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Chapter 07
Uploaded: A year ago
Contributor: rebmetpes
Category: Physics
Type: Solutions
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Description
Physics For Scientists And Engineers 6E
Transcript
7 Energy and Energy Transfer CHAPTER OUTLINE ANSWERS TO QUESTIONS 7.1 Systems and Environments 7.2 Work Done by a Constant Q7.1 The force is perpendicular to every increment of displacement. Force 7.3 The Scalar Product of Two Therefore, Fr⋅=∆ 0. Vectors 7.4 Work Done by a Varying Q7.2 (a) Positive work is done by the chicken on the dirt. Force 7.5 Kinetic Energy and the (b) No work is done, although it may seem like there is. Work-Kinetic Energy Theorem (c) Positive work is done on the bucket. 7.6 The Non-Isolated System—Conservation of Energy (d) Negative work is done on the bucket. 7.7 Situations Involving Kinetic Friction (e) Negative work is done on the person’s torso. 7.8 Power 7.9 Energy and the Automobile Q7.3 Yes. Force times distance over which the toe is in contact with the ball. No, he is no longer applying a force. Yes, both air friction and gravity do work. Q7.4 Force of tension on a ball rotating on the end of a string. Normal force and gravitational force on an object at rest or moving across a level floor. Q7.5 (a) Tension (b) Air resistance (c) Positive in increasing velocity on the downswing. Negative in decreasing velocity on the upswing. Q7.6 No. The vectors might be in the third and fourth quadrants, but if the angle between them is less than 90° their dot product is positive. Q7.7 The scalar product of two vectors is positive if the angle between them is between 0 and 90°. The scalar product is negative when 90 180°<< °θ . Q7.8 If the coils of the spring are initially in contact with one another, as the load increases from zero, the graph would be an upwardly curved arc. After the load increases sufficiently, the graph will be linear, described by Hooke’s Law. This linear region will be quite large compared to the first region. The graph will then be a downward curved arc as the coiled spring becomes a completely straight wire. As the load increases with a straight wire, the graph will become a straight line again, with a significantly smaller slope. Eventually, the wire would break. Q7.9 kk′=2. To stretch the smaller piece one meter, each coil would have to stretch twice as much as one coil in the original long spring, since there would be half as many coils. Assuming that the spring is ideal, twice the stretch requires twice the force. 191 192 Energy and Energy Transfer Q7.10 Kinetic energy is always positive. Mass and squared speed are both positive. A moving object can always do positive work in striking another object and causing it to move along the same direction of motion. Q7.11 Work is only done in accelerating the ball from rest. The work is done over the effective length of the pitcher’s arm—the distance his hand moves through windup and until release. Q7.12 Kinetic energy is proportional to mass. The first bullet has twice as much kinetic energy. Q7.13 The longer barrel will have the higher muzzle speed. Since the accelerating force acts over a longer distance, the change in kinetic energy will be larger. Q7.14 (a) Kinetic energy is proportional to squared speed. Doubling the speed makes an object's kinetic energy four times larger. (b) If the total work on an object is zero in some process, its speed must be the same at the final point as it was at the initial point. Q7.15 The larger engine is unnecessary. Consider a 30 minute commute. If you travel the same speed in each car, it will take the same amount of time, expending the same amount of energy. The extra power available from the larger engine isn’t used. Q7.16 If the instantaneous power output by some agent changes continuously, its average power in a process must be equal to its instantaneous power at least one instant. If its power output is constant, its instantaneous power is always equal to its average power. Q7.17 It decreases, as the force required to lift the car decreases. Q7.18 As you ride an express subway train, a backpack at your feet has no kinetic energy as measured by you since, according to you, the backpack is not moving. In the frame of reference of someone on the side of the tracks as the train rolls by, the backpack is moving and has mass, and thus has kinetic energy. Q7.19 The rock increases in speed. The farther it has fallen, the more force it might exert on the sand at the bottom; but it might instead make a deeper crater with an equal-size average force. The farther it falls, the more work it will do in stopping. Its kinetic energy is increasing due to the work that the gravitational force does on it. Q7.20 The normal force does no work because the angle between the normal force and the direction of motion is usually 90°. Static friction usually does no work because there is no distance through which the force is applied. Q7.21 An argument for: As a glider moves along an airtrack, the only force that the track applies on the glider is the normal force. Since the angle between the direction of motion and the normal force is 90°, the work done must be zero, even if the track is not level. Against: An airtrack has bumpers. When a glider bounces from the bumper at the end of the airtrack, it loses a bit of energy, as evidenced by a decreased speed. The airtrack does negative work. Q7.22 Gaspard de Coriolis first stated the work-kinetic energy theorem. Jean Victor Poncelet, an engineer who invaded Russia with Napoleon, is most responsible for demonstrating its wide practical applicability, in his 1829 book Industrial Mechanics. Their work came remarkably late compared to the elucidation of momentum conservation in collisions by Descartes and to Newton’s Mathematical Principles of the Philosophy of Nature, both in the 1600’s. Chapter 7 193 SOLUTIONS TO PROBLEMS Section 7.1 Systems and Environments Section 7.2 Work Done by a Constant Force P7.1 (a) WFr== °=∆ cos . . cos. .θ afaf160 220 250 319 N m J (b), (c)The normal force and the weight are both at 90° to the displacement in any time interval. Both do 0 work. (d) ∑W =++=319 00319.. J J P7.2 The component of force along the direction of motion is Fcos . cos. .θ=°=af350 250317 N N. The work done by this force is 3 WF r== =×afafafcos . . .θ∆ 317 500 15910 N m J. P7.3 Method One. Let φ represent the instantaneous angle the rope makes with the vertical as it is swinging up from φi =0 to φf =°60 . In an incremental bit of motion from angle φ to φφ+d , the definition of radian measure implies that ∆rd=af12 m φ. The angle θ between the incremental displacement and the force of gravity is θφ=°+90 . Then cos cos sinθφφ=°+=−bg90 . The work done by the gravitational force on Batman is FIG. P7.3 f φ=°60 WFdrmg d==−cos sinθφφ12 m zz bgaf i φ=0 60° 2 60° =− =− −mg da12 80 98 12 m kg ms mf sin . cosφφ φ z bgeja fbg 0 0 =− − °+=−× 3 afafaf784 12 601 47010 N m Jcos . Method Two. 2 The force of gravity on Batman is mg==bg80 98 784 kg ms Nej. down. Only his vertical displacement contributes to the work gravity does. His original y-coordinate below the tree limb is –12 m. His final y-coordinate is af−°=−12 60 6 m mcos . His change in elevation is −−−=6126 m m maf . The work done by gravity is WFr== °=−∆ cos cos .θ afaf784 6 180 470 N m kJ . 194 Energy and Energy Transfer −−52 P7.4 (a) Wmgh==× =×ej33510 980100 32810...afaf J J −2 (b) Since Rmg= , Wair resistance =−×32810. J Section 7.3 The Scalar Product of Two Vectors P7.5 A=500. ; B=900. ; θ=°500. AB⋅= = °=ABcos . .cos. .θ afaf500900 500 289 P7.6 AB i j k i j k⋅=++⋅++ejejAAABBB xyzxyz AB ii ij ik⋅= ⋅+ ⋅+ ⋅AB AB AB xx xy xzejejej +⋅+⋅+⋅AB AB AB yx yy yzejejejji jj jk +⋅+⋅+⋅AB AB ABzx zy zzejejejki kj kk AB⋅= + +ABABABxx yy zz P7.7 (a) WFxFy=⋅=+= ⋅+− ⋅=Fr∆ xy afafafaf600300 200100 160.. .. . Nm Nm J −−Fr⋅∆ 16 (b) θ= 11 cos cosF I G J = =°369. H Fr∆ K 2222 ejafaf600 200 300 100....+− +ejafaf P7.8 We must first find the angle between the two vectors. It is: θ=°−°−°−°=°360118900132200.. Then Fv⋅= = °Fvcos . . cos.θ af328 0173 200 N msbg Nm⋅ J or Fv⋅=533 533 533...== W s s FIG. P7.8 P7.9 (a) Aij=−300200. . −−11AB⋅ 120800..+ Bij=−400400. . θ=cos cos= =°113. AB afaf130320.. (b) Bijk=−+300400200. . . AB⋅ −−600160.. Aij=− +200400. . cosθ= = θ=°156 AB afaf200290.. (c) Ai j k=− + 200200. . −−11AB⋅ F −+600800.. I Bjk=+300400. . θ=cos cosF I G J = =°823. H AB K G J H 900250..⋅ K Chapter 7 195 P7.10 AB ijk i j k−= +−−−++ejej300 200500. . . AB ij k−=−−400 600. . CAB j k ij k⋅−= − ⋅ −− =+− ++ =a f ejej200300 400 600 0 200 180 160. . . . a ...f a f Section 7.4 Work Done by a Varying Force f P7.11 WFdx== area under curve from x to x z i f i (a) xi =0 xf =800. m 1 W = area of triangle ABC AC=F I G J × altitude, H2K 1 W F I 08→ =G J××=800 600 240... m N J H2K FIG. P7.11 (b) xi =800. m xf =100. m 1 W = area of ∆CDE CE=F I G J × altitude, H2K 1 W F I 810→ =G J××−=−afaf200 300 300... m N J H2K (c) WWW010 08 810→→→=+=+−=240 300 210...af J P7.12 Fxx =−af816 N (a) See figure to the right −a200 160.... m N m Nfa f a100 800fa f (b) Wnet = +=−120. J 2 2 FIG. P7.12 196 Energy and Energy Transfer P7.13 WFdx=z x and W equals the area under the Force-Displacement curve (a) For the region 0500≤≤x . m, afaf300 500.. N m W == 750. J 2 (b) For the region 500 100..≤≤x , FIG. P7.13 W ==afaf300 500 150.. . N m J (c) For the region 100 150..≤≤x , a300 500.. N mfa f W == 750. J 2 (d) For the region 0150≤≤x . W =++=af750750150 300.... J J f 5 m P7.14 Wdxydx=⋅= + ⋅Fr ij i43 N zzej i 0 5 m 2 5 m x bgbg404 Nm Nm Jxdx+= = 500. z 0 2 0 F Mg afaf400980.. N P7.15 k=== − =× 3 15710. Nm y y 2 25010. × m mg afaf150980.. (a) For 1.50 kg mass y== = 0938. cm k 15710. × 3 1 2 (b) Work = ky 2 1 32− 2 Work =×⋅×=ejej15710 40010 125... Nm m J 2 P7.16 (a) Spring constant is given by Fkx= F af230 N k== = 575 Nm x a0400. mf 1 (b) Work == =Fxavg afaf230 0400 460 N m J.. 2 Chapter 7 197 *P7.17 (a) F kxkxkxkxyapplied leaf helper=+=+−hhbg0 55 5N N 510 52510 36010 05×=×+× − N ...xx bg m m m 6810. × 5 N x= 5 = 0768. m 88510. × Nm 1 1 1 N 2 1 N 2 (b) Wkxkx=+=×22 5F I 5 hh G52510 0768.... Ja m f +×36010 0268a mf 2 2 2H mK 2 m =× 5 16810. J f P7.18 (a) Wd=⋅Fr z i 0600. m Wxxdx=+− °2 2 15000 10000 25000 0 N Nm Nm cos zej 0 23 0600. m 10000xx25000 Wx=+−15000 2 3 0 W =+−=900 180 180 900.... kJ kJ kJ kJ (b) Similarly, 2 2 3 bg100 100.. kNm maf ej250 100.. kNm maf W =+ −afaf150 100.. kN m 2 3 W = 117. kJ, larger by 29.6% 1 2 P7.19 400.. J m= ka0100 f 2 ∴=k 800 Nm and to stretch the spring to 0.200 m requires 1 2 ∆W =−=a8000200 400 120fa ...f J J 2 P7.20 (a) The radius to the object makes angle θ with the horizontal, so its weight makes angle θ with the negative side of the x-axis, when we take the x–axis in the direction of motion tangent to the cylinder. ∑Fmaxx= Fmg−=cosθ 0 Fmg= cosθ FIG. P7.20 f (b) Wd=⋅Fr z i We use radian measure to express the next bit of displacement as drRd= θ in terms of the next bit of angle moved through: π2 π2 WmgRdmgR==cos sinθθ θ z 0 0 WmgR mgR=−=a10f 198 Energy and Energy Transfer *P7.21 The same force makes both light springs stretch. (a) The hanging mass moves down by mg mg F 11 I xxx=+=+=+12 mg k k G J 12 12Hkk K F 1 m 1 m I − =+2 2 15.. kg 9.8 ms G N 1800 NJ =×20410 m H1200 K (b) We define the effective spring constant as −1 F mg F 11 I k== =+ x mgk k11+ G J bg12 Hkk12K −1 F 11 m m I =+G J = 720 Nm H1200 N 1800 NK *P7.22 See the solution to problem 7.21. F 11 I (a) xmg=+G J Hkk12K −1 F 11 I (b) k=+G J Hkk12K 2 F N kgms⋅ kg P7.23 k =L O== = 2 NMxQP m m s Section 7.5 Kinetic Energy and the Work-Kinetic Energy Theorem Section 7.6 The Non-Isolated System—Conservation of Energy 1 2 P7.24 (a) KA ==bgbg0600 200 120.. . kg ms J 2 1 2 2K afaf2750. (b) mv K B BB= : vB == = 500. ms 2 m 0600. 1 22 (c) ∑WKKKmvv==−= −=−=∆ BA BAej 750120 630... J J J 2 1 2 1 2 P7.25 (a) Kmv== =bgbg0300 150 338... kg ms J 2 2 1 221 (b) K====afafafafafaf0300300.... .030015044338 135 J 2 2 Chapter 7 199 P7.26 viji =−=ej600200. . ms 22 (a) vvviixiy=+= 400. ms 1 2 1 22 Kmvii== =bg300 400 600.. . kg ms J 2 2 ej (b) vijf =+800400. . 2 22 vfff=⋅=+=vv 640160800... ms 1 300. ∆KKKmvv=−= −= −=22 fi fi a800600 600... Jf 2 ej 2 P7.27 Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. Let d=5.00 m represent the distance over which the driver falls freely, and h=012. m the distance it moves the piling. 1 221 ∑WK=∆ : W W mv mvgravity beam+=−fi 2 2 so bgmghd Fdaf+°+ °=−cos cos018000.diaf 2 bgmghdaf+ bg2100 980 512 kg ms mej..af 5 Thus, F= ==×87810. . The force on the pile N d 0120. m driver is upward. 1 2 P7.28 (a) ∆KKKmv W=−=−==fi f 0 ∑ (area under curve from x=0 to x=500. m) 2 22750afafarea J. vf === 194. ms m 400. kg 1 2 (b) ∆KKKmv W=−=−==fi f 0 ∑ (area under curve from x=0 to x=100. m) 2 22225afafarea J. vf === 335. ms m 400. kg 1 2 (c) ∆KKKmv W=−=−==fi f 0 ∑ (area under curve from x=0 to x=150. m) 2 22300afafarea J. vf === 387. ms m 400. kg 1 2 P7.29 (a) KWKmviff+==∑ 2 1 −3 2 0+=× =∑W ej15010 780 456.. kg ms kJbg 2 3 W 45610. × J (b) F== = 634. kN ∆rcosθ a0.720 mfcos0° 22 2 vvfi− bg780 0 ms − 2 (c) a= = = 422 kms 2xf 20720af. m −33 2 (d) ∑Fma==× × =ejej1510 42210 634 kg ms kN. 200 Energy and Energy Transfer 87 P7.30 (a) vf =×=×0096310 28810..ej ms ms 1 23171 −−2 16 Kmvff==× × =×91110 28810 37810.. . kg ms J 2 2ejej (b) KWKif+= :0+=Fr K∆ cosθ f −16 Faf0028 037810.cos. m J°= × −14 F=×13510. N ∑F − 13510. × 14 N +16 2 (c) ∑Fma= ; a== =×14810. ms m −31 91110. × kg 716 2 (d) vvatxf xi x=+ 28810 014810..×=+× ms msejt −9 t=×19410. s 1 Check: xxvvtf i xi xf=+ +di 2 1 7 0028 0.. m ms=++×ej028810 t 2 −9 t=×19410. s Section 7.7 Situations Involving Kinetic Friction P7.31 ∑Fmayy= : n−=392 0 N n=392 N fnkk== =µ afaf0300392 118. N N (a) WFrF == °=∆ cos .cosθ afaf130500 0 650 J (b) ∆∆Efxint == =k a118500 588fa . f J FIG. P7.31 (c) Wnrn == °=∆ cos .cosθ afaf392500 90 0 (d) Wmgrg == −°=∆ cos .cosθ a392500 90 0fa f a f (e) ∆∆KKKW E=−= −fi ∑ other int 1 2 mvf −= − ++=065058800620 J J J. 2 2Kf 2620af. J (f) vf == = 176. ms m 400. kg Chapter 7 201 1 22 21 1 − 2 P7.32 (a) Wkxkxsif=−= ×−=af50050010 00625.. J 2 2 2 ej 1 2221 1 Wmvmvmvsfif=−=−0 2 2 2 2ch∑W 20625af. so vf == = ms ms0791. m 200. 1 221 (b) mv fxW mvik s f−+=∆ 2 2 1 2 0035020098000500 0625−+=a .... .fa fa fbg J J mvf 2 1 2 0282.. J kg= bg200 vf 2 FIG. P7.32 20282a . f vf == ms ms0531. 200. P7.33 (a) Wmgg =°+cos.af900 θ 2 Wg =°=−bg100 980 500 110 168.. .cos kg ms m Jdia f (b) fnmgkkk==µµθcos ∆Efmgint ==kkµθcos ∆Eint =°=afafafaf500 0400100980 200 184....cos. m J (c) WFF == = a100500 500fa . Jf FIG. P7.33 (d) ∆∆∆KWEWWE=−=+−=∑ other int intFg 148 J 1 221 (e) ∆Kmvmv=−fi 2 2 22148af af∆K 2 2 vfi=+=+=v af150 565.. ms m 100. P7.34 ∑Fmayy= : n+°−=af700 200147 0.sin. N N n=123 N fnkk== =µ 0300123 369..af N N (a) WFr== °=∆ cos . . cos.θ a700 500 200 329 N m Jfa f (b) WFr== °=∆ cos . cos.θ afaf123 500 900 0 N m J FIG. P7.34 (c) WFr== °=∆ cos . cos.θ a147 500 900 0 N mfa f (d) ∆∆EFxint == =afaf369 500 185.. N m J (e) ∆∆KKKWE=−=−=−=+fi ∑ int 329 185 144 J J J 202 Energy and Energy Transfer P7.35 vi =200. ms µk =0100. 1 KfxWKik f−+=∆ 2 other : mv fxik−=∆ 0 2 2 2 1 2 v bg200. ms mv mgxik=µ ∆ ∆x== =i 204. m 2 2µkg 20100980afaf.. Section 7.8 Power 2 W K 2 f mv 0875 0620.. kg msbg *P7.36 Pav ==== − = 801. W ∆∆∆t t 2 t 22110ej× 3 s W mgh afaf700 100 N m. P7.37 Power= P== = 875 W t t 800. s P7.38 A 1 300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine is equal to its final kinetic energy, 1 2 bgbg1300 246 390 kg ms kJ. = 2 390000 J 4 with power P= ~ around 30 horsepower.10 W 15.0 s P7.39 (a) ∑WK=∆ , but ∆K=0 because he moves at constant speed. The skier rises a vertical distance of af600 300300.sin.. m m°= . Thus, 2 4 WWin =−= = × =g bg700 98 300 20610 206.. .. . kg ms m J kJejaf . (b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus, W 20610. × 4 J Pinput == ==686 0919 W hp.. ∆t 30.0 s P7.40 (a) The distance moved upward in the first 3.00 s is 0175+ . ms ∆yvt==L O = M Paf300 263.. s m. N 2 Q The motor and the earth’s gravity do work on the elevator car: 1 221 mvW mgy mvif++ °=motor ∆ cos180 2 2 1 2 =−+=× 4 Wgmotor bgbgbg650 175 0650 263 17710 kg ms kg m J...af 2 W 17710. × 4 J 3 Also, Wt=P so P== =×=59110 792.. W hp. t 3.00 s (b) When moving upward at constant speed bgv=175. ms the applied force equals the ==×2 3 weight kg ms Nbg650 980 63710ej. . . Therefore, P==× =× =34 Fv ej63710 175 11110 149.... N ms W hpbg . Chapter 7 203 P7.41 energypowertime=× For the 28.0 W bulb: Energy used =×=⋅4 af280 10010 280.. W h kilowatthrsej total cost =+ =$17. $0. $39.00280 080 40af kWh kWhbg For the 100 W bulb: 43 Energy used =×=×⋅af100 10010 10010 W h kilowatthrsej.. 4 10010. × h # bulb used = =133. 750 hbulb 3 total cost =+× =133 420 10010 080 60.$0. . $0. $85.bg ej kWh kWhbg Savings with energy-efficient bulb =−=$85. $39. $46.60 40 20 F454 gIF9 kcalI 4186 J 7 *P7.42 (a) Burning 1 lb of fat releases energy 1 lb F I G J G J =×17110.. J H 1 lb KGH 1 g JKH1 kcalK ×=7 The mechanical energy output is 17110020ej..cos J a f nFr∆ θ. ×= °6 Then 34210 0.cos J nmgy∆ 6 34210 50 98 80 0150...×= 2 J kg ms steps mnbgejbgaf ×=×63 34210 58810.. J Jnej 6 34210. × J where the number of times she must climb the steps is n= 3 = 582 . 58810. × J This method is impractical compared to limiting food intake. (b) Her mechanical power output is 3 W 58810. × J 1 hp P== ==905 905.. . W WF I G J = 0121 hp. t 65 s H746 WK 8 1 3 1 13010 h mi F kcalIF . × JI *P7.43 (a) The fuel economy for walking is F I G JG JG J = 423 migal . 220 kcalH h KH4186 JKH 1 gal K 8 1 10 13010 h mi F1 kcalIF . × JI (b) For bicycling F I G J = 776 migal . 400 kcalH h KGH4186 JJG J KH 1 gal K 204 Energy and Energy Transfer Section 7.9 Energy and the Automobile P7.44 At a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 =183. kW to the wheels. If an additional load of 350 kg is added to the car, a larger output power of PP21=+ (power input to move 350 kg at speed v) will be required. The additional power output needed to move 350 kg at speed v is: ∆∆Pout ==bgbgfv mgvµr . Assuming a coefficient of rolling friction of µr =00160. , the power output now needed from the engine is 2 PP21=+ = +bgbg00160350 980 268 183 147..... kg ms ms kW kWejbg . With the assumption of constant efficiency of the engine, the input power must increase by the same factor as the output power. Thus, the fuel economy must decrease by this factor: FPI 183. bgbgbgfuel economy fuel economy kmL1 F I 2 =G PJ 1 =G J 640. H 2K H183147..+ K or bgfuel economy kmL2 = 592.. 1 2 1 2 1 2 2mv mv mvfi f− 2 2 −0 P7.45 (a) fuel needed= = useful energy per gallon eff.energy content of fuel×bg 1 2 2bgbg900 246 kg ms. − = 8 =× 2 13510. gal af015013410..ej× Jgal (b) 738. 1 gal 8 550 100 13410... mi F hIF × JI (c) power =F IF I G JG J af0150 808..= kW H380. miKH 1.00 h KG JG J H3600 sKH 1 gal K Additional Problems P7.46 At start, vij=°+°bgbg400 300 400 300.cos. ms ms .sin. At apex, viji=°+=bg bg400 300 0 346.cos. ms ms. 1 2 1 2 And Kmv== =bgbg0150 346 900... kg ms J 2 2 Chapter 7 205 1 step P7.47 Concentration of Energy output =⋅0600 600... Jkgstep kg F I bgbgG J =240 Jm H1.50 mK F=⋅=bgbg240 1 240.. Jm NmJ N P=Fv 700 240.. W N=afv v= 292. ms P7.48 (a) Ai⋅= aA fa1fcosα. But also, Ai⋅=Ax. A Thus, afafAA1cosα= x x or cosα= . A Ay Similarly, cosβ= A A and cosγ= z A 222 where AAAA=++xyz . 2 2 2 2 222 F I (b) cos cos cosαβγ++=FAxI Ay FA I A G J + + z G J == H A K G A J H A K 2 1 H K A 3 P7.49 (a) xt t=+200. Therefore, dx v==+ 2 1600. t dt 1 221 2 24 Kmv t tt==+=++af4001600 200240 720.. ... J 2 2 eje j dv 2 (b) a== af120. t ms dt Fma t t== =400120 480.. .afaf N 23 (c) P== + = +Fv t t t taf4801600 480288.. .ejej W 200..200 3 (d) Wdtttdt==+=P 480288 1250. J zzej 0 0 206 Energy and Energy Transfer *P7.50 (a) We write b Fax= b 1000 0129 N m=aaf. b 5000 0315 N m=aa . f b 0315. =F I = b 5 G J 244. H0129. K ln ln.5244=b ln5 bb===180. ln.244 1000 N 4 1.8 aa==×=1.80 40110. Nm a0.129 mf 025.. m 025 m 41.8N (b) WFdx xdx==×40110. zz 1.8 0 0 m 28. 025. m 28. N x N a025. mf =× =×4 4 40110. 1.8 40110. m 28. 1.8 0 m 28. = 294 J *P7.51 The work done by the applied force is fxmax 2 W F dx kxkxdx==−−+zz applied 1 2ej i 0 xxmax max 2 xxmax max3 2 x x =+=+kxdx kxdxk1 2 1 k zz 2 0 0 23 0 0 2 3 x x =+k max max 1 k2 23 P7.52 (a) The work done by the traveler is mghNs where N is the number of steps he climbs during the ride. N = (time on escalator)(n) h where aftime on escalator =vertical velocity of person and vertical velocity of person =+vnhs nh Then, N= vnh+ s mgnhhs and the work done by the person becomes Wperson = vnh+ s continued on next page Chapter 7 207 (b) The work done by the escalator is Wmgvte == =bgpowertime force exertedspeedtimea f a fbga f h where t= as above. vnh+ s mgvh Thus, We = . vnh+ s As a check, the total work done on the person’s body must add up to mgh, the work an elevator would do in lifting him. mgnhhs mgvh mghnhvbgs + It does add up as follows: ∑WW W=+=person e + = =mgh vnh+ ssvnh+ vnh+ s 1 2 P7.53 (a) ∆Kmv W=−=0 ∑ , so 2 2 2W 2W v = and v= m m W (b) WFdF=⋅=⇒=Fd xx d *P7.54 During its whole motion from y=100. m to y=−320. mm, the force of gravity and the force of the plate do work on the ball. It starts and ends at rest KWKif+=∑ 001800+°+°=Fy Fxgp∆∆cos cos mg Fbgbg100032 000320 0.. m m−=p 2 510 kg9.8 ms mejaf 5 Fp = −3 =×15310. N upward 3210. × m 2 F FF I P7.55 (a) P==+=+FvFvatFF I bgi G0 tJ = H m K G Jt H m K 2 Laf200. N O (b) P=M Pa300 240. s Wf= NM 500. kg QP 208 Energy and Energy Transfer f xxia1+ 1 2 2 1 2 *P7.56 (a) W Fdx kxdx kx x x kx xx11 1 11== =+−=+zz bgiai aai1 1ej2 1 i xi1 2 2 −+xxia2 1 2 2 1 2 (b) Wkxdxkxxxkxxx2222==−+−=−bgiai aai2 2ej2 2 −xz i2 2 2 (c) Before the horizontal force is applied, the springs exert equal forces: kxkx1122ii= kx x 11i i2 = k2 1 2 1 2 (d) WW kxkxx kxkxx121+=+ +−aai aai112 22 2 2 1 2 1 2 kx =++−kx kxkxxkx 11i 1 aaaia2 112 2 2 k2 1 =+ 2 bgkkx12 a 2 tt 23 *P7.57 (a) vadt tt tdt==−+116021 024... zzej 0 0 234 t ttt =−+=−+234 116......021 024 058 007 006ttt 2 3 4 0 At t=0, vi =0. At t=25. s, 345234 vf =−+=ej058 25 007 25 006 25 488....... ms s ms s ms s msaf ejaf ejaf KWKif+= 1 2 1 2 4 0+= = =×Wmvf 1160 488 13810 kg ms Jbg.. 2 2 (b) At t=25. s, 23 a=− + =34 52 ejej116 25 25 0240 25 534.. .... ms s0.210 ms s ms s msa f eja f . Through the axles the wheels exert on the chassis force 2 3 ∑Fma== =×1160 534 61910 kg ms N.. and inject power 34 P==× =×Fv 61910488 30210... N ms Wbg . Chapter 7 209 22 P7.58 (a) The new length of each spring is xL+ , so its extension is 22 xLL+− 22 and the force it exerts is kxLLF +− I toward its H K fixed end. The y components of the two spring forces add to zero. Their x components add to F I Fi i=− +−221kxLLF 22 I x L =− −kx FIG. P7.58 H K 22 22G J . xL+ H xL+ K f 0 F L I (b) WFdx=z x Wkx=− −21G 22 Jdx i Az H xL+ K 12 0 0 0 0 22 −12 2 x ejxL+ WkxdxkLxLxdx=− + +22 22ej Wk=− +2 kL AAzz 2 A bg12 A 22 22 22 22 WkAkLkLAL=−+ + − +022 WkLkAkLAL=+−+22 *P7.59 For the rocket falling at terminal speed we have ∑Fma= +−=RMg 0 1 2 Mg DAv= ρ T 2 (a) For the rocket with engine exerting thrust T and flying up at the same speed, ∑Fma= +−−=TMgR 0 TMg=2 The engine power is P===FvTv MgvTT2. (b) For the rocket with engine exerting thrust Tb and flying down steadily at 3vT, 1 2 RDAvMgbT==ρ bg39 2 ∑Fma= −−+=TMgMgb 90 TMgb =8 The engine power is P== =Tv Mgv Mgv8324TT . 210 Energy and Energy Transfer P7.60 (a) Fijij 1 =°+°=+af250 350 350 205143.cos. N Nejejsin. . . Fijij 2 =°+°=−+af420 150 150 364210.cos N Nejejsin . . (b) ∑FFF i j=+=−+12 ej159353. . N ∑F 2 (c) a==−+ej318707. ij. ms m 2 (d) vva ij ijfi=+=+ +−+t ejejej400250 318707 300. . ms ms s. . af. vij f =−+ej554237. . ms 1 2 (e) rrvafii=++tt2 1 2 2 rij ijf =+ + +−+0400250 300ej. . bgms s ms saf..ejej318707 300 . af. 2 ∆rr i j==−+ f ej230393. . m 1 2 1 22 2 (f) Kmvff== + =bg500 554 237 148... . kg ms kJafaf 2 2 ej 1 2 (g) Kmvfi=+⋅∑Fr∆ 2 1 22 2 Kf =++−−+bg500 400 250 159 230 353 393... .... kg ms N m N ma f a f bg a fa f a fa f 2 Kf =+=556 1426 148.. J J kJ P7.61 (a) ∑WK=∆ : WWsg+=0 1 2 kx mgxi −+ °+°=090600∆ cosa f 2 1 3 2 ej14010 0100 0200980 600 0....sin.××− °= Nm afafafaf∆x 2 ∆x= 412. m (b) ∑WKE=+∆∆ int: WWEsg+−=∆ int 0 1 2 kxmgx mg xik+°−°=∆∆cos cos150 60 0µ 2 1 3 2 14010 0100 0200980 600 02009800400 600 0....sin....cos.××− °− °= Nm afafafafafafafaf∆∆xx 2ej ∆x= 335. m Chapter 7 211 P7.62 (a) FLFLaNmmNmmf a f a f a f 2.00 15.0 14.0 112 4.00 32.0 16.0 126 6.00 49.0 18.0 149 8.00 64.0 20.0 175 10.0 79.0 22.0 190 12.0 98.0 FIG. P7.62 (b) A straight line fits the first eight points, together with the origin. By least-square fitting, its slope is 0125 2%125 2%. Nmm Nm±= ± F In Fkx= , the spring constant is k= , the same as the slope of the F-versus-x graph. x (c) Fkx== =bg125 0105 131 Nm m Na ..f P7.63 KWWKisgf++= 1 222 21 1 1 mv kx kxmgx mviif f+−+ =∆ cosθ 2 2 2 2 1 221 0+−+ °=kx mgx mvii f0100cos FIG. P7.63 2 2 1 2 1 2 bg120 500 00500 0100 980 00500 100......sin.. Ncm cm m kg ms m kgafbgbg−°=bgbg0100 v 2 ej 2 −32 015085110 00500.. . J J kg−×=bgv 0141. v== 168. ms 00500. 1 1 22 2 P7.64 (a) ∆∆ 22 EKmvvint =−=− −ejfi : ∆Eint =− − =bg0400 600 800 560... . kg ms Ja f a f bg 2 2 ej (b) ∆∆Efrmgrint ==µπ 2 k af2 : 560 0400 980 2150.... J kg ms m=µπkbgej a f Thus, µk = 0152.. (c) After N revolutions, the object comes to rest and Kf =0. 1 2 Thus, ∆∆EKKmvint =−=−+=0 ii2 1 2 or µπkimgNr mva2 f = . 2 1 2 2 mv 1 i 2bg800. ms This gives N== =2 2 228. rev . µπkmgraf2 af0152980 2150.. .ej ms mπaf 212 Energy and Energy Transfer P7.65 If positive F represents an outward force, (same as direction as r), then f rf 1313−−77 WdFrFrdr=⋅= −Fr 2 σσ zzej0 0 i ri 1312−−76 rf 2FrFrσσ W = 0 − 0 −12 6− ri 13 12 12−− −− −− 766 FrrFrr0σσejejfi fi0 − 7 13 Fσσ−− −−66 F 12 12 W = + =−−−0 rr 0 fi firr 6666 −−− −−−77 6 6 134 12 12 Wrrrr=× −−× −10310 18910..fi fi −−77 6 −−−−6 60 134 12 8 120 W =× ×−10310 18810 2.. ....441010 18910 35410 5961010×−××−× −− − W =− × + × =− ×21 21 21 24910 11210 13710... J J J a∆ 2 mvf P7.66 P∆∆tWK=== 2 ∆∆mm The density is ρ== . vol Ax∆ ∆x Substituting this into the first equation and solving for P, since =v, ∆t FIG. P7.66 ρ 3 Av for a constant speed, we get P= . 2 ρ 2 Av Also, since P=Fv, F= . 2 Our model predicts the same proportionalities as the empirical equation, and gives D=1 for the drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airflow around the object is complicated. 237. 375dx P7.67 We evaluate z 3 by calculating 128. xx+375. 3750100af. 3750100af. 3750100af. 33 3+ +… =0806. afaf128 375128...+ afaf129 375129...+ afaf236 375236...+ and 3750100af. 3750100af. 3750100af. 33 3+ +… =0791. . afaf129 375129...+ afaf130 375130...+ afaf237 375237...+ The answer must be between these two values. We may find it more precisely by using a value for ∆x smaller than 0.100. Thus, we find the integral to be 0799. Nm⋅ . Chapter 7 213 1 23 *P7.68 P= Drvρπ 2 1 3 2 3 3 (a) Pa ==×1120 15 8 21710ej.. . kgm m ms Wπaf bg 2 3 3 Pb vb F24 msI 3 (b) == ==327 P 3 G J a va H 8 ms K 34 Pb =×=×2721710 58610ej.. W W P7.69 (a) The suggested equation P∆tbwd= implies all of the v = constant following cases: n d w ∆t (1) P∆tb= F I w G Jaf2d (2) PF I F I G J =bG Jd H 2K H 22K H K fk =µ kn F ∆t d (3) PF I P w G J =bwF I F I F I G J and(4) G J∆tb= G Jd H 22K H K H 22K H K w These are all of the proportionalities Aristotle lists. FIG. P7.69 (b) For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal floor with which the object has coefficient of friction µk. ∑Fa=m implies that: +−=nw 0 and Fn−=µk 0 so that Fw=µk As the object moves a distance d, the agent exerting the force does work W WFd Fd wd==°=cos cosθµ0 and puts out power k P= ∆t This yields the equation P∆twd=µk which represents Aristotle’s theory with b=µk. Our theory is more general than Aristotle’s. Ours can also describe accelerated motion. *P7.70 (a) So long as the spring force is greater than the friction force, the block will be gaining speed. The block slows down when the friction force becomes the greater. It has maximum speed when kx f maak−==0. 0 3 −3 ej1010 40 0..×−= Nm Nxa x=−×4010. m 0 (b) By the same logic, FIG. P7.70 3 −2 ej1010 100..×− Nm N=0xb x=−×1010. m 214 Energy and Energy Transfer ANSWERS TO EVEN PROBLEMS 3 P7.2 15910. × J P7.44 592. kmL −2 −2 P7.4 (a) 32810. × J; (b) −×32810. J P7.46 90.0 J P7.6 see the solution A Ay A P7.48 (a) cosα= x; cosβ= ; cosγ= z ; A A A P7.8 5.33 W (b) see the solution P7.10 16.0 401. kN P7.50 (a) a= 1.8 ; b=180.; (b) 294 J m P7.12 (a) see the solution; (b) −120. J mgnhhs mgvh P7.52 (a) ; (b) P7.14 50.0 J vnh+ s vnh+ s P7.16 (a) 575 Nm; (b) 46.0 J 5 P7.54 15310. × N upward P7.18 (a) 9.00 kJ; (b) 11.7 kJ, larger by 29.6% P7.56 see the solution P7.20 (a) see the solution; (b) mgR P7.58 (a) see the solution; 22 22 −1 (b) 22kLkA kLAL+− + mg mg F 11 I P7.22 (a) + ; (b) + k12k GHkk J 12K P7.60 (a) Fij1 =+ej205143. . N; P7.24 (a) 1.20 J; (b) 500. ms; (c) 6.30 J Fij 2 =− +ej364210. . N; (b) ej−+159353. ij. N; P7.26 (a) 60.0 J; (b) 60.0 J (c) ej−+ 2 318707. ij. ms ; P7.28 (a) 194. ms; (b) 335. ms; (c) 387. ms (d) ej−+554237. ij . ms; −16 −14 P7.30 (a) 37810. × J; (b) 13510. × N; +16 2 (e) ej−+230393. ij. m; (f) 1.48 kJ; (g) 1.48 kJ (c) 14810. × ms ; (d) 1.94 ns P7.32 (a) 0791. ms; (b) 0531. ms P7.62 (a) see the solution; (b) 125 2% Nm± ; (c) 13.1 N P7.34 (a) 329 J; (b) 0; (c) 0; (d) 185 J; (e) 144 J P7.64 (a) 5.60 J; (b) 0.152; (c) 2.28 rev P7.36 8.01 W P7.66 see the solution 4 P7.38 ~10 W P7.68 (a) 2.17 kW; (b) 58.6 kW P7.40 (a) 5.91 kW; (b) 11.1 kW P7.70 (a) x=−40. mm; (b) −10. cm P7.42 No. (a) 582; (b) 905 0121.. W hp=
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