Q.A person trying to lose weight (dieter) lifts a $10\ \text{kg}$ mass, one thousand times, to a height of $0.5\ \text{m}$ each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

Concept: Work-Energy Principle – the work done against gravity equals the gain in gravitational potential energy, and the mechanical work required comes from the chemical energy in fat at a given efficiency.

(a) Work done per lift = gain in potential energy = $mgh$.

$m = 10\ \text{kg},\ g = 9.8\ \text{m/s}^2,\ h = 0.5\ \text{m}$

Per lift: $10 \times 9.8 \times 0.5 = 49\ \text{J}$

For 1000 lifts: $49 \times 1000 = 4.9 \times 10^{4}\ \text{J}$

(b) Fat energy content = $3.8 \times 10^{7}\ \text{J/kg}$, efficiency = $20\% = 0.20$

Useful mechanical energy per kg of fat = $0.20 \times 3.8 \times 10^{7} = 7.6 \times 10^{6}\ \text{J/kg}$ …

Concept First: Why the Work-Energy Principle is the natural tool

When the dieter lifts the mass slowly (so kinetic energy changes are negligible), the force she applies must exactly balance the weight. The work she does against gravity goes entirely into increasing the mass’s gravitational potential energy. When she lowers it, that potential energy is dissipated (as heat, sound, etc.) — so she must do fresh work on the next lift. The total work over 1000 lifts is simply 1000 times the work for one lift.

The second part is a straightforward energy conversion: only 20% of the chemical energy stored in fat becomes mechanical work. So the mechanical work we just calculated is only one-fifth of the fat energy that must be consumed.

Step-by-step solution

1. Work done in a single lift

The gravitational force on the mass is $mg$, where $m = 10\ \text{kg}$ and $g = 9.8\ \text{m/s}^2$ (standard value for such problems). Lifting it through a height $h = 0.5\ \text{m}$ against this force requires work:

2. Total work over 1000 lifts

Since the potential energy is lost each time she lowers the mass, she must do the same work again on every lift. No energy is “recovered” between cycles. So:

That answers part (a).

3. Energy available from fat …

Concept: Work Against Gravity → Chemical Energy Conversion with Efficiency

Step 1: Work done in a single lift

Step 2: Total work over 1000 lifts (PE is dissipated on every lowering, so no energy is recovered)

Step 3: Usable mechanical energy per kg of fat (only 20% is converted)

Step 4: Fat consumed …