the bird above the ground:
$U = (0.2\text{kg})(9.8\text{m/s}^2)(50\text{m}) = 98\text{J}$
Next, we can find the kinetic energy of the bird:
$K = \frac{1}{2}mv^2$
where $v$ is the speed of the bird:
$K = \frac{1}{2}(0.2\text{kg})(12\text{m/s})^2 = 14.4\text{J}$
Therefore, the mechanical energy of the bird is:
$E_f = K + U = 14.4\text{J} + 98\text{J} = 112.4\text{J}$