Calculating Work Done by Gravitational and Resistive Forces on a Rain Drop

This solution explains how to calculate the work done by gravity and air resistance acting on a falling rain drop, using the principles of physics, specifically the concepts of work and energy.

Understanding the Physics Concepts

We need to find two quantities:

• The work done by the gravitational force as the rain drop falls.
• The work done by the resistive force of air opposing the motion of the rain drop.

We are given the following information:

• Mass of the rain drop ($m$): $2 \text{ g}$
• Height of fall ($h$): $1.5 \text{ km}$
• Initial downward velocity ($v_i$): $20 \text{ m/s}$
• Final velocity ($v_f$): $70 \text{ m/s}$
• Acceleration due to gravity ($g$): $10 \text{ m/s}^2$

First, let's convert the given units to standard SI units:

• Mass: $m = 2 \text{ g} = 0.002 \text{ kg}$
• Height: $h = 1.5 \text{ km} = 1500 \text{ m}$

Calculating Work Done by Gravitational Force

The gravitational force ($F_g$) acting on the rain drop is given by $F_g = mg$. Work done by a force is calculated as the force multiplied by the distance moved in the direction of the force. Since the rain drop falls downwards and gravity also acts downwards, the angle between the force and displacement is $0^\circ$. The work done by gravity ($W_g$) is:

$W_g = \text{Force} \times \text{Distance} \times \cos(\theta)$

Here, Force is $mg$, Distance is $h$, and $\theta = 0^\circ$. So the formula becomes:

$W_g = mgh \cos(0^\circ) = mgh$

Now, let's substitute the values:

$W_g = (0.002 \text{ kg}) \times (10 \text{ m/s}^2) \times (1500 \text{ m})$

$W_g = 0.02 \times 1500 \text{ J}$

$W_g = 30 \text{ J}$

So, the work done by the gravitational force is $30 \text{ J}$. This is positive because the force and displacement are in the same direction.

Calculating Work Done by Resistive Force of Air using Work-Energy Theorem

The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy ($\Delta KE$).

$W_{net} = \Delta KE$

The net work is the sum of the work done by all forces acting on the object. In this case, the forces are gravity and air resistance.

$W_{net} = W_g + W_r$

Where $W_r$ is the work done by the resistive force of air.

The change in kinetic energy is:

$\Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Let's calculate the initial and final kinetic energies:

• Initial Kinetic Energy ($KE_i$):

  $KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2} \times (0.002 \text{ kg}) \times (20 \text{ m/s})^2$

  $KE_i = 0.001 \times 400 \text{ J} = 0.4 \text{ J}$

• Final Kinetic Energy ($KE_f$):

  $KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2} \times (0.002 \text{ kg}) \times (70 \text{ m/s})^2$

  $KE_f = 0.001 \times 4900 \text{ J} = 4.9 \text{ J}$

Now, calculate the change in kinetic energy:

$\Delta KE = KE_f - KE_i = 4.9 \text{ J} - 0.4 \text{ J} = 4.5 \text{ J}$

According to the Work-Energy Theorem:

$W_g + W_r = \Delta KE$

We know $W_g = 30 \text{ J}$ and $\Delta KE = 4.5 \text{ J}$. We can find $W_r$:

$30 \text{ J} + W_r = 4.5 \text{ J}$

$W_r = 4.5 \text{ J} - 30 \text{ J}$

$W_r = -25.5 \text{ J}$

The work done by the resistive force of air is $-25.5 \text{ J}$. The negative sign indicates that the air resistance force opposes the motion of the rain drop, acting upwards while the drop moves downwards.

Summary of Work Done

Here's a summary of the calculated work done: