## MCAT Physics and Math Review

## Chapter 2: Work and Energy

### 2.2 Work

Often, the term work is used erroneously to mean another form of energy. After all, the SI unit for work is the joule (J), which is the same SI unit for all forms of energy. Nevertheless, to say that work is just another form of energy is to miss something important: **work** is not actually a form of energy itself, but a process by which energy is transferred from one system to another. In fact, it is one of only two ways in which energy can be transferred. The other transfer of energy is called heat, which we will focus on quite a bit in Chapter 3 of *MCAT Physics and Math Review*.

**KEY CONCEPT**

Work is not energy but a measure of energy transfer. The other form of energy transfer is heat.

The transfer of energy by work or heat is the only way by which anything occurs. We are familiar with both processes from everyday life. For example, as discussed in the introduction to this chapter, every time King Sisyphus pushed the rock up the hill, the rock gained kinetic and potential energy. That energy came from Sisyphus’s muscles, in which the potential energy contained in the high-energy phosphate bonds of ATP molecules was converted to the mechanical energy of the contracting muscles, which exerted forces against the rock, causing it to accelerate and move up the hill.

On a chemical level, the potential energy in the ATP was harnessed by heat transfer. In fact, at the molecular level, this is no different from work because it involves the movement of molecules, atoms, and electrons, each of which exert forces that do work on other molecules and atoms. Like any transfer of energy, it’s not a perfectly efficient process, and some of the energy is lost as thermal energy. Our muscles quite literally warm up when we contract them repeatedly.

FORCE AND DISPLACEMENT

Energy is transferred through the process of work when something exerts forces on or against something else. This is expressed mathematically by the equation

*W* = **F ·** **d** = *Fd* cos *θ*

**Equation 2.7**

where *W* is work, *F* is the magnitude of the applied force, *d* is the magnitude of the displacement through which the force is applied, and *θ* is the angle between the applied force vector and the displacement vector. You’ll notice that work is a dot product; as such, it is a function of the cosine of the angle between the vectors. This also means that only forces (or components of forces) parallel or antiparallel to the displacement vector will do work (that is, transfer energy). We’ve already said that the SI unit for work is the joule. While this suggests that work and energy are the same thing, remember they are not: work is the process by which a quantity of energy is moved from one system to another.

PRESSURE AND VOLUME

As described above, work is a process of energy transfer. In mechanics, we think of work as application of force through some distance. We will learn in our discussion of fluids in Chapter 4 of *MCAT Physics and Math Review* that pressure can be thought of as an “energy density.” In systems of gases, we therefore approach work as a combination of pressure and volume changes. In Chapter 3 of *MCAT Physics and Math Review*, we’ll examine how these changes also relate to heat.

For a gas system contained in a cylinder with a movable **piston**, we can analyze the relationship between pressure, volume, and work. When the gas expands, it pushes up against the piston, exerting a force that causes the piston to move up and the volume of the system to increase. When the gas is compressed, the piston pushes down on the gas, exerting a force that decreases the volume of the system. We say that work has been done when the volume of the system has changed due to an applied pressure. Gas expansion and compression processes can be represented in graphical form with volume on the *x*-axis and pressure on the *y*-axis. Such graphs, as shown in Figure 2.2, are termed **P–V graphs**.

**Figure** **2.2.** **Pressure–Volume (P–V) Curves** *The work done on or by a system undergoing a thermodynamic process can be determined by finding the area enclosed by the corresponding pressure–volume curve.*

When a gas expands, we say that work was done by the gas and the work is positive; when a gas is compressed, we say that work was done on the gas and the work is negative. There are an infinite number of paths between an initial and final state. Different paths require different amounts of work. You can calculate the work done on or by a system by finding the area under the pressure–volume curve. Note that if volume stays constant as pressure changes (that is, Δ*V* = 0), then no work is done because there is no area to calculate. This is the case in Figure 2.2a above, and is called an **isovolumetric** or **isochoric process**. On the other hand, if pressure remains constant as volume changes (that is, Δ*P* = 0), then the area under the curve is a rectangle of length *P* and width Δ*V* as shown in Figure 2.2b. For processes in which pressure remains constant (**isobaric processes**), the work can be calculated as

*W* = *P*Δ*V*

**Equation 2.8**

**KEY CONCEPT**

When work is done *by* a system (the gas expands), the work is said to be positive. When work is done *on* a system (the gas compresses), the work is said to be negative. The MCAT will not expect you to calculate the integral of a P–V graph using calculus, but you are expected to be able to calculate the area under a straight-line graph if necessary.

Figure 2.2c shows a process in which neither pressure nor volume is held constant. The total area under the graph (Regions I and II) gives the work done.

Region I is a triangle with base Δ*V* and height Δ*P*, so the area is

Region II is a rectangle with base Δ*V* and height *P*_{2}, so its area is

*A*_{II} = *P*_{2}Δ*V*

The work done is the sum of the areas of regions I and II:

*W* = *A*_{I} + *A*_{II}

Figure 2.2d shows a closed cycle in which, after certain interchanges of work and heat, the system returns to its initial state. Because work is positive when the gas expands and negative when the gas is compressed, the work done is the area enclosed by the curve. Calculating the work done in this situation would require calculus, but the MCAT does not test calculus-based physics.

POWER

**Power** refers to the rate at which energy is transferred from one system to another. It is calculated with the equation

**Equation 2.9**

where *P* is power, *W* is work (which is equal to Δ*E*, the change in energy) and *t* is the time over which the work is done. The SI unit for power is the **watt** (**W**), which is equal to In Chapter 6 of *MCAT Physics and Math Review*, we will identify additional ways to calculate power in electric circuits. For now, note that many of the devices we use every day—toaster ovens, light bulbs, phones, cars, and so on—are quantified by the rate at which these appliances transform electrical potential energy into other forms, such as thermal, light, sound, and kinetic energy.

**BRIDGE**

Power is calculated in many different situations, especially those involving circuits, resistors, and capacitors. The equation for electric power is *P* = *IV*, where *P* is power, *I* is current, and *V* is electrical potential difference (voltage). This equation is discussed in Chapter 6 of *MCAT Physics and Math Review*. Power is always a measure of the rate of energy consumption, transfer, or transformation per unit time.

WORK–ENERGY THEOREM

The **work–energy theorem** is a powerful expression of the relationship between work and energy. In its mechanical applications, it offers a direct relationship between the work done by all the forces acting on an object and the change in kinetic energy of that object. The net work done by forces acting on an object will result in an equal change in the object’s kinetic energy. In other words,

*W*_{net} = Δ*K* = *K*_{f} – *K*_{i}

**Equation 2.10**

This relationship is important to understand, as it allows one to calculate work without knowing the magnitude of the forces acting on an object or the displacement through which the forces act. If one calculates the change in kinetic energy experienced by an object, then—by definition—the net work done on or by an object is the same. Pressing the brake pedal in your car puts the work–energy theorem into practice. The brake pads exert frictional forces against the rotors, which are attached to the wheels. These frictional forces do work against the wheels, causing them to decelerate and bringing the car to a halt. The net work done by all these forces is equal to the change in kinetic energy of the car.

In more general iterations, the work–energy theorem can be applied to changes in other forms of energy. In fact, the first law of thermodynamics is essentially a reiteration of the work–energy theorem, in which the change in internal energy (Δ*U*) is equal to the heat transferred into the system (*Q*) minus the mechanical work done by the system (*W*).

**Example:**

A lead ball of mass 0.125 kg is thrown straight up in the air with an initial velocity of Assuming no air resistance, find the work done by the force of gravity by the time the ball is at its maximum height.

**Solution:**

The answer could be calculated using kinematics and determining the maximum height of the ball (*W* = *Fd cos θ*), but it is simpler to use the work–energy theorem:

**MCAT Concept Check 2.2:**

Before you move on, assess your understanding of the material with these questions.

1. What are the units for work? How are work and energy different?

2. Provide three methods for calculating the work done on or by a system.

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