Respuesta :
Given:
The height of the mass is,
[tex]m=15\text{ kg}[/tex]The height the mass moves 500 times is,
[tex]h=1.5\text{ m}[/tex]The efficiency is,
[tex]\eta=18\text{ \%}[/tex]The output energy is,
[tex]\begin{gathered} E_{out}=\text{mgh}\times500 \\ =15\times9.8\times1.5\times500 \\ =110.25\times10^3\text{ J} \end{gathered}[/tex]If the energy input is,
[tex]E_i[/tex]We can write,
[tex]\begin{gathered} \eta=\frac{E_{out}}{E_i}\times100\text{ \%} \\ E_i=\frac{E_{out}}{\eta}\times100\text{ \%} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} E_i=\frac{110.25\times10^3}{18\text{ \%}}\times100\text{ \%} \\ =0.6125\times10^6\text{ J} \\ =612500\text{ J} \end{gathered}[/tex]Hence, the input chemical energy should be,
[tex]0.6125\times10^6J=612500\text{ J}[/tex]