A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m, 1000 times

Q: Why don't we calculate the work done when the person lowers the mass?

The problem explicitly states that the potential energy lost each time he lowers the mass is 'dissipated'. This implies the person isn't actively doing positive mechanical work to lower it (e.g., letting it drop or just resisting gravity passively without recovering energy). We only account for the active lifting phase.

Q: What exactly does the $20\%$ efficiency mean in this context?

It means that for every $100 \text{ Joules}$ of chemical energy released by burning fat, only $20 \text{ Joules}$ go into actually lifting the weight. The remaining $80 \text{ Joules}$ are lost, mostly as thermal energy (heat), which is why our body temperature rises during exercise.

Q: Why is the final answer so small? Did I make a calculation error?

No, your calculation is likely correct! Fat is an extremely energy-dense molecule. It stores a massive amount of energy ($3.8 \times 10^7 \text{ J/kg}$) in a very small mass. This is an evolutionary adaptation to store maximum energy with minimum weight.

Q: Can I solve this using calories instead of Joules?

While you technically could if you converted everything properly, it is highly recommended to stick to standard SI units (Joules, kilograms, meters) to avoid unnecessary conversion errors and confusion.

Q: What happens to the $80\%$ of energy that isn't used for lifting?

By the Second Law of Thermodynamics, no engine is $100\%$ efficient. The 'lost' energy is primarily converted into heat. Your body must then work to dissipate this heat through sweating and increased blood flow to the skin to maintain a safe internal temperature.

Question

A person trying to lose weight by burning fat lifts a mass of $10 	ext{ kg}$ upto a height of $1 	ext{ m}$, $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 	imes 10^7 	ext{ J}$ of energy per kg which is converted into mechanical energy with a $20\%$ of efficiency rate. (Take, $g = 9.8 	ext{ ms}^{-2}$)

Options:

(A) $2.45 	imes 10^{-3} 	ext{ kg}$

(B) $6.45 	imes 10^{-3} 	ext{ kg}$

(C) $9.89 	imes 10^{-3} 	ext{ kg}$

(D) $12.89 	imes 10^{-3} 	ext{ kg}$

Sigma Percentile · Accepted Answer

Have you ever wondered why losing weight is so incredibly difficult? Why hours at the gym seem to barely move the needle on the weighing scale? This problem is a beautiful, albeit slightly depressing, mathematical illustration of exactly that reality. We are going to calculate exactly how much fat a person burns by performing a grueling physical task: lifting a $10 	ext{ kg}$ mass to a height of $1 	ext{ m}$, a staggering $1000$ times!

### The Heavy Lifting

Let's start by analyzing the sheer mechanical work being done. Every single time the person lifts the $10 	ext{ kg}$ mass, they are fighting against the relentless pull of Earth's gravity. The work done in one single lift is stored as gravitational potential energy in the mass.

We know the formula for this is:

$$W_1 = Mgh$$

But our dedicated individual doesn't just do this once. They do it $n = 1000$ times. The problem explicitly states that the energy lost when lowering the mass is dissipated, meaning we only need to calculate the work done during the lifting phase.

So, the total mechanical work done is:

$$W = n 	imes Mgh$$

Substituting our values ($M = 10 	ext{ kg}$, $g = 9.8 	ext{ ms}^{-2}$, $h = 1 	ext{ m}$):

$$W = 1000 	imes 10 	imes 9.8 	imes 1 = 9.8 	imes 10^4 	ext{ J}$$

That is $98,000 	ext{ Joules}$ of pure mechanical work. It sounds like a massive number, doesn't it? Let's see how much fat it actually burns.

### The Body's Engine

Now, we need to look at the fuel source: human fat. The problem tells us that $1 	ext{ kg}$ of fat supplies a colossal $3.8 	imes 10^7 	ext{ J}$ of energy. Fat is an incredibly dense energy storage medium.

However, the human body is not a perfect machine. Just like a car engine loses energy as heat, our bodies are only $20\%$ efficient at converting the chemical energy stored in fat into useful mechanical work. The other $80\%$ is lost, primarily as heat (which is why you sweat when you work out!).

If we assume the person burns $m$ kilograms of fat, the total chemical energy released is $m 	imes 3.8 	imes 10^7 	ext{ J}$. But the **effective mechanical energy** available to do the lifting is only $20\%$ of this total:

$$E_{	ext{eff}} = m 	imes (3.8 	imes 10^7) 	imes \frac{20}{100}$$

### The Final Burn

To find the mass of the fat burnt, we simply equate the total mechanical work done to the effective energy provided by the fat:

$$W = E_{	ext{eff}}$$

$$9.8 	imes 10^4 = m 	imes (3.8 	imes 10^7) 	imes 0.2$$

Now, we carefully solve for $m$:

$$m = \frac{9.8 	imes 10^4}{3.8 	imes 10^7 	imes 0.2}$$

$$m = \frac{9.8}{0.76} 	imes 10^{-3}$$

$$m \approx 12.89 	imes 10^{-3} 	ext{ kg}$$

This is approximately $12.89 	ext{ grams}$.

Let that sink in. Lifting $10 	ext{ kg}$ a thousand times—a workout that would leave most people exhausted—burns less than $13 	ext{ grams}$ of fat! This perfectly illustrates the incredible energy density of fat and why diet plays such a crucial role alongside exercise in weight management. Physics doesn't just help us pass exams; it explains the everyday realities of our lives.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Work Done Formula$

$Total Work Setup$

$Calculating Total Work$

$Energy from Fat$

$Effective Energy Equation$

$Equating Energies$

$Solving for Mass$

$Final Answer$

$The Way Forward$

The Sigma Insight: Work Done by Forces

The Heavy Lifting

The Body's Engine

The Final Burn

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Work Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Work

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy from Fat

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effective Energy Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward