Contrary to Everyday Usage, the Term Work Has a Very Specific Meaning in Physics

Work Contrary to everyday usage, the term work has a very specific meaning in physics. In physics, work is related to the transfer of energy by forces. There are essentially two complementary ways to conceive of work. According to the work-kinetic energy theorem, the net work done on a object is equal to the object’s change in kinetic energy (energy associated with motion). Symbolically, Wnet = ∆T = Tf − Ti , where Tf represents the object’s final k inetic e nergy, a nd T i i ts i nitial k inetic e nergy. Complementarily, we find t hat t he t otal work d one o n a n o bject by n onconservative f orces i s e qual t o the object’s change in total energy (kinetic energy T plus potential energy U ). Wnonc. = ∆E = ∆T + ∆U = (Tf − Ti ) + (Uf − Ui ). If we analyze the units of the above equations, we see that work must have units of energy, because the right-hand sides of the equations have units of energy. Since the units of work are energy, we can see that the amount of work done by a force on an object is simply the amount of energy transferred by the force into the object. In the SI system of units the units of energy are Joules, and J = Nm, where Newtons (N) are a unit of force, and meters (m) are a unit of distance. Thus, it follows that the dimensions of work are a force times a distance. It turns out that a force only does work when it acts on an object over some distance. Think about a desk that is so heavy that it does not move no matter how hard you try to push it. Although you are exerting a force on the desk when you try to push it, the force you apply does not move the desk, so no energy is transfered to it. If we restrict ourselves to one-dimensional motion, the work done by a constant force acting on an object is given by W = F d, where F is the force and d is the displacement of the object (the amount it is moved by the force). This is consistent with our above definition o f w ork, a s t he u nits o f w ork a re s till N ewton meters (Nm), which are better known as Joules (J), the unit of energy. Gravity, electrostatic forces, and the restoring force exerted by a spring are all examples of conservative forces. When a conservative force does work on an object, the change in the object’s energy is independent of the path taken, or equivalently, the amount of work done by a conservative force as an object moves along a closed path is zero. Consider an object in a gravity field. I f the object is moved around but eventually returns to its initial position, the net work done by gravity will be zero, because gravity is a conservative force. Thus, if an object moves around in a gravity field w ith n o f orces o ther t han g ravity a cting o n i t, by t he work-kinetic e nergy t heorem i ts change in kinetic energy will be zero, because the net work will be zero. Similarly, the change in total energy and thus potential energy will be zero because there are no nonconservative forces acting. It is for this reason that the (gravitational) potential energy of an object in a gravity field only depends on the position of the object, not how it got to that position. Friction and applied mechanical forces are the most common examples of nonconservative forces. In contrast to conservative forces, work done by nonconservative forces does depend on the path taken. Imagine an object moving against the resistance of friction. Due to the force of friction energy from the object will be dissipated into heat. If the object is moved along some path so that it returns to its initial position, the amount of energy lost will depend on the path it took, because the more distance it traveled the more energy it will have lost to heat. 1 Let’s consider the example of an individual throwing a ball. To throw the ball he or she exerts a nonconservative force on the ball, which transfers kinetic energy to the ball, sending it into motion through the air. Since energy is transferred to the ball by the force of throwing the ball, the individual throwing the ball is doing work on the ball. Once the individual has thrown the ball, there is no way for him or her to recover the energy exhausted in doing so. Thus, in terms of energy, the action of throwing the ball is nonreversible. Nonconservative forces, such as the applied force in throwing the ball, can be thought of as nonreversible forces. In contrast, work done by conservative forces is in a sense recoverable. Think about lifting a box vertically to a height h. In order to lift the box we will need to exert a force opposite to the force of gravity, which is pushing the box downwards. How much force do we need to exert though? We need to exert at least as much force as gravity, in the opposite direction. In general, to calculate the amount of work required to move an object a given distance, we consider applying a force equal and opposite to the forces opposing the motion of the object. The idea is that we apply a force infinitesimally greater than the forces opposing the motion of object to set the object into motion, and then hold the applied force constantly equal to the forces opposing motion. In doing so the sum of the forces acting on the object will be zero, and consequently (by Newton’s second law) its acceleration will be zero, so it will remain in motion with a constant, albeit infinitesimal, velocity1 . Therefore, if we lift the box slowly, using a constant force about equal in magnitude to the weight of the box (the force due to gravity), the work done by lifting the box will be Wl = F d = mgh. At the same time gravity is acting with a force of −mg, so Wg = Fg d = −mgh, and so no net work is done. This means that there is no change the object’s kinetic energy. This is sensible, because after the object has been lifted to a height h it is at rest. If we consider only the nonconservative forces acting in this situation we can gain some additional information. The net work done by nonconservative forces is mgh, which corresponds to a change in total energy, but we already know there is no change in kinetic energy. Thus, this change in energy must be related to the potential energy of the box. After being lifted, the box has gravitational potential energy, related to its increased height. In this situation work is done by the person on the box, which increases the energy of the box, but the person doing work on the box expends energy in the exertion of lifting the box (thus the energy of the person is decreased). If the box is dropped, then it will begin to accelerate and have energy corresponding with motion. When it finally hits the ground energy will be transferred into heat and sound. In a more general situation, we can consider the amount of work done on an object by a force that is not constant. Think of stretching out a spring. The further the spring is stretched, the more force (and thus work) required to move it an additional distance. In order to find the total work required to stretch the spring a given distance, we can think of subdividing the distance the spring is stretched into very small subintervals. As these subintervals become small enough, or of length dx, the force required to stretch the spring the distance dx will be relatively constant, because the resistance of the spring will not change over such a small distance. To find the total work required to stretch the spring a distance b − a, we simply sum the work required to stretch the spring each 1 In a sense this is the minimum amount of energy that can be used to move the object a given distance. If we think about applying more force than this in order to move the object, the additional energy is transfered to something other than the object, for instance, heat. 2 infinitesimal distance. Thus, for a variable force, we find that the work done in moving an object from a to b is given by Z b W = F (x)dx. a Hooke’s law says when a spring is stretched or compressed a distance x from its natural (uncompressed) state, it exerts a force of Fs = −kx, where k is a spring constant, related to the specific design of the spring. This force is a a restoring force, which tends to push the spring to equilibrium. Thus, in order to stretch or compress a spring a distance x, one must exert a force of Fa = kx, so that if this force is maintained it will balance with the restoring force, and the spring will remain at a position x. Example 1. Find the work required to compress a spring from its natural length of 50 cm to a length of 40 cm if the spring constant is 20 N/m. Solution It is often easiest to convert units into meters, so that we can be certain our units will work out properly. In this case we want to compress the spring by an amount of 10 cm, which is the same as 0.1 m. Thus, we will be integrating from 0 (no compression) to -0.1 (a compression of 10 cm). We find that Z −0.1 W = Z −0.1 xdx = 10x2 kxdx = 20 0 0 −0.1 0 = 0.1 J. Example 2. Suppose a spring has a natural length of 1 m. A 24 Newton weight is placed on the end of the spring, which stretches it to a length of 1.8 m. i. Find the spring constant k. ii. How much work is required to stretch the spring 2m beyond its natural length? iii. How far will a 45 N weight stretch the spring? Solution Since we know F = kd, and we have a force of 24 Newtons, with d = 0.8, we find that k= 24 d = = 30 N/m. F 0.8 Next, we apply Hooke’s law and note that to stretch the spring 2 meters we will go from x = 0 to x = 2, so Z 2 2 xdx = 15x2 = 60 J. W = 30 0 0 Finally, we use the relationship F = kd for F = 45 N to find that d= F 45 = = 1.5 m, k 30 which means that this weight will stretch the spring to a length of 2.5 meters (because it is stretched 1.5 m beyond its natural length of 1m). 3 Example 3. A 3 kg bucket is lifted from the ground into the air by pulling in 6 meters of rope at a constant speed. The rope weighs 0.12 kg/m. How much work is required to lift the bucket and rope? Solution Since the weight of the bucket does not change, the work required to lift it is simply force times distance, which in this case is weight times distance. In order to find the weight of the object, we need to multiply its mass by the acceleration due to gravity, 9.8 m/s2 . 3 kg · 9.8 m/s2 · 6 m = 176.4 J. The weight of the rope remaining, however, varies with how far down the bucket is hanging. As more of the rope is pulled in, less of it remains hanging, and so the weight remaining decreases. As a function x of the height of the bucket, the weight of the rope is given by 0.12 kg/m · 9.8 m/s2 · (6 − x) m = 1.17(6 − x) N. Finally, we integrate from 0 to 6, finding that Z 6 h x2 Wrope = 1.17(6 − x)dx = 1.17 6x − 2 0 6 0   = 1.17 36 − 18 = 21.06 J. Thus, the total work done is 176.4 + 21.06 = 197.46 J. Using calculus we can also calculate the work required to pump liquid up out of a given container. The method we use is to think about dividing the liquid into a number of cylindrical slabs, treat the slabs as a solid object with a given weight, and then look in the limit as the height of the slabs approaches zero. Let’s suppose we have a liquid with a given weight per unit volume, ρ. The volume of a given slab will be given by A(y)dy, where A(y) is the area of a typical cross-section. Thus, the weight of a single slab will be given by ρA(y)dy, which is equal to the force required to lift the slab. Finally, we need to calculate how much distance we need to lift the slab. If the top of the container is at some height h, then we will need to lift the slab a distance (h − y), where y is the height of the slab itself. Summing up the work required to lift each individual slab leads to the integral Z W = ρA(y)(h − y)dy. Example 4. Suppose a cylindrical tank of water has a radius of 1 meter, and a height of 10 meters. Further, suppose that this tank is filled to the brim with water. Use the fact that water has a density of 1000 kg/m3 to determine how much work is required to pump all of the water up out of the tank. Solution The first thing to do is find the weight of the water per unit volume, which we do by multiplying by 9.8, the constant of acceleration due to gravity on earth at sea level. We then find 9.8 · 1000 = 980 N/m3 . 4 The volume of a typical slab is given by π22 dy = 4πdy m3 , and the work required to lift a slab is equal to its weight, so we find F (y) = 3920π N. Finally, we need to lift a slab from a height y to a height 10, so we need to lift each slab a distance (10 − y). Integrating over the limits 0 to 10 (the portions of the tank where water is present - all of it), we find the total work as Z W = 10 3920π(10 − y)dy = 3920π(10y − y 2 /2) 0 10 0 = 196000π ≈ 615752.16 N. Example 5. Consider the cone generated by rotating the line x = y/2 around the y-axis. Suppose that this forms a tank with height 4m, and that the tank is filled up to 3m with olive oil, weighing 8953.98544 Newtons per meter cubed. At the top of the tank there is a pipe which transfers the liquid in this tank to another tank. How much work is required to pump all of the liquid out of this tank into the other one? Solution We begin by finding the volume area of a cylindrical slab, given by A(y)dy = π(y/2)2 = π 2 y dy m3 . 4 The force required to lift this slab is given by its weight 8953.98544 · π 2 y dy = 2238.49636πy 2 dy N. 4 In order to transfer the slab of water to the other tank we must lift it to the rim of the tank, so that it will go through the transfer pipe. This requires our force F (y) to act through a distance of of the rim is 4 − y, so integrating from 0 to 3 (the portion of the tank containing liquid), we find that Z 3 4 y4 3 W = 2238.49636πy 2 (4 − y)dy = 2238.49636π( y 3 − ) = 110760.989 J. 3 4 0 0 5