1.1 Work: the scientific definition  (Page 3/7)

Calculating the work you do to push a lawn mower across a large lawn

How much work is done on the lawn mower by the person in [link] (a) if he exerts a constant force of $\text{75}\text{.}0\text{N}$ at an angle $\text{35}\text{º}$ below the horizontal and pushes the mower $\text{25}\text{.}0\text{m}$ on level ground? Convert the amount of work from joules to kilocalories and compare it with this person’s average daily intake of $\text{10},\text{000}\text{kJ}$ (about $\text{2400}\text{kcal}$ ) 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 $4\text{.}\text{184}\text{J}$ , while one food calorie (1 kcal) is equivalent to $\text{4184}\text{J}$ .

Strategy

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

Solution

The equation for the work is

Substituting the known values gives

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

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.