A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as mu

NCERT Exemplar · Q18

Q. A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?

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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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Solution

Concept: Work-Energy Principle applied to metabolic energy expenditure.

The mechanical energy spent climbing up the stairs is $mgh$ . Since the person burns twice as much fat going up as coming down, energy spent coming down is half that of going up.

Energy spent climbing up:

E_{up} = mgh = 60 \times 9.8 \times 10 = 5880 \text{ J}

Energy spent coming down (half of going up):

E_{down} = \frac{1}{2}E_{up} = 2940 \text{ J}

Energy per round trip:

E_{trip} = E_{up}+E_{down} = 5880+2940 = 8820 \text{ J}

Total energy needed to burn 5 kg fat:

Using 1 kcal = 4200 J: …

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

COMEDK 2026Set 2026-A1 markMCQ
$5 \mathrm{~ms}^{-1}$
›Reveal solutionSolution

Energy balance $\tfrac12 mv^2 = f x + \tfrac12 k x^2$ gives $f=\dfrac{18.75-12.5}{0.05}=125\ \text{N}$ — option (A).

The block's initial kinetic energy is spent partly on the friction work over the compression distance and partly stored in the compressed spring, until the block stops.

Initial kinetic energy:
$\tfrac12 mv^2=\tfrac12(1.5)(5)^2=18.75\ \text{J}.$
Energy stored in the spring (compression $x=5\ \text{cm}=0.05\ \text{m}$ ):

$\tfrac12 k x^2=\tfrac12(10000)(0.05)^2=12.5\ \text{J}.$ …
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KCET 2026Set C21 markMCQ
$Q. A horizontal force of 5 N is applied on a stationary body of mass 5 kg, which is initially at rest on a frictionless table. The change in kinetic energy of the body in 10 s is (A) 25 J (B) Zero (C) 125 J (D) 250 J$
›Reveal solutionSolution

A constant applied force produces constant acceleration on a frictionless surface; the resulting velocity after a given time, via the work-energy relation, gives the change in kinetic energy directly as $\tfrac12mv^2$ (since it starts from rest).

Step 1 — Find the acceleration
$a = \frac{F}{m} = \frac{5\ \text{N}}{5\ \text{kg}} = 1\ \text{m/s}^2$
Step 2 — Find the velocity after 10 s

The body starts at rest ( $u=0$ ) on a frictionless table, so:

$v = u + at = 0 + (1)(10) = 10\ \text{m/s}$ …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{R}$
›Reveal solutionSolution

There is no factor of 1/2 (that appears only for a force that varies linearly from zero, which friction does not) and no factor of 2.

Concept: kinetic friction is a constant force f = mu * R (R = normal reaction), directed opposite to the displacement. Work done against it over a straight displacement d is

W = f * d = mu R d. …
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KCET 2022Set B-31 markMCQ
$2$
›Reveal solutionSolution

The work equals the gain in gravitational PE: lift the hanging portion's centre of mass (which sits at half the hanging length) up to the table top.

Step 1 — The concept: work = change in potential energy.

The table is smooth and the chain starts and ends at rest, so no work goes into friction or kinetic energy. All the work done in pulling the chain up is stored as gravitational potential energy:
$W = \Delta U = m_{\text{hanging}}\, g\, \Delta y_{\text{cm}}$
The subtlety — and the whole point of the question — is that only the hanging portion has to be raised (the part already on the table doesn't change height), and it must be raised by the rise of its centre of mass, not by its full length.

Step 2 — Linear mass density.

The chain is uniform, so
$\lambda = \frac{M}{L} = \frac{4\ \mathrm{kg}}{2\ \mathrm{m}} = 2\ \mathrm{kg\,m^{-1}}$
Step 3 — Mass of the hanging part.

The hanging length is $60\ \mathrm{cm} = 0.6\ \mathrm{m}$ (converting to SI is essential here):
$m = \lambda \ell = 2 \times 0.6 = 1.2\ \mathrm{kg}$
Step 4 — Locate its centre of mass.

The hanging piece is a uniform segment dangling from the edge, so its centre of mass lies at its midpoint — that is, a depth
$y_{\text{cm}} = \frac{\ell}{2} = \frac{0.6}{2} = 0.3\ \mathrm{m}$
below the table top. To get the whole chain onto the table, this centre of mass must be raised through exactly $0.3\ \mathrm{m}$ . …
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KCET 2022Set B-31 markMCQ
$10^9~W$
›Reveal solutionSolution
The reactor’s power output comes from mass-energy conversion via $E = \Delta m c^2$ . The fuel consumed in one hour is 0.04 g, option (C).

The core idea here is Einstein’s mass-energy equivalence. In a nuclear reactor, the “fuel” isn’t burned chemically — it’s converted into energy through nuclear fission, where a tiny fraction of the mass disappears and reappears as heat (which is then turned into electricity). The power rating tells us how much energy is produced per second, and from that we can find the mass lost per second, then scale to one hour.

A classic trap is to forget that the power given is the electrical output, but here the problem states “delivers a power of $10^9~\text{W}$ ” — that’s the thermal power produced by the fission itself, so we can use it directly. If it were electrical output, we’d need the efficiency, but no such detail is given, so we take it as the fission power.

Let’s work it through.

Energy produced in one hour

Power $P = 10^9~\text{W} = 10^9~\text{J/s}$ .

Time $t = 1~\text{hour} = 3600~\text{s}$ .

Energy $E = P \times t = 10^9 \times 3600 = 3.6 \times 10^{12}~\text{J}$ .

Relate energy to mass lost

From $E = \Delta m \, c^2$ , where $c = 3 \times 10^8~\text{m/s}$ ,
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Q.A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?

Solution

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