On complete combustion a litre of petrol gives off heat equivalent to $3 \times 10^7$ J. In a test drive a car weighing 1200 kg. including the mass of driver, runs 15 km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.

> [!TLDR] > The energy from petrol is partly converted to useful work against friction; using efficiency and the work-energy relation, the friction force comes out to $1.0 \times 10^3$ N. The key idea here is that the engine's efficiency tells us what fraction of the chemical energy from burning petrol actually becomes mechanical work. That work is done entirely against the resistive force (friction + air drag) because the car moves at uniform speed — so there is no change in kinetic energy. The force of friction is therefore found by dividing the useful work by the distance travelled. Let’s go step by step. 1. **Energy released by one litre of petrol** Given: $3 \times 10^7$ J of heat is released on complete combustion of 1 litre. 2. **Useful work output from the engine** Efficiency $\eta = 0.5$ means only half the heat energy is converted into mechanical work. $$W_{\text{useful}} = \eta \times \text{heat released} = 0.5 \times 3 \times 10^7 = 1.5 \times 10^7 \text{ J}$$ 3. **Distance travelled per litre** The car runs 15 km per litre, i.e., $$s = 15 \text{ km} = 15\,000 \text{ m}$$ 4. **Why work equals force times distance** Since the car moves with uniform speed, the net force is zero. The engine’s forward thrust exactly balances the total resistive force $f$ (friction + air drag). The work done by the engine against this resistive force is $$W = f \times s$$ 5. **Equating useful work to work against friction** …

On complete combustion a litre of petrol gives off heat equivalent to 3 × 10⁷ J. In a test drive a car weighin

NCERT Exemplar · Q43

3 \times 10^7

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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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The energy from petrol is partly converted to useful work against friction; using efficiency and the work-energy relation, the friction force comes out to $1.0 \times 10^3$ N.

The key idea here is that the engine's efficiency tells us what fraction of the chemical energy from burning petrol actually becomes mechanical work. That work is done entirely against the resistive force (friction + air drag) because the car moves at uniform speed — so there is no change in kinetic energy. The force of friction is therefore found by dividing the useful work by the distance travelled.

Let’s go step by step.

Energy released by one litre of petrol

Given: $3 \times 10^7$ J of heat is released on complete combustion of 1 litre.
Useful work output from the engine

Efficiency $\eta = 0.5$ means only half the heat energy is converted into mechanical work.

W_{\text{useful}} = \eta \times \text{heat released} = 0.5 \times 3 \times 10^7 = 1.5 \times 10^7 \text{ J}

Distance travelled per litre The car runs 15 km per litre, i.e.,

s = 15 \text{ km} = 15\,000 \text{ m}

Why work equals force times distance Since the car moves with uniform speed, the net force is zero. The engine’s forward thrust exactly balances the total resistive force $f$ (friction + air drag). The work done by the engine against this resistive force is

W = f \times s

Equating useful work to work against friction …

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