Question

A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(A) How much work does she do against the gravitational force?
(B) Fat supplies 3.8 × 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer 1

Hint: The total work done to lift an object of mass,$m$kg against the gravitational force, the amount of upward force applied is $mg$to overcome the downward force of gravity.
If the object is lifted to the height h, the work done against the gravitational force is:
$W = F \cdot d = mg \times h$
$ = mgh$.
Where W is work done against the gravitational force by the body, F is the force applied by the body against gravity and d is the displacement due to that force, g is the acceleration due to gravity.
Here, the force applied would be mg and displacement would be h.
Work done by gravity is positive when the displacement is in the same direction of gravity and would be negative when it is in the opposite direction of gravity.

Complete step by step solution:
Step 1
(A) Express the relation of work done and lifting the mass
Total work done = $mgh \times$ $\text{no of times weight is lifted}$
 $
= 10 \times 9.8 \times 0.5 \times 1000 \\
= {\text{49000 J}} \\
 $
Step 2
(B) Energy stored in 1 kg of fat = $3.8 \times {10^7}$J (given)
Efficiency =$20\% $ (given)
Mechanical energy supplied by the person’s body=$\dfrac{{20}}{{100}} \times 3.8 \times {10^7}$J
= $\dfrac{1}{5} \times 3.8 \times {10^7}$J
Step 3
Equivalent mass of fat lost by the dieter=\[\dfrac{1}{{\dfrac{1}{5} \times 3.8 \times {{10}^3}}} \times 49 \times {10^3}\]
=$\dfrac{{245}}{{3.8}} \times {10^{ - 4}}$
=$6.45 \times {10^{ - 3}}$ kg

$\therefore$ (i) Total work done = $49000 J $
(ii) Equivalent mass of fat lost by the dieter = =$6.45 \times {10^{ - 3}} kg$

Additional information:
Gravitational energy is the potential energy linked with gravitational forces, as work is required to elevate objects opposite to Earth's gravity.
The potential energy stored due to elevated positions is called gravitational potential energy, e.g. water in an elevated reservoir or kept behind a dam.
If an object falls from one point to another point inside a gravitational field, the force of gravity will do positive work on the object, and the gravitational potential energy will decrease by the same amount.

Note:
Gravitational energy is the potential energy. The work is done,$mgh$ against it on an object of mass \[m\]kg, lifted to the height $h$, is transferred to the object.
Work is against gravity because the mass is lifted upward i.e. against the direction of gravity.
Efficiency refers to achieving maximum output with the least amount of input.