Suppose a 356–g kookaburra (a large kingfisher bird) picks up a 65.0–g snake and raises it 2.30 m from the ground to a branch. How much work did the bird do on the snake? How much work did it do to raise its own center of mass to the branch?
Work = F*d = Mg * d = 0.065*9.8 * 2.3 =
To calculate the work done, we need to use the formula W = F * d * cos(theta), where W is the work done, F is the force applied, d is the distance over which the force is applied, and theta is the angle between the force vector and the displacement vector.
For the first question, we need to calculate the work done by the kookaburra on the snake. In this case, the force applied is the weight of the snake, which is given by F = m * g, where m is the mass of the snake and g is the acceleration due to gravity (9.8 m/s^2).
So, F = 0.065 kg * 9.8 m/s^2 = 0.637 N.
The distance over which the force is applied is the height the snake is raised, which is 2.30 m.
The angle between the force vector and the displacement vector is 0 degrees because the force is applied vertically upwards, and the displacement is also vertical.
Therefore, cos(theta) = cos(0) = 1.
Plugging in these values into the formula, we have:
W = 0.637 N * 2.30 m * 1 = 1.4661 N·m = 1.47 J.
So, the kookaburra did 1.47 Joules of work on the snake.
For the second question, we need to calculate the work done by the kookaburra on itself to raise its center of mass to the branch.
The force applied is the weight of the kookaburra, which is given by F = m * g, where m is the mass of the kookaburra.
The distance over which the force is applied is the same as before, 2.30 m.
The angle between the force vector and the displacement vector is still 0 degrees, as the force is applied vertically upwards.
Using the same reasoning as before, we find:
W = (356 g * 0.001 kg/g) * 9.8 m/s^2 * 2.30 m * 1 = 7.949 N·m = 7.95 J.
So, the kookaburra did 7.95 Joules of work on itself to raise its center of mass to the branch.