P161Sp08Chapter_11

P161Sp08Chapter_11 - Chapter 11 Work Energy Approach to...

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Unformatted text preview: Chapter 11 Work Energy Approach to Problems n n The energy approach to describing motion is particularly useful when the force is not constant An approach will involve Conservation of Energy n This could be extended to biological organisms, technological systems and engineering situations Systems n A system is a small portion of the Universe n We will ignore the details of the rest of the Universe n A critical skill is to identify the system Valid System n A valid system may n n n n be a single object or particle be a collection of objects or particles be a region of space vary in size and shape Problem Solving n Does the problem require the system approach? n What is the particular system and what is its nature? n Can the problem be solved by the particle approach? n Yes, the particle approach is what we have been using to this time Environment n There is a system boundary around the system n n The boundary is an imaginary surface It does not necessarily correspond to a physical boundary n The boundary divides the system from the environment n The environment is the rest of the Universe Work (Physics not Job-Related) n The physical work, W, done on a system by an agent exerting a constant force on the system is the product of the magnitude, F, of the force, the magnitude Dr of the displacement of the point of application of the force, and cos q, where q is the angle between the force and the displacement vectors Work, cont. n W = F Dr cos q n n n The displacement Dr is along the point of application of the force A force does no work on the object if the force does not move through a displacement The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application Work Example n The normal force, n, and the gravitational force, mg, do no work on the object moving along the table n cos q = cos 90 = 0 n The force F does do work on the object equal to FDrcosq More About Work n The system and the environment must be determined when dealing with work n Work by the environment on the system or work done by the system on the environment n The sign of the work depends on the direction of F relative to Dr n n Work is positive when projection of F onto Dr is in the same direction as the displacement Work is negative when the projection is in the opposite direction Units of Work n n Work is a scalar quantity The unit of work is the Joule (J), the same as energy n n 1 joule = 1 newton . 1 meter J=Nm Work Is An Energy Transfer n n n This is important for a system approach to solving a problem If the work is done on a system and it is positive, energy is transferred to the system If the work done on the system is negative, energy is transferred from the system Work Is An Energy Transfer, cont n If a system interacts with its environment, e.g. gravity, this interaction can be described as a transfer of energy across the system boundary n This will result in a change in the amount of energy stored in the system Scalar Product of Two Vectors n The scalar product of two vectors is written as A . B n It is also called the dot product n A . B = A B cos q n n q is the angle between A and B Work = FDr Scalar Product, cont n The scalar product is commutative n A .B = B .A n n The scalar product obeys the distributive law of multiplication A . (B + C) = A . B + A . C Dot Products of Unit Vectors n n Using component form with A and B: Dot Product of a Vector n The dot product with itself is the magnitude of the vector squared ^ ^ ^ A A = (A i + A ^ + A k ) (A ^ + A ^ + A k) j i j = A +A +A x y z x y z 2 2 2 x y z A = A +A +A 2 2 x y 2 z Work Done by a Constant Force n Consider: ^ ^ F = 5.0i + 3.0 ^ & Dr = 2.0i + 4.0 ^ j j 1/ 2 1/ 2 F = (25.+ 9.0) = 5.8 & Dr = (4.0 +16.) = 4.5 W n W n W n n qF=tan-1(Fy/Fx) = 31 & qDr=tan-1(Dry/Drx) = 63 = FDr = 5x2+3x4 = 10+12 = 22 J |F|x|Dr| = 26 = |F|x |Dr| cos(qF-qDr) = 22 J Work Done by a Varying Force n Assume that during a very small displacement, Dx, F is constant For that displacement, W ~ F Dx For all of the intervals, n n Work Done by a Varying Force, cont n n Therefore, The work done is equal to the area under the F vs x curve n Work Done By Multiple Forces n If more than one force acts on a system the total work done on the system is the work done by the net force Work Done by Multiple Forces, cont. n If the system cannot be modeled as a particle, then the total work is equal to the algebraic sum of the work done by the individual forces on different parts of the system Hooke's Law n The force exerted by the spring is Fs = - kx n n x is the position of the block with respect to the equilibrium position (x = 0) k is called the spring constant or force constant and measures the stiffness of the spring n This is called Hooke's Law Hooke's Law, cont. n n n When x is positive (spring is stretched), F is negative When x is 0 (at the equilibrium position), F is 0 When x is negative (spring is compressed), F is positive Hooke's Law, final n n n The force exerted by the spring is always directed opposite to the displacement from equilibrium F is called the restoring force If the block is released it will oscillate back and forth between x and x if there is no friction Work Done by a Spring on a Block n n Identify the block as the system Calculate the work as the block moves from xi = - xmax to xf = 0 n The work done as the block moves from xi = 0 to xf = +xmax is 1 2 Ws = Fs dx = (-kx)dx = - kxmax x 0 2 xf x max i Spring with an Applied Force n n n Suppose an external agent, Fapp, slowly stretches the spring so that a~0 . The applied force is equal and opposite to the spring force Fapp = -Fs = -(-kx) = kx Work done by Fapp is equal to 1/2 kx2max Kinetic Energy & Work n Kinetic Energy is the energy of a particle due to its motion n K = 1/2 mv2 n n n K is the kinetic energy m is the mass of the particle v is the speed of the particle n A change in kinetic energy is the result of doing work to transfer energy into a system Kinetic Energy, cont n Calculating the work: dv W = Fdx = madx = m dx dt dv dx dv W = m dx = m vdx = mvdv dx dt dx 1 1 W = mv - mv 2 2 xf xf xf xi xi xi xf xf vf xi xi vi 2 2 f i Work-Kinetic Energy Theorem n n In the case in which work is done on a system and the only change in the system is in its speed, the net work done by the net force equals the change in kinetic energy of the system. W = DK = Kf - Ki n where K = 1/2 mv2 Work-Kinetic Energy Theorem n In this example the normal and gravitational forces do no work since they are perpendicular to the direction of the displacement n n n W = F Dx W = DK = 1/2 mvf2 - 0 take vi = 0, Nonisolated System n A nonisolated system is one that interacts with or is influenced by its environment, such as gravity. n An isolated system would not interact with its environment n The Work-Kinetic Energy Theorem can be applied to nonisolated systems Nonisolated System: Work Done by Man and by Gravity A man is sliding a box up the ramp. The man & box are a system. Gravity is opposing the man's effort. Both the man and gravity do work. W = Wman + Wgravity 1 2 1 = Mv f - Mvi2 2 2 Work Done by the Man Work done by the man sliding the box up the ramp of length L. Use x,y coordinates. r f = constant force due to the man r ^ f = - f cos q i + f sin q ^ j r L = -L cos q i^ + L sin q ^ j Work done by the Man r r ^ ^ Wm = f L = (- f cos q i + f sin q ^) (-L cos q i + L sin q ^) j j Wm = fL cos2 q + fL sin 2 q = fL(cos2 q + sin 2 q ) = fL ^ ^ j j Re m. i i^ = ^ ^ = 1 i ^ = ^ i^ = 0 j j Work Done by Gravity Work done by Gravity: r Fg = -mg^ j r Dr = L cos qi^ + L sin q^ j r r ^ Wg = Fg Dr = (-mg^) (L cos qi + L sin q^) j j Wg = -mgL sin q Work Done by Both the Man and by Gravity-1 W = Wman + Wgravity = fL - mgL sin q Wman + Wgravity 1 2 1 = fL - mgL sin q = Mv f - Mvi2 2 2 Note that if the man does the minimum work f = mg sinq then Wman= -Wgravitythe net work is zero, and hence vf = vi If f > mg sinq then v2f > v2i If f < mg sinq then v2f < v2i This theorem doesn't determine if v is + or - Work Done by Man and by Gravity-2 Now consider that the man lifts the box straight up the same distance as before, Dr = L sinq j i.e. does not use the incline. Assume that that the man does the minimum work to overcome the work by gravity so, that vi = vf. r r Wm = f Dr = f ^ (L sin q ^) = fL sin q j j Wg = (-mg ^) (L sin q ^) = -mgL sin q j j Wm + Wg = 0 fL sin q = mgL sin q f = mg He does the same work but exerts more force than when pushing the box up the incline , i.e. mg > mgsinq. PinBall Problem The ball launcher in a pinball machine has a spring that has a force constant of 120. N/m. The surface on which the ball moves is inclined 10.0 with respect to the horizontal. If the spring is initially compressed 5.00 mm, find the launching speed of a 0.100kg ball when the plunger is released. Friction and the mass of the plunger are negligible. Pinball Note that the x,y scale is not the same as the scale of the picture. xi = -0.005m, xf = 0.000 yi = 0.000, yf = 0.005 sin10m vi = 0.000 vf = ? Solution: Pinball Prob. Note that the spring force is variable i.e. it depends on position. The force of gravity is constant. 1 mv2 f 2 1 mv2 -2 i sf si W = Wsp + Wg = Wsp = - ksds = 1 ksi2 - 1 ks 2 = 1 k[(xi2 + yi2 )] f 2 2 2 Wg = -mg(y f - yi ) = -mgy f 1 k(x 2 + y 2 ) - mgy i i f 2 = 1 mv2 f 2 Solution: Pinball Prob. 1 k(x 2 + y 2 ) - mgy i i f 2 = 1 mv2 f 2 v2 f v2 f 2 vf k 2 = (xi ) - 2gy f m 120N /m =)[(0.005)2 ] - 2(9.80m / s 2 )0.005m sin10 o ) 0.100kg = 0.030 - 0.017 = 0.013m / s 2 2 v f = 0.114m / s Conservative Forces n The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle n Gravity and spring forces are conservative forces n The work done by a conservative force on a particle moving through any closed path is zero n A closed path is one in which the beginning and ending points are the same Work Done by Gravity A 4.00-kg particle moves from the origin to position C, having coordinates x = 5.00 m and y = 5.00 m. One force on the particle is the gravitational force acting in the negative y direction. Calculate the work done by the gravitational force in going from O to C along (a) OAC, (b) OBC, (c) OC. Your results should all be identical. Why? y Fg Wg Along OAC = mg = (4.00 kg )( 9.80 m s ) = 39.2 N 2 B C (5.00, 5.00) m O A x (a) Work along OAC = work along OA + work along AC WOA = Fg (x A - xO ) cos 90 WAC = Fg (yC - x A ) cos180 WOAC = Fg (x A - xO ) cos 90 + Fg (yC - yA ) cos180 = (39.2N)(5.00m)(0) + (39.2N)(5.00m)(-1) = -196J Wg Along OBC y B C (5.00, 5.00) m O A x (b) W along OBC = W along OB + W along BC WOB = mg(yB - y0 ) cos180 o WBC = mg(xC - x B ) cos 90 o WOBC = 39.2N(5.00m)(-1) + 39.2N(5.00m)(0) WOBC = -196.J y Wg Along OC WOC = LOCmg cos(90+45) LOC = (5.002+5.002)1/2= 7.07 cos135 = -.707 WOC = (39.2N)(7.07m)(-.707) = -196J B C (5.00, 5.00) m O A x The results are the same for all paths because gravity is a conservative force. The work only depends on the altitudes of the initial and final points. Show that the work along OBCAO = 0 Conservative Forces n Examples of conservative forces: n n Gravity Spring force n We can associate a potential energy for a system with any conservative force acting between members of the system n n This can be done only for conservative forces In general: Wcon. = - DU Work by Conservative Forces & Potential Energy n n The work along path OC was W = -196 J The gravitational potential energy is U = mgy At y = 0 Ui = 0 At y = 5.0 Uf =(4.0kg)(9.80m/s2)(5.0m)= 196 J W = -(Uf - Ui) = -196 J Energy Conservation n Work Energy Theorem n n n n For conservative forces W =- (Uf - Ui) If all the work is due to conservative forces W = - (Uf - Ui) The work energy theorem can be rewritten as n n W = Kf - Ki - (Uf - Ui) = Kf - Ki Ki +Ui = Kf + Uf or Energy is conserved Nonconservative Forces n n A nonconservative force depends on the path taken, not on the end points alone Nonconservative forces acting in a system cause a change in the mechanical energy of the system Friction - A Nonconservative Force n n The work done against friction is greater along the red path than along the blue path Because the work done depends on the path, friction is a nonconservative force Work Done by Friction B y C (5.00, 5.00) m Consider a 4.00kg object moved in the horizontal plane, i.e. the force of gravity is into the screen. The coeff. of friction k = 0.100. The frictional force O 2 = -3.92 N. is f = -mg = -0.100(4.00kg)9.80m/s The work done by friction in going from O to C by the path OAC is WOAC = WOA +WAC = f (xA-xO) + f (yC-yA) = (-3.92N)10.0m = -39.2J Work done by going the direct route OC WOC = (-3.92N) 7.07 m = -27.2J The work done by friction is path dependent. A x Energy and Non-Conservative Forces The work done by conservative forces can be related to a change in potential energy because the work only depends on the beginning and end points. Wcon=-(Uf-Ui) The work done by non-conservative forces cannot be related to a potential energy change because potential energy depends on the beginning and end points while the n-c work depends on the path taken between the beginning and end. Mechanical Energy and Nonconservative Forces n In general, if friction is acting in a system: n Wcon +Wnon-con= DK n n n n DU is the change in all forms of potential energy If friction is zero, this equation becomes the same as Conservation of Mechanical Energy, DEmech = 0 If friction is present Wnon-con= -fd n We defined DEmech = DK + DU -fd = DEmech = DK + DU Problem Solving Strategies Nonconservative Forces n n Define the isolated system and the initial and final configuration of the system Identify the configuration for zero potential energy n These are the same as for Conservation of Energy n The difference between the final and initial energies is the change in mechanical energy due to friction Start 4/6 1) Only Conservative Forces: gravity, springs Wcon = DK or since Wcon= -DU Mechanical Energy Conservation Kf + Uf = Ki + Ui 2) Non conservative (friction) + Conservative forces Wfr = DEMech Wfr = (Kf + Uf) - (Ki + Ui) Work Done with Friction A 6.0 kg block is pulled from rest by a force F = 12 N over a surface with a coefficient of kinetic friction 0.17 at an angle of 5. Find the speed of the block after it has been moved 3.0 m. Work Done by Friction W fr = fDx cos180 = - fDx f = mk n y : n - mg + F sin q = 0 n = mg - F sin q Note that n 0. W fr = -m k (mg - F sin q )Dx If Fsinq > mg the cube will be lifted off the surface. Since n 0 the frictional work is negative and will reduce the amount of KE imparted to the block. Work Done by F n Work done by the Applied force F is a conservative force: WF = Fcosq(x-xi) n We could say WF = -DU = (Fcosqx-Fcosqxi) U = -Fcosqx DU = -FcosqDx Wf + WF = DK Wf = DK + DU Alternatively we could just put Wf + FcosqDx = DK Total Work Done Work Done by the Applied Force WF = FDx cos q Work Kinetic Energy Theorem 1 2 1 2 W fr + WF = mv f - mvi 2 2 1 2 1 2 [-m k (mg - F sin q ) + F cos q ]Dx = mv f - mvi 2 2 Work Done with Friction m = 6.0kg F = 12.N q = 5 o m k = 0.17 Dx = 3.0m [-m k (mg - F sin q ) + F cos q ]Dx vf = m /2 [-.17(58.8 -12 sin 5) +12 cos 5]3.0 vf = = 1.5m / s 3.0 w/o friction v = 3.5 m/s Water Slide w/friction vi = 0 h = 2.00m vf = 3.00m/s Uf = 0 m = 20.0kg What is Wfr? DEmech = DK + DU DEmech = (Kf Ki )+(Uf - Ui) DEmech= (1/2)mvf2mgh DEmech = 90J - 392J = - 302J Wfr = -fd = -302J Note that it is not easy to determine the friction f because n is not always in the same direction. Spring Compression A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction of 0.250. The object has a speed of vi = 3.00 m/s when it makes contact with a light spring that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring's unstretched position. Finally the object comes to rest a distance D to the left of the unstretched spring. Find (a) the distance of compression d, (b) the speed v at the unstretched position when the object is moving to the left, and (c) the distance D where the object comes to rest. Spring Compression a) Find d. Between the second and the third picture, DE mech = DK + DU = - fd K i = 1 mvi2 K f = 0 2 U i = mgy U f = mgy + 1 kd 2 2 -mmgd = - 1 mvi2 + 1 kd 2 2 2 1 1 -0.250(1.00kg)(9.80m / s 2 )d = - (1.00kg)(3.00m / s) 2 + (50.0N /m)d 2 2 2 25.0d 2 + 2.45d - 4.50 = 0 -2.45 (2.45) 2 - 4(25.0)(-4.50) (-2.45 21.35)N d= = = 0.378m 50.0 50.0N /m Spring Compression b) Find v going to the left at the un-stretched x. Between picture two and picture four, DE mech = DK + DU = - f 2d 1 1 2 -f (2d) = mv - mv2 i 2 2 2 (2.45 N) (2) (0.378 m) v = ( 3.00 m s) ( 1.00 kg ) 2 = 2.30 m s Spring Compression c) Find the rest position. For the motion from picture two to five: DE mech = DK + DU = - f (D + 2d) 1 K i = mvi2 K f = 0 DU = 0 2 1 2 -f (D + 2d) = - ( 1.00 kg ) (3.00 m s ) 2 9.00 J D= - 2( 0.378 m) = 1.08 m 2 2( 0.250) (1.00 kg ) 9.80 m s ( ) Internal Energy n n n The energy associated with an object's temperature is called its internal energy, Eint In this example, the surface of the book is the boundary of the system The friction does work and increases the internal energy of the surfaces Missing Energy? Where did the mechanical energy go? Internal Energy of One Atom At the microscopic level atoms move very rapidly. In solids they vibrate back and forth at a high speed, typically 500 m/s. The mass of an iron atom is about 9x10-26 kg. The kinetic energy of one vibrating atom is (1/2)mv21 x 10-20J. Not very much. Total Internal Energy The atomic mass of iron is 56. 56 grams of iron has Avogadro's number of atoms (NA = 6.02x1023). An iron ball with m=0.5 kg has 9NA 5x1024 atoms. Therefore the microscopic internal kinetic energy of an iron ball is about 50,000J. A considerable amount. This internal energy is called thermal energy. The higher the kinetic energy the higher the temperature. By comparison the macroscopic energy of a 0.5 kg iron ball moving at 10m/s is 25 J. Thermal Energy The microscopic internal energy of atoms and molecules in matter is called internal energy Ethermal = Kmicro + Umicro We have seen before that when friction is present mechanical energy isn't conserved. Where did it go? It went into thermal energy causing the atoms to vibrate even faster. Total energy is conserved. Internal Energy n There are different types of internal potential energy n n chemical nuclear n The internal kinetic energy is called temperature. Usually measured empirically in degrees not Joules. Ways to Transfer Energy Into or Out of A System n n n Work transfers by applying a force and causing a displacement of the point of application of the force Mechanical Waves allow a disturbance to propagate through a medium Heat is driven by a temperature difference between two regions in space More Ways to Transfer Energy Into or Out of A System n n n Matter Transfer matter physically crosses the boundary of the system, carrying energy with it Electrical Transmission transfer is by electric current Electromagnetic Radiation energy is transferred by electromagnetic waves Conservation of System Energy n Energy is conserved n n This means that energy cannot be created or destroyed If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer Conservation of Energy, cont. n Mathematically, SEsystem = ST n n Esystem is the total energy of the system T is the energy transferred across the system boundary n Established symbols: Twork = W and Theat = Q n The Work-Kinetic Energy theorem is a special case of Conservation of Energy Power n n The time rate of energy transfer is called power The average power is given by when the method of energy transfer is work Instantaneous Power n The instantaneous power is the limiting value of the average power as Dt approaches zero n This can also be written for constant F as Power Generalized n n Power can be related to any type of energy transfer In general, power can be expressed as n dE/dt is the rate rate at which energy is crossing the boundary of the system for a given transfer mechanism Units of Power n The SI unit of power is called the watt n 1 watt = 1 joule / second = 1 kg.m2 / s2 n A unit of power in the US Customary system is horsepower n 1 hp = 746 W = 550 ftlbs/s n Units of power can also be used to express units of work or energy n 1 kWh = (1000 W)(3600 s) = 3.6 x106 J Human Power Consider a 200lb man climbing a ladder at the rate of 1 ft/sec. P = Fv = (200lbs)(1 ft / s) = 200 ft lbs / s 1hp P = 200 ft lbs / s = 0.36hp 550 ft lbs / s 746W P = 0.36hp = 270W = 0.27kW 1hp Power Ski Lift A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30.0 slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform this task? Ski Lift: Solution (a) How much work is required to pull him a distance of 60.0 m up a 30.0 slope (assumed frictionless) at a constant speed of 2.00 m/s? W = DK, but DK = 0 because he moves at constant speed. Wmoter + Wgravity = 0 The skier rises a vertical distance of (60.0m)sin30. Thus Wmoter = - Wg = (70.0kg)(9.80m/s2)(30.0m) = 2.06104 J Ski Lift b) A motor of what power is required to perform this task? The time to travel 60.0 m at a constant speed of 2.0m/s is 30.0 s. Thus, Pmoter W 2.06 10 4 J 1hp = = = 686W = 686W = 0.920hp Dt 30.0s 746W Elevator Power An elevator car has a mass of 1,600kg and is carrying passengers having a combined mass of 200kg. A constant frictional force of 4,000N retards its motion upward. What power must the motor deliver when the speed is 3.00m/s if the elevator is accelerating at 1.00m/s2. Elevator Power Apply Newton's 2nd Law T - f - Mg = Ma T = f + M (g + a) T = 4,000N + (1800kg)(9.80m / s 2 +1.00m / s 2 ) T = 23.4 10 3 N P = Tv = (23.4 10 3 N)(3.00m / s) P = 70.2kW P = (70.2kW) 1hp = 94.1hp .746kW Start 4/9 n Frictional Work Change in mechanical energy fi Wf = DK+ DU o o Experiment - Pulling Power Experiment - Pile Driver n n Power = dW/dt = dEdt Energy and the Automobile Pile Driver EA= mgH EB= mgd+1/2mv2 EC= mgh AB Wf = 0 =EB-EA fi EB=mgH BC Wf =-f(d-h) =EC-EB f = mg(H-h)/(d-h) Automobile Drag: Snow Plow Model As it plows a parking lot, a snowplow pushes an ever-growing pile of snow in front of it. Suppose we model a car moving through the atmosphere with velocity v as a cylinder pushing a growing slug of air in front of it. The original stationary air is set into motion at the speed v. Automobile Drag: Snow Plow Model In a time interval Dt, a new disk of air of mass Dm must be moved a distance vDt and hence must be given a kinetic energy (1/2)(Dm)v2 . Using this model show that: (a) the automobile's power loss due to air resistance is (1/2)rAv3 and (b) the resistive force acting on the car is (1/2)rAv2, where r is the density of air. Compare this with the empirical expression 1/2)DrAv2 for the resistive drag force. Automobile Drag: Snow Plow Model (a) Automobile's power loss due to air drag W = Work done by the car on the air by pushing a slug of air in front of the car 1 W = DK = (Dm)v 2 2 1 2 DK = (Dm)v is the change in K of the air. 2 How much air is pushed in time Dt? Automobile Drag: Snow Plow Model m Recall that definition of density is r for any vol size volume. So the mass scooped up in time Dt can be expressed as Dm = r D(vol.). The volume of air scooped up in Dt is D(vol.) = AvDt, A is the area of the front of the car, and vDt is the length of the air column, hence Dm = rAvDt Automobile Drag: Snow Plow Model The work done by the car on the air is then 1 1 W = Dmv2 = (rAvDt)v2 2 2 1 W = rAv 3Dt 2 W The power needed is P = Dt 1 P = rAv 3 2 Automobile Drag: Snow Plow Model (b) Since we also can say P = Fv, 1 F = rAv2 2 This model predicts the same proportionalities as the empirical equation for the drag force, D = C (1/2)rAv2 and gives C=1 for the drag coefficient. Air actually slips around the moving object instead of accumulating in front of it. For this reason C is not necessarily 1. Typically C<1, although in some cases C >1 if the air flow is complicated. Energy and the Automobile n n n n n The concepts of energy, power, and friction help us to analyze automobile fuel consumption About 67% of the energy available from the fuel is lost in the engine About 10% is lost due to friction in the transmission, drive shaft, bearings, etc. Another 6% goes to internal energy and 4% to operate the fuel and oil pumps and accessories This leaves about 13% to actually propel the car. Friction in a Car n The magnitude of the total friction force is the sum of the rolling friction force and the air drag n n n t = r + a At low speeds, rolling friction predominates At high speeds, air drag predominates r= rn a = (1/2)Crv2A Automotive Power F is the driving force on the wheels derived from the engine. f = (218 + 0.70v2 ) is an empiracly determined rolling and air drag force. F = F - f - mg sin q = ma x F = ma + mg sin q + f f = (218 + 0.70v 2 ) P = Fv = mva + mgv sin q + 218v + 0.70v 3 Power to the Wheels for a = 0 3,000 lb car Power to Wheels(hp) vs v(mph) 100 90 80 70 Power(hp) 60 50 40 30 20 10 0 0 20 40 v(mph) 60 80 P(hp) 5 P(hp) 0 Friction in a Car, cont Note that the normal force changes with speed. This is due to lift associated with air over and under the car. Gas Consumed by a 3,000lb Car going from v = 0 to v = 65 mph A car has a mass of 1371 kg. Its efficiency is rated at 13%, i.e. 13 % of the available fuel energy is delivered to the wheels. How much gasoline is used to accelerate the car to 27 m/s (60mph)? The energy equivalent of gasoline is 1.3x108 J/gal. Gas Consumed by a 3,000lb Car going from v = 0 to v = 65 mph The energy of the car at 27m/s (60mph) is K = 1/2 mv2 =(1/2)(1371kg)(27m/s)2= 5.0x105J Since the car is only 13% efficient each gallon of gas yields (0.13)(1.3x108J/gal) = 1.7x107J/gal Hence the number of gallons needed is 5.0 10 J No. of Gallons = = 0.029gal 7 1.7 10 J / gal 5 Gas Consumed by a 3,000lb Car going from v = 0 to v = 65 mph Suppose it takes 10 s to go from 0 to 60mph. During this time the car goes a distance of 0 + 27m / s 1mi Dx = v ave Dt = 10s = 135m = .08mi 2 1600m The car consumed .029 gallons of gas to go .08 mi. This comes out to 2.8mi/gal during the acceleration time. Gas Consumed by a 3,000lb Car going at a constant speed of 65 mph From the previous graphs w/a=0 and v = 65mph the required power to the wheels is 25hp (746W/hp) = 18.5kW. For a 13% efficiency in delivering power from the engine to the wheels The .13Peng = Pwheels means Peng = 18.5kW/.13 = 142kW Gas Consumed by a 3,000lb Car going at a constant speed of 65 mph The energy equivalent of gasoline is 1.38 108J/gal. J J 1gal To get 142,000 we need to deliver 142,000 s s 1.38J gal 3600s gal .001 = 3.6 . For a car going 65mph the s 1hr hr mi 65 hr = 18mi / gal means we will get gal 3.6 hr Compare to the acceleration mileage of 2.8mi/gal. ...
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This note was uploaded on 05/19/2008 for the course PHYS 161 taught by Professor Hammer during the Spring '07 term at Maryland.

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