Question: (a) Calculate the energy in kJ used by a 55.0-kg woman who does 50 deep knee bends in which her center of mass is lowered and raised 0.400 m. (She does work in both directions.) You may assume her efficiency is 20%.

(b) What is the average power consumption rate in watts if she does this in 3.00 min?

The efficiency of a device tells us how much efficient it is. In other words, the efficiency tells us that what fraction of total input energy is being used by the device to carry out the work.

Mathematically,

\(\eta = \frac{{{\rm{Useful output}}}}{{{\rm{Total input}}}} \times 100\% \)

(a)

The work done by the women one knee bends is,

\({W_1} = mgd\)

Here, m is the mass of the women \(\left( {m = 55.0{\rm{ kg}}} \right)\), g is the acceleration due to gravity\(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and d is the travelled in 1-knee bend (\(d = 2h = 0.8{\rm{ m}}\), h is the height the center of mass is lowered and raised).

The work done by the women to complete 50 knee bends is,

\(W = 50mgd\)

Putting all known values,

\(\begin{aligned} W &= 50 \times \left( {55.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {0.8{\rm{ m}}} \right)\\ &= 21560{\rm{ J}}\end{aligned}\)

Since, the women have the efficiency of 20%. Therefore, the total input energy consumption is,

\(E = \frac{W}{{\eta \% }} \times 100\% \)

Putting all known values,

\(\begin{aligned} E &= \frac{{21560{\rm{ J}}}}{{20\% }} \times 100\% \\ &= 107800{\rm{ J}} \times \left( {\frac{{1{\rm{ kJ}}}}{{1000{\rm{ J}}}}} \right)\\ &= 107.8{\rm{ kJ}}\end{aligned}\)

Therefore, the required energy used is \(107.8{\rm{ kJ}}\).

(b)

If the women do knee bend for 3 min, her average power consumption rate is,

\(P = \frac{E}{t}\)

Putting all known values,

\(\begin{aligned} P &= \frac{{107.8{\rm{ kJ}}}}{{3{\rm{ min}}}}\\ &= \frac{{\left( {107.8{\rm{ kJ}}} \right) \times \left( {\frac{{1000{\rm{ J}}}}{{1{\rm{ kJ}}}}} \right)}}{{\left( {3{\rm{ min}}} \right) \times \left( {\frac{{60{\rm{ sec}}}}{{1{\rm{ min}}}}} \right)}}\\ &= 598.89{\rm{ W}}\end{aligned}\)

Therefore, the required average power consumption rate is \(598.89{\rm{ W}}\).