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

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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?

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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.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$12\ cm^2$
›Reveal solutionSolution
This tests the energy-rate (power) approach to fluid flow with an inefficient motor: compute the mechanical power delivered to the water (PE rate + KE rate), then scale up by the efficiency to get the electrical/input power the motor draws. Answer: (A).

Concept and Intuition

The motor does two useful things to the water every second: it lifts a certain mass through height $h=3.2\ m$ (giving it gravitational PE), and it accelerates it up to the exit speed $v=6\ m/s$ (giving it KE). The rate at which this useful energy is delivered is the motor's output power. But the motor is only 80% efficient — it must actually draw more input power than this useful output, because some of the input is wasted (as heat, friction, etc.). So input power $=$ output power $/$ efficiency.

Step-by-Step Solution

Mass flow rate: water leaves the pipe of area $A=12\ cm^2=12\times10^{-4}\ m^2$ at speed $v=6\ m/s$ , so the volume flow rate is $Av=7.2\times10^{-3}\ m^3/s$ , and with density $\rho=1000\ kg/m^3$ ,

$\dot m=\rho A v = 7.2\ kg/s.$

Useful (output) power: each kilogram of water gains PE $=gh=10\times3.2=32\ J$ and KE $=\tfrac12v^2=\tfrac12\times36=18\ J$ , total $50\ J/kg$ . So

$P_{out}=\dot m\times50 = 7.2\times50=360\ W.$

Motor input power: since the motor is 80% efficient, $P_{out}=0.8\,P_{in}$ , so $P_{in}=\frac{360}{0.8}=450\ W.$ …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$5\ kg$
›Reveal solutionSolution
This tests the work-energy theorem with friction: net work done on the block equals its gain in kinetic energy. The answer is 900 J.

Concept and Intuition

The work-energy theorem says the net work done by all forces on an object equals its change in kinetic energy. Here two horizontal forces act along the direction of motion: the applied 100 N push (doing positive work) and kinetic friction (doing negative work, since it always opposes relative sliding). The block starts from rest, so its final KE is exactly the net work done over the 10 m displacement.

Step-by-Step Solution

Work done by the applied force over 10 m: $W_F = F \times d = 100 \times 10 = 1000\ J$ .

Normal force on a horizontal plane equals weight: $N = mg = 5 \times 10 = 50\ N$ .

Kinetic friction force: $f = \mu N = 0.2 \times 50 = 10\ N$ . …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$Q. An object is sliding (without rolling) from the top of the smooth inclined plane of height h from rest as shown and it just completes a vertical circle of diameter 20 cm. Then the minimum height h of smooth inclined plane is [FIGURE: a right-triangular smooth inclined plane of height h, with a small block sliding down from the top and a circular loop of diameter 20 cm at the bottom of the incline] (A) 0.25 m (B) 0.2 m (C) 0.5 m (D) 2.5 m$
›Reveal solutionSolution
The minimum-height problem for a ball that slides down a smooth incline and then "just completes" a vertical circular loop: combine the "just completes the loop" condition ( $v_{top}^2=gr$ ) with energy conservation from the starting height to the top of the loop, giving $h = 2.5r$ .

Concept and Intuition

"Just completes" a vertical circular loop means that at the very top of the loop, gravity alone is exactly enough to supply the centripetal force — the normal force from the track drops to zero there. Any less speed and the object would leave the track before reaching the top. This gives the minimum speed condition $mg = \dfrac{mv_{top}^2}{r} \Rightarrow v_{top}^2 = gr$ . Everything else follows from energy conservation on a frictionless (smooth) surface, since no energy is lost.

Step-by-Step Solution

Loop diameter $d = 20\,cm = 0.2\,m \Rightarrow$ radius $r = 0.1\,m$ .

Height of the top of the loop above the ground $= 2r = 0.2\,m$ (since the loop sits on the ground and the object must reach the very top of the loop).

Minimum speed at the top of the loop (gravity supplies all the centripetal force): $v_{top}^2 = gr$ .

Energy conservation from rest at height $h$ (top of incline, ground level = reference) to the top of the loop: …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Q. A force of 250 N is required to lift a mass of 75 kg through a pulley system. In order to lift this mass through 3 m, the rope has to be pulled through 12 m. The efficiency of the system is (A) 50 % (B) 75 % (C) 33% (D) 90%$
›Reveal solutionSolution
Efficiency of a machine is output (useful) work over input work; here it comes out to 75%. Answer: (B).

Concept and Intuition

A pulley system trades force for distance: you pull the rope through a larger distance to lift a load through a smaller one, with the mechanical advantage set by the number of supporting rope segments. No real machine is loss-free, so the efficiency compares the useful work actually delivered (lifting the load) to the work you actually put in (pulling the rope) — the shortfall is lost to friction in the pulleys.

Step-by-Step Solution

Useful output work $= mgh = 75\ \text{kg} \times 10\ \text{ms}^{-2} \times 3\ \text{m} = 2250\ \text{J}$ .

Input work supplied by the pulling force $= F\times d = 250\ \text{N} \times 12\ \text{m} = 3000\ \text{J}$ . …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$20\ \mathrm{N\,C^{-1}}$
›Reveal solutionSolution
Uses the work-energy theorem: the kinetic energy gained equals the work done by the electric force over the distance moved. Answer: 8 J.

Concept and Intuition

A charged particle released from rest in a uniform electric field experiences a constant force $F=qE$ , and by the work-energy theorem, the kinetic energy it gains equals the work done by this force as it moves: $KE = F\times d$ . The mass value given is actually irrelevant to computing the kinetic energy directly via this route (it would only be needed to find velocity or acceleration separately).

Step-by-Step Solution

Electric force on the charge: $F = qE = 2\ \text{C}\times20\ \text{N/C} = 40\ \text{N}$ .

Distance moved: $d = 20\ \text{cm} = 0.2\ \text{m}$ .

By the work-energy theorem, kinetic energy gained $= F\times d = 40\times0.2 = 8\ \text{J}$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$10\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
The work done against gravity to raise a mass to a height equals $mgh$ ; with $g=10\ \mathrm{m\,s^{-2}}$ this gives 20000 J.

Concept and Intuition

Lifting a body at (essentially) constant velocity means the applied force just balances gravity, so the work done by that force equals the gain in gravitational potential energy, $mgh$ .

Step-by-Step Solution

Given $m = 100$ kg, $h = 20$ m, $g = 10\ \mathrm{m\,s^{-2}}$ (as stated in the question).

$W = mgh = 100 \times 10 \times 20 = 20000$ J.

Common Mistakes …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$Q. The work done in lifting a body of mass 2 kg from the surface of the earth to a height 10 m is (A) 98 J (B) 196 J (C) 147 J (D) 288 J$
›Reveal solutionSolution
With $g=9.8\ \mathrm{ms^{-2}}$ , lifting a 2 kg mass by 10 m requires work $W=mgh=196$ J.

Concept and Intuition

For heights small compared to Earth's radius, the gravitational field can be treated as uniform, so the work done in raising a mass against gravity equals the increase in gravitational potential energy, $mgh$ , with no need for the more general $-GMm/r$ expression.

Step-by-Step Solution

Recall $W = mgh$ for lifting a mass through height $h$ near Earth's surface.

Substitute $m=2$ kg, $g=9.8\ \mathrm{ms^{-2}}$ , $h=10$ m.

$W = 2\times9.8\times10 = 196$ J.

This matches option (B).

Common Mistakes …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$20 \, \mathrm{ms^{-1}}$
›Reveal solutionSolution
Using $\Delta KE = \tfrac12 m(v_f^2 - v_i^2)$ with $m=10$ kg, $v_i=20$ , $v_f=30\ \mathrm{ms^{-1}}$ gives an increase of 2500 J.

Concept and Intuition

Kinetic energy is $KE = \tfrac12 m v^2$ . When speed changes from $v_1$ to $v_2$ , the change in kinetic energy is the difference between the final and initial kinetic energies: $\Delta KE = \tfrac12 m v_2^2 - \tfrac12 m v_1^2 = \tfrac12 m (v_2^2 - v_1^2)$ .

Step-by-Step Solution

Initial KE: $\tfrac12 \times 10 \times 20^2 = 5 \times 400 = 2000$ J.

Final KE: $\tfrac12 \times 10 \times 30^2 = 5 \times 900 = 4500$ J.

Increase in KE: $4500 - 2000 = 2500$ J. …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$kg$
›Reveal solutionSolution
By energy conservation, the KE lost (250 J) converts entirely to gravitational PE, giving a height of 10 m.

Concept and Intuition

As the body rises, kinetic energy converts to gravitational potential energy (mechanical energy is conserved, ignoring air resistance): $KE_0=KE(h)+mgh$ .

Step-by-Step Solution

Initial KE $=500\,J$ ; at height $h$ , KE is 50% of that $=250\,J$ .

Energy conservation: $mgh=KE_0-KE(h)=500-250=250\,J$ .

$h=\dfrac{250}{mg}=\dfrac{250}{2.5\times10}=\dfrac{250}{25}=10\,m$ . …
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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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