A rain drop of radius $2\ \text{mm}$ falls from a height of $500\ \text{m}$ above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10\ \text{m s}^{-1}$?

A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration

Exercises · 5.13

2\ \text{mm}

Yanam BieapTextbookSubjective· 3mImportance★★★★★est

30% · 25/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

Start your 14-day free trial to unlock the full solution →

The work done by gravity depends only on the vertical displacement, not on the path or speed. For each half of the 500 m fall, gravity does $8.2 \times 10^{-2}\ \text{J}$ . The resistive force does negative work equal to the difference between the total gravitational work and the gain in kinetic energy: $\approx -0.162\ \text{J}$ for the whole journey.

The key idea here is the Work–Energy Theorem: the net work done on an object equals its change in kinetic energy. But we must carefully separate the work done by gravity (a conservative force) from the work done by air resistance (a non-conservative force). Gravity's work is path-independent — it depends only on the vertical drop. Air resistance, however, does work that depends on the actual path and speed.

Let's break the problem into clear steps.

1. Find the mass of the raindrop

The drop is spherical, radius $r = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$ .

Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (8 \times 10^{-9}) = \frac{32\pi}{3} \times 10^{-9}\ \text{m}^3$ .

Density of water $\rho = 1000\ \text{kg/m}^3$ , so

m = \rho V = 1000 \times \frac{32\pi}{3} \times 10^{-9} = \frac{32\pi}{3} \times 10^{-6}\ \text{kg}.

Numerically:

m \approx \frac{32 \times 3.1416}{3} \times 10^{-6} \approx 33.51 \times 10^{-6}\ \text{kg} = 3.351 \times 10^{-5}\ \text{kg}.

Tip

You can keep $m$ in symbolic form with $\pi$ until the final calculation — it often cancels or simplifies.

2. Work done by gravity in each half

Gravity is a conservative force. The work done by gravity depends only on the vertical displacement, not on how the drop moves (accelerating, constant speed, or zigzag).

For a displacement $h$ downward,

W_g = mgh.

First half: drop from $500\ \text{m}$ to $250\ \text{m}$ , so $h = 250\ \text{m}$ .

W_{g1} = m g (250) = (3.351 \times 10^{-5})(9.8)(250) \approx 8.21 \times 10^{-2}\ \text{J}.

Second half: drop from $250\ \text{m}$ to ground, also $h = 250\ \text{m}$ .

W_{g2} = m g (250) \approx 8.21 \times 10^{-2}\ \text{J}.

So gravity does equal work in both halves:

W_{g1} = W_{g2} \approx 8.2 \times 10^{-2}\ \text{J}.

Watch out

A common mistake is to think gravity does more work in the first half because the drop accelerates more there. But work by gravity is $mgh$ — it doesn't care about speed or acceleration.

3. Work done by resistive force over the whole journey

We now use the Work–Energy Theorem:

W_{\text{net}} = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2.

Initial speed $v_i = 0$ (drop starts from rest).

Final speed on ground $v_f = 10\ \text{m/s}$ .

\Delta K = \frac{1}{2} m (10)^2 = 50 m.

Numerically:

$\Delta K = 50 \times 3.351 \times 10^{-5} = 1.6755 \times 10^{-3}\ \text{J}.$

The net work is the sum of work by gravity and work by resistance:

W_{\text{net}} = W_g + W_r.

Total gravitational work over the whole $500\ \text{m}$ :

W_g = m g (500) = (3.351\times10^{-5})(9.8)(500) = 3.351\times10^{-5} \times 4900 = 0.1642\ \text{J}.

Thus

W_r = W_{\text{net}} - W_g = 1.6755\times10^{-3} - 0.1642 \approx -0.1625\ \text{J} \approx -0.162\ \text{J}.

Note

The negative sign means the resistive force does negative work — it removes energy from the drop, converting mechanical energy into heat.

4. Check consistency with the terminal speed segment …

Start your 14-day free trial to unlock the full solution →

←Previous question5.12 — An electron and a proton are detected in a cosmic ray experiment, the …→Next question5.14 — A molecule in a gas container hits a horizontal wall with speed 200m s…

1. Find the mass of the raindrop

2. Work done by gravity in each half

3. Work done by resistive force over the whole journey

4. Check consistency with the terminal speed segment …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days