Question

3.Assume chemical energy from food can be extracted through digestion and respiration within the body and the efficiency in converting this energy into mechanical energy is 18%. How much chemical energy from food should a student consume in order to lift a 15 kg mass 1.5 m and repeat this activity 500 times?

Answer 1

Given:

The height of the mass is,

`m=15\text{ kg}`

The height the mass moves 500 times is,

`h=1.5\text{ m}`

The efficiency is,

`\eta=18\text{ \%}`

The output energy is,

`\begin{gathered} E_(out)=\text{mgh}*500 \\ =15*9.8*1.5*500 \\ =110.25*10^3\text{ J} \end{gathered}`

If the energy input is,

`E_i`

We can write,

`\begin{gathered} \eta=(E_(out))/(E_i)*100\text{ \%} \\ E_i=(E_(out))/(\eta)*100\text{ \%} \end{gathered}`

Substituting the values we get,

`\begin{gathered} E_i=\frac{110.25*10^3}{18\text{ \%}}*100\text{ \%} \\ =0.6125*10^6\text{ J} \\ =612500\text{ J} \end{gathered}`

Hence, the input chemical energy should be,

`0.6125*10^6J=612500\text{ J}`