Suppose the average mass of raindrops is $3.0 \times 10^{-5}$ kg and their average terminal velocity 9 m s$^{-1}$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

Suppose the average mass of raindrops is 3.0 × 10⁻⁵ kg and their average terminal velocity 9 m s⁻¹. Calculate

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3.0 \times 10^{-5}

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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) …

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Treat all the rain landing on $1\,\text{m}^2$ in a year as one mass $M$ arriving at terminal speed $v_t$ . The energy it delivers is its kinetic energy $\tfrac12 M v_t^2 = 4.05\times10^{4}\,\text{J}$ .

Concept

Rain falls at (essentially constant) terminal velocity, so each drop reaches the ground with kinetic energy $\tfrac12 m v_t^2$ , which is transferred to the surface on impact. Summed over all the water that lands on one square metre, the transferred energy is $\tfrac12 M v_t^2$ , where $M$ is the total mass of that water. (The individual drop mass is not needed for the total.)

Step-by-step solution

1. Mass of rain per square metre. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set D-41 markMCQ
$-3$
›Reveal solutionSolution
Use $W = q\,\Delta V = q\,(V_{\text{final}} - V_{\text{initial}}) = q(V_A - V_B)$ , being careful to convert millicoulombs to coulombs and to subtract the potentials in the right order (B → A).

Step 1 — The concept: potential difference is work per unit charge

The potential difference between two points is defined as the work done by an external agent in carrying a unit positive charge from one to the other, slowly (without changing its kinetic energy):
$V_A - V_B = \frac{W_{B \to A}}{q}$
Rearranged, this gives the working formula:
$\boxed{W_{B \to A} = q\,(V_A - V_B)}$
Equivalently, $W$ equals the change in electrostatic potential energy, $W = \Delta U = U_A - U_B = q V_A - q V_B$ . Since electrostatic force is conservative, this depends only on the endpoints — the path taken is irrelevant.

Step 2 — Identify the initial and final points (order matters!)

The charge is carried from B to A. So:

Initial point = B, with $V_B = +5$ V

Final point = A, with $V_A = -3$ V

The potential difference we need is $V_{\text{final}} - V_{\text{initial}} = V_A - V_B$ — not the other way round. Reversing this is the single most common error here and simply flips the sign.
$V_A - V_B = (-3) - (+5) = -8~\text{V}$
Step 3 — Convert the charge to SI units
$q = 5~\text{mC} = 5\times 10^{-3}~\text{C}$
(The prefix milli = $10^{-3}$ . Forgetting this conversion gives $-40$ J — option (D), a deliberate trap. Mistaking mC for μC would give $-4\times10^{-5}$ J.)

Step 4 — Substitute
$W = q\,(V_A - V_B) = \left(5\times 10^{-3}~\text{C}\right) \times \left(-8~\text{V}\right)$ $W = -40 \times 10^{-3}~\text{J} = -0.04~\text{J}$ $\boxed{W = -0.04~\text{J}}$
Step 5 — Why is the sign negative? (Check the physics)

The negative sign is not an accident — it carries real meaning. …
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COMEDK 2025Set 2025-E1 markMCQ
$60 \mu \mathrm{C}$
›Reveal solutionSolution
The kinetic energy gained equals the work done by the electric field: $qEd$ . Substituting the given values yields $1.2 \times 10^{-1} \, \text{J}$ , so $m = 12$ . The correct option is (A).

Concept & Intuition

When a charged particle is released from rest in a uniform electric field, the field exerts a constant force $F = qE$ on it. This force does work as the particle moves, and that work is entirely converted into kinetic energy (since no other forces act). The work done by a constant force over a displacement $d$ is simply $Fd$ . So the kinetic energy after moving distance $d$ is $K = qEd$ . No need to involve mass or acceleration — the energy approach is direct.

Step-by-step solution

Identify the given quantities

Mass $m = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg}$ (not needed for energy, but given)

Charge $q = 60 \, \mu\text{C} = 60 \times 10^{-6} \, \text{C} = 6 \times 10^{-5} \, \text{C}$

Electric field $E = 10^5 \, \text{N/C}$

Distance moved $d = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}$

Work done by the electric field

The force on the particle is $F = qE$ . Since the field is uniform, the force is constant. The work done over displacement $d$ is:

$W = Fd = qEd$
Substitute the values:
$W = (6 \times 10^{-5}) \times (10^5) \times (2 \times 10^{-2})$

Simplify step by step First, $6 \times 10^{-5} \times 10^5 = 6 \times 10^{0} = 6$ . Then multiply by $2 \times 10^{-2}$ :
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COMEDK 2025Set 2025-M1 markMCQ
$24 \mathrm{Nm}^{-1}$
›Reveal solutionSolution

Total fall $=2+0.5=2.5\,$ m, so gravity does $mgh=12.5\,$ J while $\tfrac12kx^2=3\,$ J is stored in the spring; the net work done in the process is $12.5-3=9.5\,$ J — option (D).

Concept. As the ball drops onto the spring and compresses it, gravity does positive work over the whole fall, and part of that energy is stored as elastic potential energy in the spring. The net work done in the process is the work delivered by gravity minus the energy locked up in the spring.

Step 1 — Total vertical drop.

The ball starts $2\,$ m above the free (upper) end of the spring and the spring is compressed by $50\,\text{cm}=0.5\,$ m, so the ball descends
$h = 2 + 0.5 = 2.5\ \text{m}.$
Step 2 — Work done by gravity.

With $m=0.5\,$ kg and $g=10\,\text{m s}^{-2}$ ,

$W_{\text{grav}} = mgh = 0.5\times 10 \times 2.5 = 12.5\ \text{J}.$ …
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COMEDK 2024Set 2024-E1 markMCQ
$7 \mathrm{~ms}^{-1}$
›Reveal solutionSolution
Using the work–energy theorem, the braking force does work equal to the loss in kinetic energy. The computed resistance is one‑quarter of the scooter’s weight, so the correct option is (B).

The key idea is that the brakes apply a constant resisting force (the total resistance to motion) over the stopping distance. The work done by that force equals the change in kinetic energy. Because the weight $W$ is given, we can express the mass as $m = W/g$ and then solve for the force in terms of $W$ .

Identify the known quantities

Initial speed: $u = 7\ \mathrm{m/s}$

Stopping distance: $s = 10\ \mathrm{m}$

Final speed: $v = 0$

Weight of scooter: $W$ (in newtons)

Acceleration due to gravity: $g = 9.8\ \mathrm{m/s^2}$ (we will keep it symbolic initially).

Relate force to acceleration using kinematics

From $v^2 = u^2 + 2as$ , we get

$0 = (7)^2 + 2a(10) \quad\Rightarrow\quad 49 + 20a = 0 \quad\Rightarrow\quad a = -\frac{49}{20} = -2.45\ \mathrm{m/s^2}.$
The negative sign means deceleration. The magnitude of the acceleration is $a = 2.45\ \mathrm{m/s^2}$ .

Apply Newton’s second law The total resisting force $F$ (assumed constant) is

$F = m a,$
where $m$ is the mass. Since weight $W = mg$ , we have $m = W/g$ . Thus
$F = \frac{W}{g} \cdot a.$

Substitute numbers Using $g = 9.8\ \mathrm{m/s^2}$ and $a = 2.45\ \mathrm{m/s^2}$ :
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Q.Suppose the average mass of raindrops is $3.0 \times 10^{-5}$ kg and their average terminal velocity 9 m s $^{-1}$ . Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

The Intuition First

The Precise Statement

Why "Net" Work Matters

A Simple Derivation (for constant force)

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

Concept

Step-by-step solution

Concept: Kinetic Energy Delivered by Falling Rain

Step 1 — The concept: potential difference is work per unit charge

Step 2 — Identify the initial and final points (order matters!)

Step 3 — Convert the charge to SI units

Step 4 — Substitute

Step 5 — Why is the sign negative? (Check the physics)

Q.Suppose the average mass of raindrops is 3.0×10−53.0 \times 10^{-5}3.0×10−5 kg and their average terminal velocity 9 m s−1^{-1}−1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

Concept

Step-by-step solution

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Q.Suppose the average mass of raindrops is $3.0 \times 10^{-5}$ kg and their average terminal velocity 9 m s $^{-1}$ . Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

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