Calculating Work

Examples of work. (a) The work done by the force \(\mathbf{F}\) on this lawn mower is \(\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \). Note that \(F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) is the component of the force in the direction of motion. (b) A person holding a briefcase does no work on it, because there is no displacement. No energy is transferred to or from the briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by carrying it up stairs at constant speed, because there is necessarily a component of force \(\mathbf{F}\) in the direction of the motion. Energy is transferred to the briefcase and could in turn be used to do work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because \(\mathbf{F}\) and \(\mathbf{d}\) are in opposite directions.

How much work is done on the lawn mower by the person in figure (a) above if he exerts a constant force of \(\text{75}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{N}\) at an angle \(\text{35}\text{º}\) below the horizontal and pushes the mower \(\text{25}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m}\) on level ground? Convert the amount of work from joules to kilocalories and compare it with this person’s average daily intake of \(\text{10},\text{000}\phantom{\rule{0.25em}{0ex}}\text{kJ}\) (about \(\text{2400}\phantom{\rule{0.25em}{0ex}}\text{kcal}\)) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by \(1\text{º}\text{C}\), and is equivalent to \(4\text{.}\text{184}\phantom{\rule{0.25em}{0ex}}\text{J}\), while one food calorie (1 kcal) is equivalent to \(\text{4184}\phantom{\rule{0.25em}{0ex}}\text{J}\).

Strategy

We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation \(W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \). The force, angle, and displacement are given, so that only the work \(W\) is unknown.

Solution

The equation for the work is

\(W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta .\)

Substituting the known values gives

\(\begin{array}{lll}W& =& \left(\text{75.0 N}\right)\left(\text{25.0 m}\right)\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{35.0º}\right)\\ & =& \text{1536 J}=\text{1.54}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}\)

Converting the work in joules to kilocalories yields \(W=\left(\text{1536}\phantom{\rule{0.25em}{0ex}}\text{J}\right)\left(1\phantom{\rule{0.25em}{0ex}}\text{kcal}/\text{4184}\phantom{\rule{0.25em}{0ex}}\text{J}\right)=0\text{.}\text{367}\phantom{\rule{0.25em}{0ex}}\text{kcal}\). The ratio of the work done to the daily consumption is

\(\cfrac{W}{\text{2400}\phantom{\rule{0.25em}{0ex}}\text{kcal}}=1\text{.}\text{53}×{\text{10}}^{-4}\text{.}\)

Discussion

This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to do work. Even when we “work” all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to thermal energy or stored as chemical energy in fat.