answer7 - Answer Key Homework 7 Rubin H Landau 1 Solution...

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Unformatted text preview: Answer, Key Homework 7 Rubin H Landau 1 Solution : This print-out should have 8 questions. Check that it is complete before leaving the printer. The work done by the force F results in the Also, multiple-choice questions may continue change of potential energy of the system. The nal energy is the potential energy PEfinal = on the next column or page: nd all choices mgx; the initial energy is zero. Thus, before making your selection. This HW 7 should be done *Before* assignE = mgx ment HW 6. Yes, this means they are out of order. However, since HW 7 is short and HW = 888 N 22:3 m  6 is long, you are advised to start HW 6 early. = 19802:4 J : Block and Tackle 08:02, trigonometry, numeric, 1 min. 004 A weight of 888 N is raised by a two-pulley arrangement as shown in the gure. Assume that the pulleys are weightless, the rope does not stretch, and the system moves at a constant speed which is slow enough that the kinetic energy is negligible. How much work is done by the agent force F to raise the weight by a vertical distance of 22:3 m? F m ∆x 005 Which of the following is the SI unit of the force? 1. kg m s2 correct 2. W 3. kg 4. m s2 5. Nm 6. kg m s 7. m kg s 8. J 9. J s 10. N s Explanation: Correct answer: 19802:4 J. The SI unit of force is kg m s2. This combination of units is called a Newton N. with concepts that can be tested in quiz problems a What is the magnitude of force F ? b What is the distance that the force F move? Block and Tackle 08:02, trigonometry, numeric, 1 min. 006 Explanation: Optional questions not graded, but deal Basic Concepts: W =Fs Ugrav: =X mgh Fnet = Fi i The change in potential energy of the weight is . 1. 2mgx 2. mgx correct Answer, Key Homework 7 Rubin H Landau 2 final initial 3. F x 4. 1 2F , mgx 2 5. 1 F , mgx 2 6. F , 2mgx P T T' d T' P d/2 7. 3 F x 2 8. F , mgx 9. 2F , mgx 10. 2F + mgx Explanation: See Part 1. Optional problems a If we look at the forces on the pulley we have T T d T mg mg Note that the length of the cord between points T and T0 is d, and initially both points are a distance d=2 above the moving assembly of the pulley, which is of negligible size. When the cord moves a distance d, the point T moves a distance d above the its initial position and the point T0 stays where it is it is of course still a distance d from T; but now the assembly has reached the point T0, which means the assembly has moved up a distance d=2. Algorithm mg As the weight is lifted with no acceleration we have, by force balance, 2T = mg or T = F = mg 2 b If d is the distance the force F acts, then the work done is W = Fd = mg d. Using the 2 mg result of Part 1, W = mgx = 2 d and so d = 2x Let us examine the geometry of the pulley in order to convince ourselves that this results is the correct answer, namely that if the mass is lifted a distance d=2 then the force F acts over a distance d.  b = 888 N 200 900 5 a = 22:3 m 25 d = 2:0 a = 2:0 h22:3i = 44:6 m hmi = hi hmi F = 2b0 : = h888i 2:0 = 444 N hNi = hNi hi W =F d = h444i h44:6i = 19802:4 J hJi = hNi hmi AP M 1993 MC 18 08:02, calculus, numeric, 1 min. 007 1 2 3 units 4 units 5 units Answer, Key Homework 7 Rubin H Landau 3 When an object is moved from rest at point 99:6 km=h. From what height would the car A to rest at point B in a gravitational eld, have to be dropped to have the same kinetic the net work done by the eld depends on the energy? mass of the object and Correct answer: 39:0533 m. 1. the positions of A and B only. correct 2. the path taken between A and B only. 3. both the positions of A and B and the path taken between them. 4. the velocity of the object as it moves between A andB . 5. the nature of the external force moving the object from A to B . Explanation: Gravitational force is a conservative force, so in addition to the mass, only the positions are needed. 008 A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high must a new ramp be? 1. 4 correct 2. 2 3. 3 4. 1 5. 5 6. 6 Explanation: The gain in kinetic energy, proportional to the square of the block's speed at the bottom of the ramp, is equal to the loss in potential energy. This, in turn, is proportional to the height of the ramp. 009 Consider a compact car that is being driven at Explanation: Assume the car is dropped from the height h. By conservation of energy, KEo + PEo = KEf + PEf PEo = KEf Thus to attain the same kinetic energy as a car of the same mass driven at a speed of v, 1 mgh = 2 mv2 v2 h = 2g The velocity must be in m s. Algorithm m hkmi = 1000 m=km hhsi = 0:000277778 h=s 1 2 3 4 5 g = 9:8 m=s2  80 v = 99:6 km=h 120 m vmps = vhkmihhi s = h99:6ih1000ih0:000277778i = 27:6667 m=s hm=si = hkm=hihm=kmihh=si units v2:0 h = 2mps 6 :0g 6667 2:0 = h27::0h9:8ii 2 = 39:0533 m hm= i2:0 hmi = hihms=s2i units Sliding Down a Plane 08:02, calculus, multiple choice, 1 min. 012 A 3:51 kg block starts at a height of 51:3 cm on a plane that has an inclination angle of 51:1 as in gure. Answer, Key Homework 7 Rubin H Landau 1:0 , m θ Upon reaching the bottom, the block slides along a horizontal surface. The coe cient of friction on both surfaces is  = 0:14 How far does the block slide on the horizontal surface before coming to rest? Correct answer: 3:25035 m. Explanation: From the conservation of energy for the part of the motion on the inclined plane 1 m v2 = m g h , W 2 end where the work done is Z h= sin  , tan h51:1i h3:1415926 i 180:0 = = 3:25035 m h hmi = hi , h0:14i 4 h0:01i h51:3i h0:14i hi ,i tan h hihi hi hm=cmi hcmi units Rising Elevator 08:04, arithmetic, multiple choice, 1 min. 013 An elevator is rising at constant speed. Consider the following statements: I. The upward cable force is constant. II. The kinetic energy of the elevator is constant. III. The gravitational potential energy of the earth-elevator system is constant. IV. The acceleration of the elevator is zero. V. The mechanical energy of the earth-elevator system is constant. Which of the statements are true? 1. all ve are true 2. only II and V are true 3. only I, II, and IV are true correct 4. only I, II, and III are true 5. only IV and V are true Explanation: Basic Concepts: Potential Energy, Ki- ,f d x h =  m g cos  sin  From the conservation of energy on the horizontal plane: m g h ,  m g h cot =  m g x Note that the mass cancels out, therefore netic Energy x = h1 ,  cot = We will consider these statements one at a 0:01 m=cm51:3 cm1 , 0:14 cot51:1  time. I. Since the elevator is moving at con= 0:14 stant speed, the net force on the elevator = 3:25035 m must be zero. The tension in the cable must be equal and opposite to the weight of the Algorithm the is m 1 elevator, and since cable weight alsoconstant, hcmi = 0:01 m=cm the tension in the must be con h = 51:3 cm 24 2 stant. II. The kinetic energy depends only 52 3:1 m = 3:51 kg 3:9 3 on the mass of the elevator and its velocity 1 48 = 51:1 55 4 KE = 2 mv2 . Since the mass and velocity   = 0:14 0::12 5 are both constant, the kinetic energy is also 0 32 constant. III. The gravitational potential  mih energy of the earth-elevator system is increas1:0 , tan   hcm 180:0 x= 6 ing, because the distance between the elevator  W= 0 Answer, Key Homework 7 Rubin H Landau and the earth is increasing. The potential energy depends only on the mass of the earth, the mass of the elevator, and the distance between them. IV. Since the elevator is rising at constant speed, its acceleration is zero. V. The mechanical energy of the earth-elevator system is not constant, because the potential energy is increasing see explanation of III while the kinetic energy is constant. Since the mechanical energy is the sum of the kinetic and potential energies, the mechanical energy is increasing. Thus only I, II, and IV are true. 5 ...
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