A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s$^{-1}$. Calculate (a) the loss of P.E. of the drop. (b) the gain in K.E. of the drop. (c) Is the gain in K.E. equal to loss of P.E.? If not why. Take $g = 10$ m s$^{-2}$

A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s⁻¹. Calculate (a

NCERT Exemplar · Q38

^{-1}

(a) the loss of P.E. of the drop.

(b) the gain in K.E. of the drop.

g = 10

m s

^{-2}

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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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A falling raindrop loses gravitational potential energy but gains only part of it as kinetic energy; the rest is dissipated by air resistance. Loss of PE = 10 J, gain in KE = 1.25 J, and the 8.75 J difference goes into heating the air and the drop.

When an object falls through the atmosphere, we might naively expect all its gravitational potential energy to convert into kinetic energy. That would be true in a vacuum. But a raindrop falls through air, and air resistance does negative work on it, siphoning energy away. The Work-Energy Theorem tells us that the net work done on the drop equals its change in kinetic energy. Here, two forces act: gravity (doing positive work) and air drag (doing negative work). The difference between the PE lost and the KE gained reveals exactly how much energy air resistance has stolen.

Let's convert units first. The mass is $m = 1.00 \text{ g} = 1.00 \times 10^{-3} \text{ kg}$ , and the height is $h = 1 \text{ km} = 1000 \text{ m}$ .

Solution

1. Loss of potential energy

The raindrop starts at height $h$ and falls to the ground (taking ground as the zero PE reference). The loss in gravitational potential energy is simply

\Delta PE = mgh = (1.00 \times 10^{-3} \text{ kg})(10 \text{ m s}^{-2})(1000 \text{ m}) = 10 \text{ J}.

2. Gain in kinetic energy

The drop starts from rest (or very nearly so; raindrops form and begin falling with negligible initial speed) and reaches the ground with speed $v = 50 \text{ m s}^{-1}$ . Its kinetic energy at impact is

KE = \frac{1}{2}mv^2 = \frac{1}{2}(1.00 \times 10^{-3})(50)^2 = \frac{1}{2}(1.00 \times 10^{-3})(2500) = 1.25 \text{ J}.

Since it started with zero KE, the gain is $\Delta KE = 1.25 \text{ J}$ .

3. Comparing the two and explaining the difference

The gain in kinetic energy (1.25 J) is not equal to the loss in potential energy (10 J). The shortfall is

$10 - 1.25 = 8.75 \text{ J}.$ …

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Q.A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s $^{-1}$ . Calculate

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) …

Solution

Concept: Comparing Energy Loss with the Work–Energy Theorem

Q.A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s−1^{-1}−1. Calculate

Solution

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Q.A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s $^{-1}$ . Calculate

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