Natasha is being constantly scolded by her mother for being unfit. She decides to loose weight by lifting a mass of 10 kg by 0.5 m 100 times. Assuming that the potential energy lost each time during the lowering of mass is dissipated.

Calculate :

(a) work done by her against the gravitational force.

(b) If fat stored in her body supplies 3.8 x 10⁷ J of energy per kg, and is converted to mechanical energy with 20% efficiency, how much fat is used by Natasha

Given,

• m = 10 kg
• h = 0.5 m
• Frequency = 100

When Natasha lifts the mass, she does work against the gravitational force which is equal to the gain in gravitational potential energy and is given by,

W = mgh

(a) Work done in one lift = mgh = 10 × 9.8 × 0.5 = 49 J

She lifts the mass 100 times, so total work done = 49 × 100 = 4900 J

Thus, the total work done is 4900 J.

(b) Given,

• Energy supplied by 1 kg of fat = 3.8 x 10⁷ J
• Efficiency = 20%

Only 20% of the chemical energy from fat is converted into mechanical work.

If E is the chemical energy used then

$20\% \text { of E} = 4900\ \text J \\[1em] \Rightarrow 0.20 \times \text E = 4900 \\[1em] \Rightarrow \text E = \dfrac {4900}{0.20} \\[1em] \Rightarrow \text E = \dfrac {4900}{\dfrac{2}{10}} \\[1em] \Rightarrow \text E = \dfrac {4900 \times 10}{2} \\[1em] \Rightarrow \text E = \dfrac {49000}{2} \\[1em] \Rightarrow \text E = 24500\ \text J \\[1em] \text {Fat used} = \dfrac{\text E}{\text {Energy supplied by 1 kg of fat}} \\[1em] = \dfrac {24500}{3.8\times 10^{7}} \\[1em] \approx 6.45 \times 10^{-4} \text { kg}$

Hence, 6.45 x 10^-4 kg of fat is used by Natasha.