Work Energy Power

 2013 Yeow Kok Han Page 1 WWoorrkk EEnneerrggyy PPoowweerr 11 IInnttrroodduuccttiioonn In physics, ‘work’ has a meaning different from normal everyday usage. In physics, w ork is a method of energy transfer. Doing positive work on a system means transferring energy into the system while doing negative work means transferring energy out of the system. Energy comes in many different forms - kinetic, potential, thermal, electromagnetic radiation - even mass is a form of energy as put forth by Einstein’s famous E = mc2 equation. Some of these energies are always in a state of flux or flow. Thus a system’s total energy can change because of work done or the energy flows. That naturally leads to the concept of power which is the rate of flow or transfer or transformation of energy. 22 WWoorrkk DDeeffiinniittiioonn ooff wwoorrkk Consider a box of mass m on a horizontal frictionless surface. In itially, the box wa s stationary. A constant force F applied on the box caused it to accelerate and acquire a final velocity v and KE = ½ mv2. Of the three forces acti ng on the box, the normal force N and the weight W do not contribute to the acceleration and gain in KE. Let’s work out the connection between F and the gain in KE:  F is a net force F = ma  F constant  a also constant. Straight line motion with constant a implies v2 = u2 + 2as  v2 = 2as since u = 0  v2 = 2(F/m)d   ½ mv2 = Fd The product Fd is the work done by F. Formally, work done WD is defined as follows BBaassiicc ccaasseess ooff wwoorrkk Case 1 - . F & d in same direction, WD = Fd. This case has been considered above. F & d are magnitudes. Case 2 - . F & d in opposite direction, WD =  Fd. In Fig. 2.2, F is to the left on a box with initial velocity v to the right, causing deceleration. The box loses KE and the work done by F is negative. Work done(scalar) by a force on a system is defined as the product of the force and the displacement in the direction of the force. Basic cases: 1 F & d in same direction, WD = Fd 2 F & d in opposite direction, WD =  Fd. 3 F & d are perpendicular, WD = 0 d N W F v Fig. 2.2 Work done by a force is defined as the product of the force and the displacement in the direction of the force. N W F d v Fig. 2.1

 2013 Yeow Kok Han Page 2 Case 3 - . F & d are perpendicular, WD = 0. Referring to Fig. 2.2, if F = N vertically upwards or F = W vertically downwards, it does not contribute to the change in KE of the box and WD = 0. GGeenneerraall ccaasseess ooff wwoorrkk In general, F can be at any angle to d . However, the force can always be resolved into a component parallel to d and another component perpendicular to d as shown in Fig. 2.3 and 2.4. The component perpendicular to d does no work (see basic case 3). Work is only done by the component parallel to d . In Fig.2.4, since -(F cos)d = Fd cos, the general formula for work done: where  is the angle between F and d . Fd cos can be thought of as the product of the force component magnitude ( F cos) and the displacement magnitude d. It can also be thought of as the product of dis placement component magnitude (d cos) and the force magnitude F. It is important to note that when the vector arrows are joined such that they both point away or towards the joining point as shown in Fig. 2.5, the angle  is the angle between them. In Fig. 2.6, w hen one vector points towards but the other points away from the joining point, the angle  is not the angle between the vectors . In this case, the angle between the vectors is given by 180 - . General formula for work = Fd cos where  is the angle between F and d . Fig. 2.6 d F  d F  d F  d  d  Fig. 2.5 d F F F  cosWD F d Fd    d F Fig. 2.3  F sin F cos . Work = (F cos)d.. See basic case 1 d Fig. 2.4 F  F sin F cos . Work = -(F cos)d.. Negative because horizontal force component is opposite to d (see basic case 2). 

 2013 Yeow Kok Han Page 3 WWoorrkk DDoonnee bbyy NNoonn--ccoonnssttaanntt FFoorrccee The formula WD = Fd cos is only applicable for constant magnitude F. Consider an object that is initially at rest being acted upon by a single force only. The force is initially + F but its magnitude decreases to zero before increasing again in the opposite direction. The force is then removed when it reaches -F. In this case, the work done by the non -constant force is given by the area under the force-displacement graph. For the first half of the journey, the work done is positive while negative for the second half. Fur thermore, the total work done is zero. ++ aanndd -- WWoorrkk DDoonnee In Fig. 2.7, t he object accelerates to the right until D/2. Thereafter it decelerates to a stop. In other words, the kinetic energy of the object increases until D/2, then decreases to zero at the end. This can be accounted for by looking at the work done on the object . For the first half of the journey, positive work done causes the kinetic energy to increase while in the second half an equal amount of negative work done causes the kinetic energy to drop to zero. In general, work done tends to cause the total energy of a system to change. Since the total energy may consist of various forms of energy such as kinetic energy, potential energy, internal energy(therma l) etc, one or more forms of energy possessed by the system may change as a result of work done. In understanding the relationship between work done and energy, it is critical to have a clear idea of the system in consideration. Consider two spheres connected by a light compre ssed spring and isolated from its surrounding. When released: WD = Fd cos is only applicable for constant magnitude F. Work done for a non-constant force can be found from area under F-d graph. + and - work done by a force on a system corresponds to the transfer of energy to the system and out of the system respectively. Conclusion about relationship between work and energy depends on the chosen system. For a rigid system, total work done on it = change in its KE Positive and negative work done by a force on a system corresponds to the transfer of energy to the system and out of the system respectively. displacement force 0 D F -F D +F -F Taking to the right as positive. Fig. 2.7 system F F The net force on system = 0. No work done on the system, so total energy of the system is constant. Internally, there is conversion of elastic PE to KE of the spheres. system - sphere only F F The net force on system = F. F does work on the system, so KE of system increases. Sphere by itself does not have elastic PE. Fig. 2.8

 2013 Yeow Kok Han Page 4 WWoorrkk DDoonnee OOnn//BByy GGaass In Fig. 2.9, if the gas expands, the force exerted by the gas on the piston and external atmosphere is in the direction of the displacement. Thus positive work is done on the piston and atmosphere or energy is transferred from the gas to them. If on the other hand, the atm osphere outside pushes the piston inwards, the force by the gas does negative work on the external environment which means that energy is removed from the environment. However, the environmen t does positive work on the gas and we can say that there is energy transfer from the environment to the gas. When the expansion or compression happens while the pressure P of the gas is constant, then the absolute value of the work done on or