72. Work: The Scientific DefinitionLearning Objectives
What It Means to Do WorkThe scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to energy—whenever work is done, energy is transferred. For work, in the scientific sense, to be done, a force must be exerted and there must be motion or displacement in the direction of the force. Formally, the where $W$ is work, $\mathbf{d}$ is the displacement of the system, and $\theta $ is the angle between the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$, as in Figure 1. We can also write this as $$W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta .$$To find the work done on a system that undergoes motion that is not oneway or that is in two or three dimensions, we divide the motion into oneway onedimensional segments and add up the work done over each segment. What is Work?:The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts. For oneway motion in one dimension, this is expressed in equation form as $$W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \text{,}$$where $W$ is work, $F$ is the magnitude of the force on the system, $d$ is the magnitude of the displacement of the system, and $\theta $ is the angle between the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$. To examine what the definition of work means, let us consider the other situations shown in Figure 1. The person holding the briefcase in Figure 1(b) does no work, for example. Here $d=0$, so $W=0$. Why is it you get tired just holding a load? The answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the “briefcaseEarth system”—see Gravitational Potential Energy for more details). There must be motion for work to be done, and there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase on level ground in Figure 1(c) does no work on it, because the force is perpendicular to the motion. That is, $\text{cos}\phantom{\rule{0.25em}{0ex}}\text{90}\text{\xba =}\phantom{\rule{0.25em}{0ex}}0$, and so $W=0$. In contrast, when a force exerted on the system has a component in the direction of motion, such as in Figure 1(d), work is done—energy is transferred to the briefcase. Finally, in Figure 1(e), energy is transferred from the briefcase to a generator. There are two good ways to interpret this energy transfer. One interpretation is that the briefcase’s weight does work on the generator, giving it energy. The other interpretation is that the generator does negative work on the briefcase, thus removing energy from it. The drawing shows the latter, with the force from the generator upward on the briefcase, and the displacement downward. This makes $\theta =\text{180}\text{\xba}$, and $\text{cos 180}\text{\xba}=\mathrm{\u20131}$; therefore, $W$ is negative. Calculating WorkWork and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI units, work and energy are measured in Example 1: Calculating the Work You Do to Push a Lawn Mower Across a Large LawnHow much work is done on the lawn mower by the person in Figure 1(a) if he exerts a constant force of $\text{75}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{N}$ at an angle $\text{35}\text{\xba}$ 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 $1\text{\xba}\text{C}$, 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 $$\begin{array}{lll}W& =& \left(\text{75.0 N}\right)\left(\text{25.0 m}\right)\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\text{35.0\xba}\right)\\ & =& \text{1536 J}=\text{1.54}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}$$Converting the work in joules to kilocalories yields $W=(\text{1536}\phantom{\rule{0.25em}{0ex}}\text{J})(1\phantom{\rule{0.25em}{0ex}}\text{kcal}/\text{4184}\phantom{\rule{0.25em}{0ex}}\text{J})=0\text{.}\text{367}\phantom{\rule{0.25em}{0ex}}\text{kcal}$. The ratio of the work done to the daily consumption is $$\frac{W}{\text{2400}\phantom{\rule{0.25em}{0ex}}\text{kcal}}=1\text{.}\text{53}\times {\text{10}}^{4}\text{.}$$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. Section Summary
Conceptual QuestionsExercise 1Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or changed in form in your example? If so, explain how this is accomplished without doing work. Exercise 2Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work. Exercise 3Describe a situation in which a force is exerted for a long time but does no work. Explain. Problems & ExercisesExercise 1How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories. Show/Hide Solution Solution$$3\text{.}\text{00}\text{J}=7\text{.}\text{17}\times {\text{10}}^{4}\text{kcal}$$Exercise 2A 75.0kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. Exercise 3(a) Calculate the work done on a 1500kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the lift by the gravitational force in this process? (c) What is the total work done on the lift? Show/Hide Solution Solution(a) $5\text{.}\text{92}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J}$ (b) $5\text{.}\text{88}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J}$ (c) The net force is zero. Exercise 4Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Section 77 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s? Exercise 5Calculate the work done by an 85.0kg man who pushes a crate 4.00 m up along a ramp that makes an angle of $\text{20}\text{.}0\text{\xba}$ with the horizontal. (See Figure 2.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp. Show/Hide Solution Solution$$3\text{.}\text{14}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J}$$Exercise 6How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 3? Assume no friction acts on the wagon. Exercise 7A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction $\text{25}\text{.}0\text{\xba}$ below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart? Show/Hide Solution Solution(a) $\text{700}\phantom{\rule{0.25em}{0ex}}\text{J}$ (b) 0 (c) 700 J (d) 38.6 N (e) 0 Exercise 8Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a $\text{60}\text{.}0\text{\xba}$ slope at constant speed, as shown in Figure 4. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?
