A) We know that efficiency in a system is defined by the equation
[tex]\eta = \frac{P_{out}}{P_{in}}[/tex]
Where P is the Power in a system. Then,
[tex]P_{out} = P_{in}* \eta[/tex]
[tex]P_{out} = 700*3\%[/tex]
[tex]P_{out} = 21W[/tex]
Therefore the output power is 21W
B) Work is defined by,
[tex]W=F*d[/tex]
[tex]W=mg*d=2700(9.8)*1.2 = 31.75kJ[/tex]
To calculate the power we know that is the rate of work in a determined time, then
[tex]P=\frac{W}{t}[/tex]
Re-arrange for t,
[tex]t=\frac{W}{P} = \frac{31.75*10^3}{21}=1511s[/tex]
c) Finally we have that the amount of heat wasted depends on the power, then,
[tex]Q= (700-21)*1511[/tex]
[tex]Q=1.02*10^6J[/tex]
