WorkContrary 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 two different ways to conceiveof work, and the usefulness of each approach depends on the situation. According to thework-kinetic energy theorem, the net work done on a object is e qual to the object’s changein kinetic energy (energy associated with motion).Wnet= ∆KE = KEf− KEiSimilarly, we can consider the work done only by nonconservative forces in a system, and wefind that the total work done on an object by nonconservative forces is equal to the object’schange in total energy (kinetic plus potential energy).Let us apply these ideas in a few situations in order to get a better understanding of them.For example, we can think of throwing a ball. When an individual throws a ball he or sheexerts a (nonconservative) force on the ball, which sends the ball into motion through theair. Energy is transfered in terms of kinetic energy, related to the motion of the ball throughair. Since energy is transfered into the ball by the force of throwing the ball, the individualthrowing the ball is doing work on the ball.If we restrict ourselves to one-dimensional motion, we find the work done by a constant forceacting on an object to be given byW = F dwhere F is the force and d is the displacement of the object (the amount it is moved by theforce). It follows that the units of work are Newton meters (Nm), which are better knownas Joules (J), the unit of energy. Since the units of work are energy, we can se e that theamount of work done by a force on an object is simply the amount of energy transfered bythe force into the object.It is worth noting that simply because a force acts on a given object, it does not necessarilydo work on the given object. Think about a situation where a desk is so heavy, that nomatter how hard you try to push it, it does not move. In this situation you are clearlyexerting a force on the desk, yet you are doing no work on the desk because the desk haszero displacement. Thus, in pushing on the desk this way no energy is transfered into thedesk.For a third example, we can think about lifting a box vertically, to say a height h. In orderto lift the box we will need to exert a force opposite to that of gravity which is pushing thebox downward. If we lift the box slowly, using a constant force about equal in magnitude tothe weight of the box (the force due to gravity), the work done by lifting the box will beWl= F d = mghHowever, at the same time gravity is acting with a force of −mg, soWg= Fgd = −mghand so no work is done overall. This means that there is no change the object’s kineticenergy, which makes sense, because after the object has been lifted a height h it is at rest.However, if we consider only the nonconservative forces acting in this situation, we can gainsome additional information. Gravity is a conservative force, so we neglect the work doneby it. Then, we see that the net work done by nonconservative forces is mgh, which corre-sponds to a change in total energy, but not kinetic energy. Thus, this change in energy mustbe related to the potential energy of the box. After being lifted, the box has gravitationalpotential energy, related to its increased height. In this situation work is done by the personon the box, which increases the energy of the box, but in doing work the energy of theperson is decreased. If the box is dropped, then it will begin to accelerate and have energycorresponding with motion. When it finally hits the ground energy will be transfered intoheat and sound.In a more general situation, we can consider the amount of work done on an obje ct by a forcethat 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 findthe total work required to stretch the spring a given distance, we can think of subdividing thedistance the spring is stretched into very small subintervals. As these subintervals becomesmall enough, or of length dx, the force required to stretch the spring the distance dx willbe relatively constant, because the resistance of the spring will not change over such a smalldistance. To find the total work required to stretch the spring a dis tance b − a, we simplysum the work required to stretch the spring each infinitessimal distance. Thus, for a variableforce, we find that the work done in moving an object from a to b is given byW =ZbaF (x)dxHooke’s law says when a spring is stretched or compressed a distance x from its natural(uncompressed) state, it exerts a force ofFs= −kxwhere k is a spring constant, related to the specific design of the spring. This force is a arestoring force, which tends to push the spring to equilibrium. Thus, in order to stretch orcompress a spring a distance x, one must exert a force ofFa= kxso that if this force is maintained it will balance with the restoring force, and the spring willremain at a position x.Example 1 Find the work required to compress a spring from its natural length of 50 cmto 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 unitswill 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 (acompression of 10 cm). We find thatW =Z0.10kxdx = 20Z0.10xdx = 10x20.10= 0.1 JExample 2 Suppose a spring has a natural length of 1 m. A 24 Newton weight is placedon 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 findthatk =dF=240.8= 30 N/mNext, we apply Hooke’s law and note that to stretch the spring 2 meters we will go fromx = 0 to x = 2, soW = 30Z20xdx = 15x220= 60 JFinally, we use the relationship F = kd for F = 45 N to find thatd =Fk=4530= 1.5 mwhich means that this weight will stretch the spring to a length of 2.5 meters.Example 3A 3 kg bucket is lifted from the ground into the air by pulling in 6 meters ofrope at a constant speed. The rope weighs 0.12 kg/m. How much work is required to liftthe bucket and rope?Solution