An adult weighing 600N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length

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Q. An adult weighing 600N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.

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Concept understanding — Work Energy Principle

The Work-Energy Principle: From Intuition to Precision

Imagine pushing a heavy box across a rough floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — between the effort you put in (force × distance) and the change in the box's motion — is exactly what the Work-Energy Principle captures.

The Intuition First

Think of work as "energy transferred by a force." When you do work on an object, you're essentially pumping energy into it. That energy has to go somewhere — and in the simplest case, it shows up as a change in the object's speed. The object's kinetic energy (energy of motion) increases by exactly the amount of work you did.

Note

This is why a car's brakes get hot: the work done by friction removes kinetic energy, turning it into thermal energy. The principle holds even when energy changes form.

The Precise Statement

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

Where:

$W_{\text{net}}$ = net work done on the object (total work from all forces combined)
$K$ = kinetic energy = $\frac{1}{2}mv^2$
$m$ = mass, $v$ = speed

The net work done on an object equals the change in its kinetic energy.

Why "Net Work" Matters

If you push a box forward while friction pulls it backward, only the net force matters. Suppose you push with 50 N and friction opposes with 30 N over 2 m:

Work done by you: $50 \times 2 = 100$ J
Work done by friction: $-30 \times 2 = -60$ J (negative because force opposes motion)
Net work: $100 - 60 = 40$ J

That 40 J is exactly the increase in the box's kinetic energy. The individual works don't matter — only the sum.

A Common Trap

Watch out

The Work-Energy Principle applies to net work, not work done by a single force. A force can do positive work while the object slows down (if another force does even more negative work). Always find the total work from all forces.

When Does It Hold?

The principle works for:

Any constant or varying force
Straight-line or curved paths
Objects that don't rotate (for now)

It fails if:

The object deforms permanently (like crumpling a car) …

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Each step lifts the body's centre of gravity, requiring work against gravity. Over 6 km of jogging with 6000 steps, the mechanical energy used is 0.9 MJ; accounting for 10% conversion efficiency, the food energy required is 9 MJ.

The Work-Energy Principle tells us that the energy spent in any activity equals the work done against all forces. Here, the jogger repeatedly lifts his centre of gravity by 0.25 m with each step. Gravity pulls downward with force 600 N, so every step demands work $W = mgh = 600 \times 0.25$ J to raise the body. Over thousands of steps, this adds up. Because the body converts only a fraction of food energy into mechanical work, the actual food intake must be much larger.

Step-by-step solution

Find the number of steps in 6 km.

Total distance jogged: $d = 6 \text{ km} = 6000 \text{ m}$ .

Length of each step: $L = 1 \text{ m}$ .

Number of steps:

n = \frac{6000}{1} = 6000.

Calculate the work done in one step.

The jogger raises his centre of gravity by $h = 0.25 \text{ m}$ against his weight $W = 600 \text{ N}$ .

Work per step:

W_{\text{step}} = W \cdot h = 600 \times 0.25 = 150 \text{ J}.

Find the total mechanical energy used in jogging.

Over 6000 steps, the total work done is:

E_{\text{mech}} = n \times W_{\text{step}} = 6000 \times 150 = 900\,000 \text{ J} = 0.9 \text{ MJ}.

Watch out

A common mistake is to forget that the horizontal distance (1 m per step) does not directly contribute to gravitational work; only the vertical displacement (0.25 m) matters when calculating work against gravity.

Account for the body's conversion efficiency.

The body converts only 10% of food energy into mechanical work. If $E_{\text{food}}$ is the energy intake from food, then: …

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