Q.It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass $1.00\ \text{g}$ falling from a height $1.00\ \text{km}$. It hits the ground with a speed of $50.0\ \text{m s}^{-1}$.

Concept: Work-Energy Theorem

The work-energy theorem states that the net work done on an object equals its change in kinetic energy: $W_{\text{net}} = \Delta KE$.

Step 1: Calculate work done by gravity.

Gravitational force is conservative, so $W_g = mgh = (1.00 \times 10^{-3}\,\text{kg})(10\,\text{m/s}^2)(1000\,\text{m}) = 10.0\,\text{J}$.

Step 2: Find the change in kinetic energy.

The drop starts from rest and reaches $v = 50.0\,\text{m/s}$, so

Step 3: Apply the work-energy theorem.

$W_g + W_{\text{resistive}} = \Delta KE$ …

The Work-Energy Theorem tells us that the net work done on an object equals its change in kinetic energy. When multiple forces act, each does its own work, and their sum gives the total energy change. Here, gravity tries to accelerate the drop while air resistance opposes the motion, draining energy from the system.

The beauty of this approach is that we don't need to know the exact form of the resistive force (even though we're told it's proportional to speed). Work is a scalar quantity that depends only on the force magnitude, displacement, and the angle between them—so we can calculate the work done by each force independently.

Step-by-step solution

1. Calculate the work done by gravity

Gravity is a conservative force that does work $W_g = mgh$ when an object falls through height $h$. The force and displacement are in the same direction (downward), so the work is positive.

Given:

• Mass $m = 1.00\ \text{g} = 1.00 \times 10^{-3}\ \text{kg}$
• Height $h = 1.00\ \text{km} = 1000\ \text{m}$
• $g = 10\ \text{m s}^{-2}$ (the standard rounded value used in this example)

2. Determine the initial and final kinetic energies

The drop starts from rest at height $h$, so its initial kinetic energy is:

It hits the ground with speed $v = 50.0\ \text{m s}^{-1}$, giving a final kinetic energy:

3. Apply the Work-Energy Theorem

The net work done by all forces equals the change in kinetic energy:

where $W_r$ is the work done by the resistive force. Substituting:

$W_r = 1.25 - 10.0 = -8.75\ \text{J}$ …

Concept: Work-Energy Theorem — Work by Gravity and an Unknown Resistive Force

Step 1: Compute the work done by gravity.

Gravity is conservative: $W_g = mgh = (1.00\times10^{-3}\,\text{kg})(10\,\text{m/s}^2)(1000\,\text{m}) = 10.0\,\text{J}$.

Step 2: Compute the change in kinetic energy.

The drop starts from rest and hits the ground at $v = 50.0\,\text{m/s}$:

$\Delta KE = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(1.00\times10^{-3})(50.0)^2 = 1.25\,\text{J}$ …