Work PDF

## This Chapter is devoted to the very important concepts of work and energy. These two quantities are scalars and so have no direction associated with them, which often makes them easier to work with than vector quantities. ### 6-1 Work Done by a Constant Force The word work has a variety of mea...

## This Chapter is devoted to the very important concepts of work and energy. These two quantities are scalars and so have no direction associated with them, which often makes them easier to work with than vector quantities. ### 6-1 Work Done by a Constant Force The word work has a variety of meanings in everyday language. But in physics, work is given a very specific meaning to describe what is accomplished when a force acts on an object, and the object moves through a distance. Specifically, the work done on an object by a constant force (constant in both magnitude and direction) is defined to be the product of the magnitude of the displacement times the component of the force parallel to the displacement. In equation form, we can write W = F \cdot d. where F is the component of the constant force F parallel to the displacement d. We can also write W = Fd cos \theta. where F is the magnitude of the constant force, d is the magnitude of the displacement of the object, and \theta is the angle between the directions of the force and the displacement (Fig. 6-1). The cos \theta factor appears in Eq. 6-1 because F cos (\theta) is the component of F that is parallel to d. Work is a scalar quantity - it has only magnitude, which can be positive or negative. Let us first consider the case in which the motion and the force are in the same direction, so \theta = 0 and cos \theta = 1; in this case, W = Fd. For example, if you push a loaded grocery cart a distance of 50 m by exerting a horizontal force of 30 N on the cart, you do 30 N x 50 m = 1500 N \cdot m of work on the cart. As this example shows, in SI units work is measured in newton-meters (N \cdot m). A special name is given to this unit, the joule (J): 1 J = 1 N \cdot m. In the cgs system, the unit of work is called the erg and is defined as 1 erg = 1 dyne-cm. In British units, work is measured in foot-pounds. It is easy to show that 1 J = 10^7 erg = 0.7376 ft-lb. **FIGURE 6-1** A person pulling a crate along the floor. The work done by the force F is W = Fd cos \theta, where d is the displacement. **FIGURE 6-2** The person does no work on the bag of groceries since F_r is perpendicular to the displacement d. **CAUTION** A force can be exerted on an object and yet do no work. For example, if you hold a heavy bag of groceries in your hands at rest, you do no work on it. You do exert a force on the bag, but the displacement of the bag is zero, so the work done by you on the bag is W = 0. You need both a force and a displacement to do work. You also do no work on the bag of groceries if you carry it as you walk horizontally across the floor at constant velocity, as shown in Fig. 6-2. No horizontal force is required to move the bag at a constant velocity. The person shown in Fig. 6-2 does exert an upward force F_y on the bag equal to its weight. But this upward force is perpendicular to the horizontal displacement of the bag and thus has nothing to do with that motion. Hence, the upward force is doing no work. This conclusion comes from our definition of work, Eq. 6-1: W = 0, because \theta = 90° and cos 90° = 0. Thus, when a particular force is perpendicular to the displacement, no work is done by that force. (When you start or stop walking, there is a horizontal acceleration and you do briefly exert a horizontal force, and thus do work on the bag). **CAUTION** When we deal with work, as with force, it is necessary to specify whether you are talking about work done by a specific object or done on a specific object. It is also important to specify whether the work done is due to one particular force (and which one), or the total (net) work done by the net force on the object. **EXAMPLE 6-1 Work done on a crate.** A person pulls a 50-kg crate 40 m along a horizontal floor by a constant force F = 100 N, which acts at a 37° angle as shown in Fig. 6-3. The floor is rough and exerts a friction force F_f = 50 N. Determine (a) the work done by each force acting on the crate, and (b) the net work done on the crate. **APPROACH** We choose our coordinate system so that d can be the vector that represents the 40-m displacement (that is, along the x axis). Four forces act on the crate, as shown in Fig. 6-3: the force exerted by the person F; the friction force F_f due to the floor, the crate's weight mg, and the normal force F_n, exerted upward by the floor. The net force on the crate is the vector sum of these four forces **SOLUTION** (a) The work done by the gravitational and normal forces is zero, since they are perpendicular to the displacement d (\theta = 90° in Eq. 6-1): W_g = mgd cos 90° = 0 W_n = F_nd cos 90° = 0 The work done by F is W_F = Fd cos \theta = (100 N)(40 m) cos 37° = 3200 J. The work done by the friction force is W_f = F_fd cos 180 = (50 N)(40 m) (-1) = -2000 J. The angle between the displacement and the force F_f is 180° because they point in opposite directions. Since the force of friction is opposing the motion (and cos 180° = -1), the work done by friction on the crate is negative. (**FIGURE 6-3** Example 6-1. A 50-kg crate is pulled along a floor.) (b) The net work can be calculated in two equivalent ways: (1) The net work done on an object is the algebraic sum of the work done by each force, since work is a scalar: W = W_g + W_n + W_F + W_f = 0 + 0 + 3200 J - 2000 J = 1200 J. (2) The net work can also be calculated by first determining the net force on the object and then taking its component along the displacement: (F_net)_x = F cos \theta - F_f. Then the net work is W_net = (F_net)_x \cdot d = (100 N cos 37° - 50 N)(40 m) = 1200 J In the vertical (y) direction, there is no displacement and no work done. In Example 6-1 we saw that friction did negative work. In general, the work done by a force is negative whenever the force (or the component of the force, F) acts in the direction opposite to the direction of motion. Also, we can see that when the work done by a force on an object is negative, that force is trying to slow the object down (and would slow it down if that were the only force acting). When the work is positive, the force involved is trying to speed up the object. **EXERCISE A** A box is dragged across a floor by a force F, which makes an angle \theta with the horizontal as in Fig. 6-1 or 6-3. If the magnitude of F is held constant but the angle \theta is increased, the work done by F (a) remains the same; (b) increases; (c) decreases; (d) first increases, then decreases. **PROBLEM SOLVING Work** 1. Draw a free-body diagram showing all the forces acting on the object you choose to study. 2. Choose an xy coordinate system. If the object is in motion, it may be convenient to choose one of the coordinate directions as the direction of one of the forces, or as the direction of motion. [Thus, for an object on an incline, you might choose one coordinate axis to be parallel to the incline.] 3. Apply Newton's laws to determine any unknown forces. 4. Find the work done by a specific force on the object by using W = Fd cos \theta for a constant force. Note that the work done is negative when a force tends to oppose the displacement. 5. To find the net work done on the object, either (a) find the work done by each force and add the results algebraically, or (b) find the net force on the object, F_net, and then use it to find the net work done, which for constant net force is: W_net = F_net d cos \theta. **FIGURE 6-4** Example 6-2. **EXAMPLE 6-2 Work on a backpack**. (a) Determine the work a hiker must do on a 15.0-kg backpack to carry it up a hill of height 10.0 m, as shown in Fig. 6-4a. Determine also (b) the work done by gravity on the backpack, and (c) the net work done on the backpack. For simplicity, assume the motion is smooth and at constant velocity (i.e., acceleration is negligible). **APPROACH** We explicitly follow the Problem Solving Box step by step. **SOLUTION** 1. Draw a free-body diagram. The forces on the backpack are shown in Fig. 6-4b: the force of gravity, mg, acting downward; and F, the force the hiker must exert upward to support the backpack. Since we assume there is negligible acceleration, horizontal forces on the backpack are negligible. 2. Choose a coordinate system. We are interested in the vertical motion of the backpack, so we choose the y coordinate as positive vertically upward. 3. Apply Newton's laws. Newton's second law applied in the vertical direction to the backpack gives F - mg = ma = 0. Hence, F = mg = (15.0 kg)(9.80 m/s²) = 147 N. 4. Find the work done by a specific force. (a) To calculate the work done by the hiker on the backpack, we write Eq. 6-1 as W = F(d cos \theta), and we note from Fig. 6-4a that d cos \theta = h. So the work done by the hiker is W = F(d cos \theta) = Fh = mgh = (147 N)(10.0 m) = 1470 J. Note that the work done depends only on the change in elevation and not on the angle of the hill, \theta. The hiker would do the same work to lift the pack vertically the same height 4. (b) The work done by gravity on the backpack is (from Eq. 6-1 and Fig. 6-4c) W_g = F_g d cos (180° - \theta). Since cos(180°) = - cos \theta, we have W_g = F_g d (- cos \theta) = mg(-d cos \theta) = -mgh = -(15.0 kg)(9.80 m/s²)(10.0 m) = -1470 J. **NOTE** The work done by gravity (which is negative here) doesn't depend on the angle of the incline, only on the vertical height h of the hill. This is because gravity acts vertically, so only the vertical component of displacement contributes to work done. 5. Find the net work done. (a) The net work done on the backpack is W_net = 0, since the net force on the backpack is zero (it is assumed not to accelerate significantly). We can also determine the net work done by adding the work done by each force: W_net = W + W_g = 1470 J + (-1470 J) = 0. **NOTE** Even though the net work done by all the forces on the backpack is zero, the hiker does do work on the backpack equal to 1470 J. **FIGURE 6-5** Example 6-3. **CONCEPTUAL EXAMPLE 6-3** Does the Earth do work on the Moon? The Moon revolves around the Earth in a nearly circular orbit, kept there by the gravitational force exerted by the Earth. Does gravity do (a) positive work, (b) negative work, or (c) no work on the Moon? **RESPONSE** The gravitational force exerted by the Earth on the Moon (Fig. 6-5) acts toward the Earth and provides its centripetal acceleration, inward along the radius of the Moon's orbit. The Moon's displacement at any moment is tangent to the circle, in the direction of its velocity, perpendicular to the radius and perpendicular to the force of gravity. Hence the angle \theta between the force F_g and the instantaneous displacement of the Moon is 90°, and the work done by the Earth's gravity on the Moon as it orbits is therefore zero (cos 90° = 0). This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of fuel; no net work needs to be done against the force of gravity. **6-2 Work Done by a Varying Force** If the force acting on an object is constant, the work done by that force can be calculated using Eq. 6-1. But in many cases, the force varies in magnitude or direction during a process. For example, as a rocket moves away from Earth, work is done to overcome the force of gravity, which varies as the inverse square of the distance from the Earth's center. Other examples are the force exerted by a spring, which increases with the amount of stretch, or the work done by a varying force in pulling a box or cart up an uneven hill. **FIGURE 6-6** The work done by a force F can be calculated by taking (a) the sum of the areas of the rectangles: (b) the area under the curve of F vs d. The work done by a varying force can be determined graphically. The procedure is like that for determining displacement when the velocity is known as a function of time (Section 2-8). To determine the work done by a variable force, we plot F(d cos \theta), the component of F parallel to the direction of motion at any point) as a function of distance d, as in Fig. 6-6a. We divide the distance into small segments \Delta d. For each segment, we indicate the average of F by a horizontal dashed line. Then the work done for each segment is \Delta W = F \Delta d, which is the area of a rectangle \Delta d wide and F high. The total work done to move the object a total distance d = \sum \Delta d is the sum of the areas of the rectangles (like in the case shown in Fig. 6-6a). Usually, the average value of F for each segment must be estimated, and a reasonable approximation of the work done can then be made. If we subdivide the distance into many more segments, \Delta d can be made smaller and our estimate of the work done would be more accurate. In the limit as \Delta d approaches zero, the total area of the many narrow rectangles approaches the area under the curve, Fig. 6-6b. That is, the work done by a variable force in moving an object between two points is equal to the area under the F(cos \theta) vs. d curve between those two points. ### 6-3 Kinetic Energy, and the Work-Energy Principle Energy is one of the most important concepts in science. Yet we cannot give a simple general definition of energy in only a few words. Nonetheless, each specific type of energy can be defined fairly simply. In this Chapter, we define translational kinetic energy and some types of potential energy. In later Chapters, we will examine other types of energy, such as that related to heat (Chapters 14 and 15). The crucial aspect of all the types of energy is that the sum of all types, the total energy, is the same after any process as it was before: that is, the quantity "energy" is a conserved quantity. For the purposes of this Chapter, we can define energy in the traditional way as "the ability to do work". This simple definition is not very precise, nor is it really valid for all types of energy. It is valid, however, for mechanical energy which we discuss in this Chapter, and it serves to underscore the fundamental Energy associated with heat is often not available to do work, as we will discuss in Chapter 15. **FIGURE 6-7** A constant net force F_net accelerates a bus from speed v_i to speed v_f over a displacement of d. The net work done is W_net = F_net d. A moving object can do work on another object it strikes. A flying cannon-ball does work on a brick wall it knocks down; a moving hammer does work on a nail it drives into wood. In either case, a moving object exerts a force on a second object which undergoes a displacement. An object in motion has the ability to do work and thus can be said to have energy. The energy of motion is called kinetic energy, from the Greek word kinetikos, meaning "motion". To obtain a quantitative definition for kinetic energy, let us consider a rigid object of mass m that is moving in a straight line with an initial speed v_i. To accelerate it uniformly to a speed v_f, a constant net force F_net is exerted on it parallel to its motion over a displacement d, Fig. 6-7. Then the net work done on the object is W_net = F_net d. We apply Newton's second law, F_net=ma, and use Eq. 2-11c, which we now write as v_f² = v_i² + 2ad, with v_i as the initial speed and v_f the final speed. We solve for a in Eq. 2-11c, a = (v_f² - v_i²)/2d then substitute this into F_net=ma, and determine the work done: W_net = F_net d = mad = m d (v_f² - v_i²)/2d or W_net = 1/2 mv_f² - 1/2 mv_i² We define the quantity 1/2 mv² to be the translational kinetic energy (KE) of the object: KE = 1/2 mv² (We call this "translational" kinetic energy to distinguish it from rotational kinetic energy, which we will discuss in Chapter 8.) Equation 6-2, derived here for one-dimensional motion with a constant force, is valid in general for translational motion of an object in three dimensions and even if the force varies. We can rewrite Eq. 6-2 as: W_net = \Delta KE = KE_f - KE_i or W_net = \Delta KE = 1/2 mv_f² - 1/2 mv_i² Equation 6-4 (or Eq. 6-2) is an important result known as the ***work-energy principle***. It can be stated in words: ***The net work done on an object is equal to the change in the object's kinetic energy.*** Notice that we made use of Newton's second law, F_net = ma, where F_net is the net force - the sum of all forces acting on the object. Thus, the work-energy principle is valid only if W_net is the net work done on the object - that is, the work done by all forces acting on the object. The work-energy principle is a very useful reformulation of Newton's laws. It tells us that if (positive) net work W_net is done on an object, the object's kinetic energy increases by an amount W_net. The principle also holds true for the reverse situation: if the net work W_net done on an object is negative, the object's kinetic energy decreases by an amount W_net. That is, a net force exerted on an object opposite to the object's direction of motion decreases its speed and its kinetic energy. An example is a moving hammer (Fig. 6-8) striking a nail. The net force on the hammer (F_net in Fig. 6-8, where F is assumed constant for simplicity) acts toward the left, whereas the displacement d of the hammer is toward the right. So the net work done on the hammer, W_net = (F_net) (d) (cos 180°) = -F_net d, is negative and the hammer's kinetic energy decreases (usually to zero). **FIGURE 6-8** A moving hammer strikes a nail and comes to rest. The hammer exerts a force F on the nail, the nail exerts a force -F on the hammer (Newton's third law). The work done on the nail by the hammer is positive (W = Fd > 0), The work done on the hammer by the nail is negative (W = -Fd). Figure 6-8 also illustrates how energy can be considered the ability to do work. The hammer, as it slows down, does positive work on the nail: if the nail exerts a force -F on the hammer to slow it down, the hammer exerts a force +F on the nail (Newton's third law) through the distance d. Hence the work done on the nail by the hammer is W = (+F)(+d) = Fd and is positive. We also see that W = -W_net: the work done on the nail W equals the negative of the work done on the hammer. That is, the decrease in kinetic energy of the hammer is equal to the work the hammer can do on another object - which is consistent with energy being the ability to do work. Whereas the translational kinetic energy (1/2 mv²) is directly proportional to the mass of the object, it is proportional to the square of the speed. Thus, if the mass is doubled, the kinetic energy is doubled. But if the speed is doubled, the object has four times as much kinetic energy and is therefore capable of doing four times as much work. Let us summarize the relationship between work and kinetic energy (Eq. 6-4): if the net work W_net done on an object is positive, then the object's kinetic energy increases. If the net work W_net done on an object is negative, its kinetic energy decreases. If the net work done on the object is zero, its kinetic energy remains constant (which also means its speed is constant). Because of the direct connection between work and kinetic energy (Eq. 6-4), energy is measured in the same units as work: joules in SI units, ergs in the cgs, and foot-pounds in the British system. Like work, kinetic energy is a scalar quantity. The kinetic energy of a group of objects is the sum of the kinetic energies of the individual objects. **EXAMPLE 6-4 KE and work done on a baseball.** A 145-g baseball is thrown so that it acquires a speed of 25 m/s. (a) What is its kinetic energy? (b) What was the net work done on the ball to make it reach this speed, if it started from rest? **APPROACH** We use the definition of kinetic energy, Eq. 6-3, and then the work-energy principle, Eq. 6-4. **SOLUTION** (a) The kinetic energy of the ball after the throw is KE = 1/2 (0.145 kg) (25 m/s)² = 45.1 J. (b) Since the initial kinetic energy was zero, the net work done is just equal to the final kinetic energy, 45.1 J. **EXAMPLE 6-5 Work on a car, to increase its KE.** How much net work is required to accelerate a 1000-kg car from 20 m/s to 30 m/s (Fig. 6-9)? **FIGURE 6-9** Example 6-5. **APPROACH** To simplify a complex situation, let us treat the car as a particle or simple rigid object. We can then use the work-energy principle. **SOLUTION** The net work needed is equal to the increase in kinetic energy: W_net = \Delta KE = KE_f - KE_i = (1/2)(1000 kg)(30 m/s)² - (1/2)(1000 kg)(20 m/s)² = 2.5 × 10^5 J. **NOTE** You might be tempted to work this Example by finding the force and using Eq. 6-1. That won't work, however, because we don't know how far or for how long the car was accelerated. In fact, a large force could be acting for a small distance, or a small force could be acting over a long distance; both could give the same net work **FIGURE 6-10** Example 6-6. **CONCEPTUAL EXAMPLE 6-6 Work to stop a car.** A car traveling 60 km/h can brake to a stop within a distance d of 20 m (Fig. 6-10a). If the car is going twice as fast, 120 km/h, what is its stopping distance (Fig. 6-10b)? Assume the maximum braking force is approximately independent of speed. **RESPONSE** Since the stopping force F is approximately constant, the work needed to stop the car, Fd, is proportional to the distance traveled. We apply the work-energy principle, noting that F and d are in opposite directions and that the final speed of the car is zero: W_net = - Fd = \Delta KE = 1/2 mv_f² - 1/2 mv_i² Then -Fd = 0 -1/2 mv_i² Thus, since the force and mass are constant, we see that the stopping distance, d, increases with the square of the speed: d \propto v_i². If the car's initial speed is doubled, the stopping distance is (2)² = 4 times as great, or 80 m. **EXERCISE B** Can kinetic energy ever be negative? ### 6-4 Potential Energy We have just discussed how an object is said to have energy by virtue of its motion, which we call kinetic energy. But it is also possible to have potential energy, which is the energy associated with forces that depend on the position or configuration of an object (or objects) relative to the surroundings. Various types of potential energy (PE) can be defined, and each type is associated with a particular force. The spring of a wind-up toy is an example of an object with potential energy. The spring acquired its potential energy because work was done on it by the person winding the toy. As the spring unwinds, it exerts a force and does work to make the toy move. Perhaps the most common example of potential energy is ***gravitational potential energy***. A heavy brick held high in the air has potential energy because of its position relative to the Earth. The raised brick has the ability to do work, for if it is released, it will fall to the ground due to the gravitational force, and can do work on, say, a stake, driving it into the ground. Let us seek the form for the gravitational potential energy of an object near the surface of the Earth. For an object of mass m to be lifted vertically, an upward force at least equal to its weight, mg, must be exerted on it, say by a person's hand. **FIGURE 6-11** A person exerts an upward force F = mg to lift a brick from y_1 to y_2. To lift it without acceleration a vertical displacement of height h, from position y_1 to y_2 in Fig. 6-11 (upward direction chosen positive), a person must do work equal to the product of the needed external force, F = mg upward, and the vertical displacement h. That is, W_g = Fd cos 0° = mgh = mg(y_2 - y_1) Gravity is also acting on the object as it moves from y_1 to y_2, and does work on it equal to W_g = Fd cos \theta = mgh cos 180° where \theta = 180° because F_g and d point in opposite directions. So W_g = -mgh = -mg(y_2 - y_1) If we now allow the object to start from rest and fall freely under the action of gravity, it acquires a velocity given by v² = 2gh (Eq. 2-11c) after falling a height h. It then has kinetic energy 1/2 mv² = m(2gh) = mgh, and if it strikes a stake it can do work on the stake equal to mgh (work-energy principle). Thus, to raise an object of mass m to a height h requires an amount of work equal to mgh (Eq. 6-5a). And once at height h, the object has the ability to do an amount of work equal to mgh. We therefore define the ***gravitational potential energy*** of an object, due to Earth's gravity, as the product of the object's weight mg and its height y above some reference level (such as the ground): PE = mgy The higher an object is above the ground, the more gravitational potential energy it has. We combine Eq. 6-5a with Eq. 6-6: W_g = mg(y_2 - y_1) W_g = \Delta PE = APE. That is, the work done by an external force to move the object of mass m from point 1 to point 2 (without acceleration) is equal to the change in potential energy between positions 1 and 2. Alternatively, we can write the change in potential energy, \Delta PE, in terms of the work done by gravity itself, starting from Eq. 6-5b, we obtain W_g = -mg(y_2 - y_1) = - \Delta PE That is, the work done by gravity as the object of mass m moves from point 1 to point 2 is equal to the negative of the difference in potential energy between positions 1 and 2. **CAUTION** Potential energy belongs to a system, and not to a single object alone. Potential energy is associated with a force, and a force on one object is always exerted by some other object. Thus potential energy is a property of the system as a whole. For an object raised to a height y above the Earth's surface, the change in gravitational potential energy is mgy. The system here is the object plus the Earth, and properties of both are involved: object (m) and Earth (g). **GRAVITATIONAL PE depends on vertical height** Gravitational potential energy depends on the vertical height of the object above some reference level (Eq. 6-6). In some situations, you may wonder from what point to measure the height y. The gravitational potential energy of a book held high above a table, for example, depends on whether we measure y from the top of the table, from the floor, or from some other reference point. **CAUTION** What is physically important in any situation is the change in potential energy, \Delta PE, because that is what is related to the work done, Eqs. 6-7; and it is \Delta PE that can be measured. We can thus choose to measure y from any reference point that is convenient, but we must choose the reference point at the start and be consistent throughout a calculation. The change in potential energy between any two points does not depend on this choice. An important result we discussed earlier (see Example 6-2 and Fig. 6-4) concerns the gravity force, which does work only in the vertical direction: the work done by gravity depends only on the vertical height \Delta y, and not on the path taken, whether it be purely vertical motion or, say, motion along an incline. Thus, from Eqs. 6-7 we see that changes in gravitational potential energy depend only on the change in vertical height ^and not on the path taken. **FIGURE 6-12** Example 6-7. **EXAMPLE 6-7 Potential energy changes for a roller coaster.** A 1000-kg roller-coaster car moves from point 1, Fig. 6-12, to point 2 and then to point 3. (a) What is the gravitational potential energy at 2 and 3 relative to point 1? That is, take y=0 at point 1. (b) What is the change in potential energy when the car goes from point 2 to point 3? (c) Repeat parts (a) and (b), but take the reference point (y = 0) to be at point 3. **APPROACH** We are interested in the potential energy of the car-Earth system. We take upward as the positive y direction, and use the definition of gravitational potential energy to calculate PE. **SOLUTION** (a) We measure heights from point 1, which means initially that the gravitational potential energy is zero. At point 2, where y = 10 m, PE_2 = mgy_2 = (1000 kg)(9.8 m/s²)(10 m) = 9.8 x 10^4 J. At point 3, y = -15 m, since point 3 is below point 1. Therefore, PE_3 = mgy_3 = (1000 kg)(9.8 m/s²)(-15 m) = -1.5 x 10^5 J. (b) In going from point 2 to point 3, the potential energy change (\Delta PE) is \Delta PE = (PE_3) - (PE_2) = (-1.5 × 10^5 J) - (9.8 ×10^4 J) = -2.5 × 10^5 J. The gravitational potential energy decreases by 2.5 × 10^5 J. (c) In this instance, y = + 15 m at point 1, so the potential energy initially (at point 1) is PE_1 = (1000 kg)(9.8 m/s²)(15 m) = 1.5 × 10^5 J. At point 2, y = +25 m, so the potential energy is PE_2 = (1000 kg)(9.8 m/s²)(25 m) = 2.5 × 10^5 J. At point 3, y = 0, so the potential energy is zero. The change in potential energy going from point 2 to point 3 is \Delta PE = PE_3 - PE_2 = 0 - 2.5 × 10^5 J = -2.5 × 10^5 J, which is the same as in part (b). **PE defined in general** There are other kinds of potential energy besides gravitational. Each form of potential energy is associated with a particular force, and can be defined analogously to gravitational potential energy. In general, the change in potential energy associated with a particular force is equal to the negative of the work done by that force if the object is moved from one point to a second point (as in Eq. 6-7b for gravity). Alternatively, because of Newton's third law, we can define the change in potential energy as the work required of an external force to move the object without acceleration between the two points, as in Eq. 6-7a. **TABLE 6-1 Conservative and Nonconservative Forces** | Conservative Forces | Nonconservative Forces | | ----------------------|--------------------------| | Gravitational | Friction | | Elastic | Air resistance | | Electric | Tension in cord | | | Motor or rocket propulsion | | | Push or pull by a person | **FIGURE 6-16** A crate is pushed across the floor from position 1 to position 2 via two paths, one straight and one curved. The friction force is always in the direction exactly opposed to the direction of motion. Hence, for a constant magnitude friction force, W = Fd, so if d is greater (as for the curved path), then W is greater. The work done does not depend only on points