Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brou

NCERT Exemplar · Q32

Q. Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?

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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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When equal retarding forces act on bodies with equal kinetic energy, the work done to stop each body is the same. Since work equals force times distance, both bodies travel the same distance before coming to rest.

The Work-Energy Theorem tells us that the net work done on an object equals its change in kinetic energy. This is the key to understanding why mass doesn't matter here, even though the bodies are unequal in mass.

When a force acts on a moving body and brings it to rest, the work done by that force must equal the initial kinetic energy of the body (in magnitude). Since work is $W = F \cdot d$ for a constant force acting along the direction of motion, we can connect the stopping distance directly to the kinetic energy and the retarding force.

Let's denote the two bodies as 1 and 2, with masses $m_1$ and $m_2$ where $m_1 \neq m_2$ .

Given information:

Both bodies have equal kinetic energy: $KE_1 = KE_2 = KE$ (say)
Both experience the same magnitude of retarding force: $F_1 = F_2 = F$
Both come to rest, so final kinetic energy is zero

Step-by-step reasoning:

Apply the Work-Energy Theorem to body 1:

The retarding force does negative work (opposes motion), so:

W_1 = -F \cdot d_1

The change in kinetic energy is:

\Delta KE_1 = 0 - KE = -KE

By the Work-Energy Theorem:

-F \cdot d_1 = -KE

d_1 = \frac{KE}{F}

Apply the Work-Energy Theorem to body 2:

Similarly:

W_2 = -F \cdot d_2

\Delta KE_2 = -KE

Therefore:

-F \cdot d_2 = -KE

d_2 = \frac{KE}{F}

Compare the distances:

Since both expressions are identical:

$d_1 = d_2 = \frac{KE}{F}$ …

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Q.Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?

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: Work–Energy Theorem

Q.Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?

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