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Physics WO Calc I

by: Nyasia Waelchi

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Physics WO Calc I PHY 2053

Nyasia Waelchi
FIU
GPA 4.0

Staff

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This 15 page Class Notes was uploaded by Nyasia Waelchi on Monday October 12, 2015. The Class Notes belongs to PHY 2053 at Florida International University taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/221805/phy-2053-florida-international-university in Physics 2 at Florida International University.

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Date Created: 10/12/15
CHAPTER 6 WORKAND ENERGY CONCEPTUAL QUESTIONS 2 REASONING AND SOLUTION The force P acts along the displacement therefore it does positive work Both the normal force FN and the weight mg are perpendicular to the displacement therefore they do zero work The kinetic frictional force fk acts opposite to the direction of the displacement therefore it does negative work 4 REASONING AND SOLUTION The sailboat moves at constant velocity and therefore has zero acceleration From Newton39s second law we know that the net external force on the sailboat must be zero a There is no work done on the sailboat by a zero net external force b Work is done by the individual forces that act on the boat namely the wind that propels the boat forward and the water that resists the motion of the boat Since the wind propels the boat forward it does positive work on the boat Since the force of the water is a resistive force it acts opposite to the displacement of the boat and therefore it does negative work Since the total work done on the boat is zero each force must do an equal amount of work with one quantity being positive and the other being negative Note The answer to part a could have been deduced from the workenergy theorem as well Since the velocity of the boat is constant the kinetic energy of the boat does not change and the total work done on the boat is zero 6 REASONING AND SOLUTION The kinetic energy of an object of mass m moving with speed v is given by KE mvz The kinetic energy depends on both the mass and the speed of the object The mass of an automobile is significantly greater than the mass of a motorcycle Therefore even if an automobile is moving slowly it is possible that the product mv2 is greater for the car than it is for the fastermoving motorcycle 7 REASONING AND SOLUTION A net external force acts on a particle This net force is not zero From Newton39s second law we can conclude that the net external force causes the particle to accelerate Since the particle experiences an acceleration its velocity must change The change in velocity however may occur as a change in magnitude only a change in direction only or a change in both magnitude and direction a This is sufficient information to conclude that the velocity of the particle changes however there is not sufficient information to determine exactly how the velocity changes b There is not suf cient information to determine if the kinetic energy of the particle changes In terms of the workenergy theorem the kinetic energy will change if the net external force does work on the particle But without knowing the direction of the net external force with respect to the particle39s displacement we cannot know if work is done c There is not sufficient information to determine if the speed of the particle changes Kinetic energy is mv2 so the speed v will change if the kinetic energy changes But as explained in part b there is insufficient information to determine whether the kinetic energy changes N REASONING AND SOLUTION If the total mechanical energy of an object is conserved then the sum of the kinetic energy and the potential energy must be constant a If the kinetic energy decreases the gravitational potential energy must increase by the same amount that the kinetic energy decreases b If the potential energy decreases the kinetic energy must increase by the same amount that the potential energy decreases c If the kinetic energy does not change the potential energy cannot change either PROBLEMS 3 work is REASONING AND SOLUTION According to Equation 61 W Fs cos 639 the a W 940 N350 m cos 2500 2980 J b W 940 N350 m cos 00 3290 J 9 REASONING AND SOLUTION According to Equation 61 the work done by the husband and wife are respectively Husband WH FH cos 6H 3 Wife WW FW cos 19Ws Since both the husband and the wife do the same amount of work FH cos 6H s FW cos 19W 3 Since the displacement has the same magnitude 3 in both cases the magnitude of the force exerted by the wife is cos19 cos 58 F F 7H 67N 45N W Hcos 19W cos 38 15 REASONING AND SOLUTION The work required to bring each car up to speed is from the workenergy theorem W KEf KEO meZ mvg Therefore WB 1mv v3 020 gtlt103 kg400 nis2 0 nis2 960 gtlt105 J WB v3 000 gtlt103 kg400 nis2 0 rns 160 gtlt106 J The additional work required to bring car B up to speed is therefore WB WA16gtlt106 J7960gtlt105 J 64x105 J 20 REASONING Since the person has an upward acceleration there must be a net force acting in the upward direction The net force ZFy is related to the acceleration ay by Newton s second law ZFy may where m is the mass ofthe person This relation will allow us to determine the tension in the cable The work done by the tension and the person s weight can be found directly from the definition of work Equation 61 SOL U T I ON a The freebody diagram at the right shows the two forces that act on the person Applying Newton s second law we have S 3 y T mg may H ZFy Solving for the magnitude of the tension in the cable yields 7 7 2 2 7 2 T7mayg779 kg070 n s 980 ms 7 83gtlt10 N b The work done by the tension in the cable is WT T cos 6s83 X 102 N cos 0011m 61 c The work done by the person s weight is WW mg cos 6s 79 kg98 msz cos 180 11 m 61 d The workenergy theorem relates the work done by the two forces to the change in the kinetic energy of the person The work done by the two forces is W WT WW w w i 2 i 2 T W 2mvf 2mvo 63 W Solving this equation for the nal speed of the person gives vf v02 WT WW 0ms2 2 79kg 26 REASONING The work done by the weight of the basketball is given by Equation 61 as W F cos 6 s where F mg is the magnitude ofthe weight 91x103 J 85gtlt103 J 19 is the angle between the weight and the displacement and s is the magnitude of S the displacement The drawing shows that the weight and displacement are parallel so that 19 0 The potential energy of the basketball is given by Equation 65 as m PE mgh where h is the height of the ball above the ground g SOL U T I ON a The work done by the weight of the basketball is W F cos 6s mg cos 0 h0 hf 060 kg980 ms26l m 15 m b The potential energy of the ball relative to the ground when it is released PEO mgle 060 kg980 ms26l m 65 c The potential energy of the ball relative to the ground when it is caught is PEf mgbf 060 kg980 ms2l5 m 65 d The change in the ball s gravitational potential energy is APEPEf PE088J736J 27J We see that the change in the gravitational potential energy is equal to 727 J where Wis the work done by the weight of the ball see part a 37 REASONINGANDSOLUHON Mechanical energy is conserved if no friction acts Ef E0 or 1 2 1 2 Emvf mgh Emvo Taking h0 0 m and rearranging V0 110 ms vf 1vg Zgh 110 ms2 2980 ms250 m 48 ms 4l REASONING Friction and air resistance are being ignored The normal force from the slide is perpendicular to the motion so it does no work Thus no net work is done by nonconservative forces and the principle of conservation of mechanical energy applies yields SOLUTION Applying the principle of conservation of mechanical energy to the swimmer at the top and the bottom of the slide we have Emvz i m hm Emvz rm f E0 If we let h be the height of the bottom of the slide above the water hf h and ho H Since the swimmer starts from rest v0 0 ms and the above expression becomes v gh gH Solving for H we obtain 2 H h V f 2g Before we can calculate H we must nd Vf and h Since the velocity in the horizontal direction is constant Ax 500 W m 100ms At 0500 s The vertical displacement of the swimmer after leaving the slide is from Equation 35b with down being negative y ayt2 i980 ms20500 s2 123 m Therefore h 123 m Using these values of Vf and h in the above expression for H we nd 2 2 100w H hV f 123 m S 633m g 2 2980 ms2 44 REASONING AND SOLUTION If air resistance is ignored the only nonconservative force that acts on the person is the normal force exerted on the person by the surface Since this force is always perpendicular to the direction of the displacement the work done by the normal force is zero We can conclude therefore that mechanical energy is conserved l l Emvmgho 3mvgmghf 1 where the nal state pertains to the position where l the person leaves the surface Since the person starts from r 1111111111111111111111111 u rest v0 0 ms Since the radius of the surface is r 9f r trees 9f h0 r and hf r cos efwhere Gfis the angle at which the person leaves the surface Equation 1 becomes 1 2 mgr 3mvf mgr cos 9f 2 In general as the person slides down the surface the ET two forces that act on him are the normal force FN and the weight mg The centripetal force required to keep the person moving in the circular path is the resultant of FN and the radial component of the weight mg cos 9 1 When the person leaves the surface the normal force is ngoos e 9 zero and the radial component of the weight provides the centripetal force 9 3 v gr cos 9f 3 39 mv mg cos 9f Substituting this expression for v into Equation 2 gives mgr mgr cos 6fmgr cos 6f Solving for 6f gives 6f cos71 47 REASONING AND SOLUTION a The work done by nonconservative forces is given by Equation 67b as mm AKE APE so APE mm 7 AKE Now AKE mv mv 550 kg600 ms2 180 ms2 901 J and APE 800 J7265 J7901J 1086 J b APE mg hiho so APE 71086 J h7h0 7201 m mg 550 kg980 ms 2 Thus the skater s vertical position has changed by and the skater is below the startin oint 53 REASONING AND SOLUTION According to the workenergy theorem as given in Equation 68 we have 1 2 1 2 Wno E mvf mghf E mvo mgh0 The metal piece starts at rest and is at rest just as it barely strikes the bell so that Vf v0 0 ms In addition hf h and ho 0 m while Wno 025Mv2 whereM and v are the mass and speed of the hammer Thus the workenergy theorem becomes 2 025Mv mgh Solving for the speed of the hammer we nd 2mgh 20400 kg980 ms2500 m 417M V 025M 025 900 kg 60 REASONINGAND SOLUHON a The power developed by the engine is P FV 200 X 102 N200 ms 400 X 103 w b The force required of the engine in order to maintain a constant speed up the slope is F Fa mgsin 3700 The power developed by the engine is then PFvFamg sin 3700v 611 P 200 X 102 N 250 X 102 kg980 ms2sin 3700200 ms 335 gtlt104 w 77 REASONING AND SOLUTION The conservation of energy applied between point A and the top of the trajectory gives KEA gig1A mgl where h 400 In Rearranging we nd KEA mgUJibA VA 2gh hA 2980 msz400 m 300 m 443 ms 01 CHAPTER 12 T EMPERA T URE AND HEAT CONCEPTUAL QUESTIONS REASONIN G AND SOL U T I ON A rod is hung from an aluminum frame such that there is a small gap between the rod and the oor The rod and the frame are heated uniformly both the frame and the rod will expand The linear expansion of the vertical portion of the frame and the linear expansion of the rod can be determined from Equation 122 AL aLoAT a If the rod is made of aluminum the fractional increase in length AL L0 is the same for the vertical portion of the frame and the rod Since the rod is initially shorter than the vertical portion of the frame the rod will never be as long as the vertical portion of the frame Therefore the rod will never touch the oor b If the rod is made of lead the heated rod could touch the oor The coefficient of linear expansion of lead is greater than that of aluminum Therefore the fractional change in the length of the rod will be greater than the fractional change in the length of the vertical portion of the frame If the temperature is raised high enough the length of the rod could expand enough to fill the gap and touch the oor REASONING AND SOLUTION According to Archimedes39 principle any uid applies a buoyant force to an object that is partially or completely immersed in it the magnitude of the buoyant force equals the weight of the uid that the object displaces Therefore the magnitude of the buoyant force exerted on an object immersed in water is given by FB pwater Vg where pwater is the density of water and Vis the volume displaced by the immersed object As shown in Figure 1220 the density of cold water above 4 0C is greater than the density of warm water above 4 OC Therefore cold water provides a greater buoyant force than warm water UI REASONING AND SOLUTION Two identical mugs contain hot coffee from the same pot One mug is full while the other is only onequarter full According to Equation 124 the amount of heat that must be removed from either cup to bring it to room temperature is Q cmAT where m is the mass of coffee in the cup 0 is the specific heat capacity of the coffee and AT is the temperature difference between the initial temperature of the coffee and room temperature Since the mug that is full contains a larger mass m of coffee more heat Q must be removed from the full mug than from the mug that is only onequarter full Therefore the mug that is full remains warmer longer 17 REASONING AND SOLUTION An alcohol rub has a cooling effect on a sick patient for the following reason In order for the alcohol to evaporate heat must be supplied to it The evaporating alcohol removes heat from the skin of the sick patient The evaporating alcohol therefore helps to lower the body temperature of the patient PROBLEMS 5 E REASONING AND SOLUTION The temperature of 727315 0C is 27315 Celsius degrees below the ice point ofO 0C This number of Celsius degrees corresponds to 27315 C 95 F o 49167 F Co Subtracting 49167 F0 from the ice point of 3200 0F on the Fahrenheit scale gives a Fahrenheit temperature of 745967 F 8 REASONING AND SOLUTION If the voltage is proportional to the temperature difference between the junctions then V V V 71 2 or ATZ iATI ATI AT V1 Thus 73 T2 00 0C Wu 100 0C 00 0C 475X10 V Solving for T2 yields T2 440 C 10 REASONING The change in length AL of the pipe is proportional to the coefficient of linear expansion afor steel the original length L0 of the pipe and the change in temperature AT The coefficient of linear expansion for steel can be found in Table 121 SOLUYYON The change in length of the pipe is AL aLOAT 12 x 10 5 C 165 m18 C 7 745 C W 122 17 REASONING AND SOLUTION AL XLOAT gives for the expansion of the aluminum ALA xALAAT 1 and for the expansion of the brass ALB xBLBAT 2 Adding 1 and 2 and solving for AT gives ALA ALB 13x10 3m 21C ozALA 0tBLB 23 gtlt1076 C 1 10 m19 gtlt1076 C 1 20 m AT The desired temperature is then T28 C21C 28 REASONING The change in volume of the cavity is proportional to the coef cient of volume expansion for aluminum the original volume V0 of the cavity and the change in temperature AT The original volume of the spherical cavity is V0 where r is its radius The coef cient of volume expansion for aluminum can be found in Table 121 SOL U T I ON a The change in the volume of the spherical cavity is AV V0AT 123 69 gtlt10 5 CO 1 50 cm3 282 C 23 0c b Because the temperature increases the cavity becomes larger 34 REASONING The change in volume of the cylindrical mercury column is proportional to the coefficient of volume expansion for mercury its original volume V and the change AT in temperature The volume of a cylinder of radius r and height AL is AV IFVZAL The coefficient of volume expansion for mercury can be found in Table 121 SOLUTION The change in the volume of the mercury is MAL VOAT Hm 123 AV Solving for AL gives Mzwz 182 gtlt104Co7145 mum00 7 quot l7gtlt1072 mm2 41 I REASONING Let the system be comprised only of the metal forging and the oil Then according to the principle of energy conservation the heat lost by the forging equals the heat gained by the oil or Qmetal Qo According to Equation 124 the heat lost by the forging is Qmetal cmetalmmeta1T0meta1 Teq where Teq is the final temperature of the system at thermal equilibrium Similarly the heat gained by the oil is giVen by Qoil coilmoilTeq T0011 SOL U T I 0N Qmetal Qoil cmetal mmetal TOmetal Teq coilmoilTeq T0011 Solving for Tom etal we have T coilmoi1Teq T0011 T Ometal c m eq metal metal 01 2700 JkgC 710 kg47 C 32 C T 47 C 940 C 0mm 430 Jkg C 75 kg 51 I REASONING Heat Q1 must be added to raise the temperature of the aluminum in its solid phase from 130 0C to its melting point at 660 OC According to Equation 124 Q1 cmAT Once the solid aluminum is at its melting point additional heat Q2 must be supplied to change its phase from solid to liquid The additional heat required to melt or liquefy the aluminum is Q2 Lf where Lfis the latent heat of fusion of aluminum Therefore the total amount of heat which must be added to the aluminum in its solid phase to liquefy it is Qtotal Q1 Q2 mCATLf SOL U T I ON Substituting values we obtain Qtotal045 kg 900gtlt102 JkgC 660 C7130 C40gtlt105 Jkg 39x105 J 52 REASONING AND SOLUTION The block of ice must undergo a change in temperature followed by a change in phase and another change in temperature Q Qice Qiceto water Qwater Q cmAT mLf cmAT ice water 411 X 106 J 200 X 103 JkgCquot100 kg100 CO 100 kg335 X 104 Jkg 4186 JkgCquot100 kgTf Solving for wae obtain Tf 13 C 63 REASONING AND SOLUTION If we assume that all the heat generated by kinetic friction goes into the block of ice then from energy conservation Q 39iction Q gained 1 y ice Also from energy conservation Q iction W iction From the workenergy theorem the work done by friction is given by 2 1 2 l W 39iction 3 mV E mV0 Since the ice is at 0 0C any heat that it absorbs will be used to change the phase ie melt ofthe ice Thus Equation 1 becomes A 2 i 2 2 mblockv 2 mblockVO l mmeltede 2 Solving Equation 2 for the mass of the ice that melts gives 1 1 Embmkwg v2 335 kg65 my 48 my 73 Inmelted 4 10 X10 kg Lf 335x10 Jkg

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