Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. when the raindrop reaches the ground, it won't kill you or even bruise you because its terminal velocity is just 10 m/s. calculate the work done by the air resistance on the raindrop over the course of its fall to the ground. make sure you report your answer with the correct sign. again, assume that the falling raindrop maintains its shape so that no energy is lost to the deformation of the droplet

The work done by the air resistance is -0.0782 J

Explanation:

Hi there!

The energy of the raindrop has to be conserved, according to the law of conservation of energy.

Initially, the raindrop has only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = acceleration due to gravity (9.8 m/s²)

h = height.

Let´s calculate the initial potential energy of the drop:

(convert 4 mg into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000  m

PE = 0.0784 J

When the drop starts falling, some of the potential energy is converted into kinetic energy and some energy is dissipated by the work done by the air resistance. On the ground all the initial potential energy has been either converted into kinetic energy or dissipated by the resistance of the air:

initial PE = final KE + W air

Where:

KE = kinetic energy.

W air = work done by the air resistance.

The kinetic energy when the raindrop reaches the ground is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

Then:

KE = 1/2 ·  4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can calculate the work done by the air resistance:

initial PE = final KE + W air

0.0784 J = 2 × 10⁻⁴ J + W air

W air = 0.0784 J - 2 × 10⁻⁴ J

W air = 0.0782 J

Since the work is done in the opposite direction to the displacement, the work is negative, then, the work done by the air resistance is -0.0782 J.

do you have a formula possibly