Work is the change in energy of a system (neglecting heat transfer). To examine what could


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1 Work Work s the change n energy o a system (neglectng heat transer). To eamne what could cause work, let s look at the dmensons o energy: L ML E M L F L so T T dmensonally energy s equal to a orce tmes a length. Consder a constant orce F on a block. The block genercally could be dsplaced n three ways. (Ths dsplacement could be due to other d R orces or an ntal momentum.) What s the work d U done by F on the block? Work s a scalar and so the uncton must take two vectors, F and d, and produce a scalar. It turns out that the correct way to do ths s to look at parallel components o F and d. Work s dened as W F d Fy dy Fz dzor W F d. Ths s correct only or a constant orce. Lookng at the dagram above, what s the work done durng a dsplacement d R? It s smply the product Fd R. The orce s dong postve work on the block. What about or a dsplacement d L? In ths case the orce and dsplacement are antparallel and the work s FdL. The orce reduces the energy o the block,.e. t would slow the block down and reduce the knetc energy. How about or a dsplacement that s perpendcular to the orce? Here the work done s zero. The orce does not change the energy o the block as t moves along ths perpendcular path. There s an mportant specal case that has ths perpendcular relatonshp between the orce and dsplacement. Consder your crcular orbt and the arrows that you drew. The orce (or acceleraton) was radally nward whle the velocty (nstantaneous dsplacement d v t ) s tangent to the crcle and thereore perpendcular to the orce. Ths means the work done by ths orce s zero. Another case were the work s zero s when there s no dsplacement o the object. An nterestng case s that o jumpng o the loor. What s the work done by the loor on your body? Clearly the loor places a orce on your body, but yet the loor does not move so the work done by the loor s zero. Ths may seem strange and you should ask where you get your energy to jump up when the loor does no work. We wll return to ths nterestng queston n a couple o weeks. d L F The last case wth constant orces to consder s a dsplacement at an arbtrary angle to the orce. Usng the denton or the dot product W F d Fd cos. d F
2 Now the general case or nonconstant orces along an arbtrary trajectory needs to be eamned. The red arrows show the orce that a partcle eperences along a trajectory shown n black. I the trajectory s broken up nto many small segments as shown n the gure below, then these segments wll be appromately straght and the work done s W F r F cos r where s the angle between the orce and the small dsplacement at locaton. O course, we want to take the lmt o ths so that r dr and the sum becomes an ntegral, a lne ntegral. Takng ths lmt, the general epresson or the work done by a nonconstant orce along an arbtrary trajectory s y z. You wll learn how W F dr F d F dy F dz y z y z to evaluate ths ntegral n your class on calculus o several varables. Rght now you could evaluate ths usng numercal technques. Let s nd the work done by a varable orce rom a sprng, where F k( ) ˆ. The dsplacement s along the as. Here s the stuaton: F dr
3 The work done by the sprng on the block as t moves rom to s gven by: 1 1 ˆˆ W k( ) dr k k k( ) k ( ). You need to be careul when dong ths calculaton to make sure that F dr has the correct sgn. Qute oten t s convenent to place the orgn o the coordnate system at the pont o equlbrum or the sprng, 1.e.. Wth ths choce the work smples to W k ( ). Note that n ths case the orce that the sprng does on the block s negatve. It s removng knetc energy rom the block. Eventually, the block wll come to rest and begn to move back towards ; durng ths part o the cycle the sprng wll do postve work on the block. I a complete cycle s consdered the sprng wll perorm zero net work and the block wll return to the orgnal pont wth eactly the same knetc energy as t had the cycle beore. Let s consder a slghtly more complcated case where the moton s not one dmensonal. A mass s launched wth an ntal velocty v ( v, v ) n a unorm gravtatonal eld and we want to calculate the work done by gravty. Frst let there be no ar resstance. Ths means that we have to calculate W F dr along the trajectory. Consderng the orce o gravty 1 F (, mg). The parabolc trajectory s gven by r( t) ( vt, vyt gt ) thereore dr( t) ( v, v gt) dt so the epresson or the work done by gravty s y t t 1 W( t) (, mg) ( v, vy gt) dt mg ( vy gt) dt mg( vyt gt ) now note that the quantty n the parenthess s just the negatve o the y component o the trajectory,.e. the negatve o the heght about the ground. So the work done by gravty s W() t mgry. Snce the orce o gravty has no component, there s no dependence on how ar laterally the partcle has moved. Only the vertcal dsplacement matters or the work done by gravty. Or more generally, only the component o the dsplacement that s (ant)parallel to the orce contrbutes to the work. Now that we know how to calculate the work done by a orce you mght ask how does ths aect the knetc energy o a body? Startng wth W F dr we consder or an nntesmal tme dt that an nntesmal amount o work s done dw dp dv orce, we can use Newton s law F net m dt dt we nd that dv dr dw m dr m dv mv dv dt dt ntegratng ths yelds, y F dr Let the orce be the net or total
4 1 1 1 So the work done by the net orce W dw mv dv mv v mv mv KE on an object s equal to the change n the knetc energy o the object. Ths s an mportant result and named the WorkEnergy Theorem. Lastly, t s nterestng to know the tme rate o change o the work done by a orce. Usng dw dr dw F dr and dvdng by dt you can see that F F v. Ths quantty s call the dt dt dw power, P F v and tells the rate at whch energy o a system s beng changed by a dt orce. The unts s joules/sec whch s called a watt. Let s end ths dscusson by eamnng a problem usng all that we have learned. Consder the same problem above, projectle n a unorm gravtatonal eld, but now let s nclude ar drag. We can no longer compute the trajectory analytcally, but numercally t s no problem. The orces on the mass are gravty and ar drag. The total orce s gven by 1 F ˆ net Fgrav Fdrag mg ACv v and dr vdt. By takng the dot product o these epressons t s straghtorward to nd the nntesmal work done by gravty and ar drag at any tme. Then a numercal ntegraton (sum) can be perormed. Ths s how the man loop looks or the mplementaton n Python:
5 You should be able to understand each lne o ths man loop. In the last block o code, ponted to by the red arrow, the rst two lnes nd the delta work done by ar and gravty durng the last tme ncrement. The last two lnes sum up the delta work done to nd the total work done by ar and gravty. Here s a pcture o the object n lght. The red arrow s the orce and the cyan s proportonal to dr. The dot produce o these vectors gves the nntesmal work done durng that tme step. Notce that the orce vector s not vertcal. Ths s due to the ncluson o ar drag. The entre trajectory looks lke ths. Notce that the net orce s much reduced. The object s approachng termnal velocty. Now let s look at the power and work beng done by the orces. The power s equal to dw dr F so t s a relecton o the dt dt nstantaneous work beng done. Below s the graph o the power. The red curve s the rate o work that gravty s dong and the black curve s the power or the ar drag. There are a couple o thngs to notce. The work beng done by
6 gravty changes sgn at 1 sec. Whereas the work beng done by ar drag s always negatve. What happens at 1 sec to change the sgn o the gravtatonal power? Ths s when the partcle has reached mamum heght and the y component o the velocty changes rom postve to negatve. Gravty does negatve works as the partcle ncreases heght,.e. t s slowng down the partcle, but ater the ape gravty does postve work and tres to accelerate the partcle downward. Now at the same tme the drag orce s always dong negatve work and reducng the velocty. You mght correctly epect that at termnal velocty the negatve power due to drag s equal to the postve power rom gravty. The object s close to reachng termnal velocty near the end o the trajectory. Notce that the two powers are appromately equal, but wth derent sgns. I the power s ntegrated then you get the total work done by the orce. Below s a graph o ths work. Agan red s gravty and black s ar drag. The cyan lne s the knetc energy o the partcle and the orange lne shows the ntal total energy o the system. The work energy theorem says that work done by the sum o the orces s equal to the change n knetc energy. In other words, the sum o the red and black lnes should equal the cyan mnus orange. Ths s true or all tmes. There are a couple other thngs to notce. Frst the total work done by gravty at the end o the trajectory s zero. The partcle has gone up and back down by the same change n heght, and gravty beng a conservatve orce only depends on the startng and endng heght so the work done must be zero. Ths s qute derent rom the work done by ar drag. Here the work s always an ncreasng negatve value. The drag s always removng mechancal energy rom the system. Lastly, notce that the knetc energy has just about reached a steady value. Ths s agan a relecton o the partcle nearng termnal velocty so the knetc energy does not change.
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