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Expert-verified Found in: Page 263 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Using data from Table 7.5, calculate the daily energy needs of a person who sleeps for 7.00 h, walks for 2.00 h, attends classes for 4.00 h, cycles for 2.00 h, sits relaxed for 3.00 h, and studies for 6.00 h. (Studying consumes energy at the same rate as sitting in class.)

The daily energy needs of a person is $$3799.62{\rm{ kcal}}$$.

See the step by step solution

## Work

Useful work: Useful work includes all the work done by a person, because these are accomplished by exerting forces (nonconservative in nature) on the outside work.

## Person’s daily energy needed

Daily energy need of a person is,

$$E = {P_{sl}} \times {T_{sl}} + {P_w} \times {T_w} + {P_{ac}} \times {T_{ac}} + {P_{cy}} \times {T_{cy}} + {P_s} \times {T_s} + {P_{st}} \times {T_{st}}$$

Here, $${P_{sl}}$$ is the power consumed in sleeping $$\left( {{P_{sl}} = 83{\rm{ W}}} \right)$$, $${T_{sl}}$$ is the time spent in sleeping $$\left( {{T_{sl}} = 7.00{\rm{ h}}} \right)$$, $${P_w}$$ is the power consumed in walking $$\left( {{P_w} = 280{\rm{ W}}} \right)$$, $${T_w}$$ is the time spent in walking $$\left( {{T_w} = 2.00{\rm{ h}}} \right)$$, $${P_{ac}}$$ is the power consumed in attending classes $$\left( {{P_{ac}} = 210{\rm{ W}}} \right)$$, $${T_{ac}}$$ is the time spent in attending classes $$\left( {{T_{ac}} = 4.00{\rm{ h}}} \right)$$, $${P_{cy}}$$ is the power consumed in cycling $$\left( {{P_{cy}} = 400{\rm{ W}}} \right)$$, $${T_{cy}}$$ is the time spent in cycling $$\left( {{T_{cy}} = 2.00{\rm{ h}}} \right)$$, $${P_s}$$ is the power consumed in sitting relaxed $$\left( {{{\rm{P}}_{\rm{s}}}{\rm{ = 125 W}}} \right)$$, $${T_s}$$ is the time spent in sitting relaxed $$\left( {{T_s} = 3.00{\rm{ h}}} \right)$$, $${P_{st}}$$ is the power consumed in studying $$\left( {{P_{st}} = 210{\rm{ W}}} \right)$$, and $${T_{st}}$$ is the time spent in studying $$\left( {{T_{st}} = 6.00{\rm{ h}}} \right)$$.

Putting all known values,

\begin{aligned}E &= \left( \begin{aligned}{l}\left( {83{\rm{ W}}} \right) \times \left( {7.00{\rm{ h}}} \right) + \left( {280{\rm{ W}}} \right) \times \left( {2.00{\rm{ h}}} \right) + \left( {{\rm{210 W}}} \right){\rm{ \times }}\left( {{\rm{4}}{\rm{.00 h}}} \right) + \\\left( {{\rm{400 W}}} \right) \times \left( {2.00{\rm{ h}}} \right) + \left( {125{\rm{ W}}} \right) \times \left( {3.00{\rm{ h}}} \right) + \left( {210{\rm{ W}}} \right) \times \left( {6.00{\rm{ h}}} \right)\end{aligned} \right)\\ &= 4416{\rm{ W}} \cdot {\rm{h}} \times \left( {\frac{{1{\rm{ J}}/{\rm{s}}}}{{1{\rm{ W}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hr}}}}} \right)\\ &= 15897600{\rm{ J}} \times \left( {\frac{{{\rm{1 kcal}}}}{{4184{\rm{ J}}}}} \right)\\ &= 3799.62{\rm{ kcal}}\end{aligned}

Therefore, the required daily energy needs of a person is $$3799.62{\rm{ kcal}}$$. ### Want to see more solutions like these? 