Chapter 5: Work, Energy, and Power

Ans: As provided, the circular area swept by the windmill \[=A\]

And the velocity of the wind is \[=v\]

And the density of the air is \[=\rho \]

The volume flow rate of wind through the windmill is\[=Av\]

And hence the mass flow rate is \[m/t=\rho Av\]

Hence the mass flowing in time t is \[m=\rho Avt\].

1. What is the kinetic energy of the air? 

Ans: Using this mass the kinetic energy is obtained as

\[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\rho A{{v}^{2}}t={{E}_{wind}}\]

This is also equal to the energy that the wind carries with itself.

1. Assume that the windmill converts \[25%\] of the wind’s energy into electrical energy, and that \[A=30{{m}^{2}}\], \[v=36km/h\] and the density of air is \[\rho =1.2kg/{{m}^{3}}\]. What is the electrical power produced?

Ans: The area of the circle swept is \[A=30{{m}^{2}}\]

And the given velocity is \[v=36km/h=10m/s\]

The density is \[\rho =1.2kg/{{m}^{3}}\]

Since the windmill converts only 25% of wind energy to electricity.

 \[{{E}_{electric}}=25%{{E}_{wind}}=\frac{1}{8}\rho A{{v}^{2}}t\]

Hence the electric power is obtained as \[{{P}_{electric}}={{E}_{electric}}/t=4.5kW\].

The electric power output from the windmill is 4.5kW.

Ans: Mass of the dumbbell \[m=10kg\]

The height up to which it is lifted \[h=0.5m\]

Hence the work done with every lift is\[mgh\]

Number of times the dumbbell is lifted\[n=1000\]

Hence the total work done against gravity is  \[=n\,mgh=49kJ\].

1. Fat supplies \[3.8\times {{10}^{7}}J\] of energy per kilogram which is converted to mechanical energy with a \[20%\] efficiency rate. How much fat will the dieter use up?

Ans: Energy supplied with every kilogram of fat is\[3.8\times {{10}^{7}}J\]

The efficiency is\[20%\]

Hence the mechanical energy supplied by the body

\[=\frac{20}{100}3.8\times {{10}^{7}}J=7.6\times {{10}^{6}}J\]

And the equivalent fat loss is \[6.45\times {{10}^{-3}}kg\]

Hence the body loses \[6.45\times {{10}^{-3}}kg\] of fat.

Ans: The given power requirement of a family is\[P=8kW=8\times {{10}^{3}}W\]

The solar energy per sq mts \[{{P}_{solar}}/A=200W\]

Efficiency in converting solar to electrical energy \[\eta =20%\]

Let the area needed to generate the required electricity be \[A\]

Total solar power is  \[{{P}_{solar}}=({{P}_{solar}}/A).A\]

And the expected electrical power output can be written as\[P=20%({{P}_{solar}})=\frac{20}{100}200\times A\]

\[\Rightarrow A=200{{m}^{2}}\]

Hence the area required is \[A=200{{m}^{2}}\].

1. Compare this area to that of the roof of a typical house.

Ans: A typical house can have a roof of the dimensions \[15\times 15\,{{m}^{2}}\], hence the area is \[\ge 200{{m}^{2}}\]. And this is more than the required area.