= W Q H. ɛ = T H T C T H = = 0.20 = T C = T H (1 0.20) = = 320 K = 47 C

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1 1. Four identical 0.18 kg masses are placed at the corners of a 4.0 x 3.0 m rectangle, and are held there by very light connecting rods which form the sides of the rectangle. What is the moment of inertia of this system about an axis which passes through one of the particles and is perpendicular to the plane of the rectangle (answer in kg m 2 )? a) * 9.0 b) 7.5 c) 8.5 d) 8.0 e) none of these The moment of inertia for point particles is given by I = i m i r 2 i. In this case the sum runs over three masses since the one through which the axis passes has 0 distance. Two of the other masses are at 4 cm and 3 cm, and the third is at the diagonal distance 5 cm. So the result is I = 0.18 ( ) = 9 kg m An engine is proposed to work off the thermal gradient found in the ocean. What would be the maximum efficiency of such an engine given that the surface temperature of the sea water is 15 C, and the coldest temperature below the surface is 5 C a) * b) 0.17 c) d) 0.67 e) none of these The maximum efficiency is the Carnot efficiency given by the Second Law of Thermodynamics: ɛ = (T H T C )/T H with the temperatures in Kelvin. So in this case ɛ = 10/288 = An automobile engine operates with an overall efficiency of 12%. How many gallons of gasoline are essentially wasted for every 10 gallons which are burned in the engine? a) 1.2 b) * 8.8 c) 6.5 d) 4.7 e) none of these Here we use the general definition of efficiency: ɛ = (Q H Q C )/Q H. What we want to find is the wasted heat (in terms of gallons of gas) which can be found by re-arranging the formula: Q C = (1 ɛ)q H = (1.12)10 = 8.8 gallons wasted. 4. An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5 milliseconds. What is the magnitude of this acceleration (answer in kilometers/s 2 )? a) 2.5 b) 0.80 c) * 1.60 d) 1.30 e) none of these Use one of the three basic kinematic equations for constant acceleration x(t) = x 0 + v 0 t + at 2 /2. In this case x 0 = 0 and v 0 = 0, so a = 2x(t)/t 2 = 2(2.0)/( ) 2 = 160, 000 cm/s = 1.60 km/s. 5. A Carnot engine whose high temperature (T H ) source is at 127 o C takes in 200 calories and expels 160 calories to the low temperature (T C ) sink. What is the value of the low temperature T C (answer in o C) a) 102 b) 87 c) 40 d) 0 e) * none of these You must first convert to Kelvin and then use the definition of efficiency and the efficiency for a Carnot engine according to the Second Law of Thermodynamics: ɛ = T H T C T H = W Q H = = 0.20 = T C = T H (1 0.20) = = 320 K = 47 C 1

2 6. The speed of a 2.0 kg object changes from 30 m/s to 40 m/s during a 5 second time interval. During this time interval the object changes is direction by 90 degrees. What is the magnitude of the average force exerted on this object? (answer in Newtons) a) 30 b) * 20 c) 40 d) 50 e) none of these Compute change in velocity according to the Pythagorean Theorem: v = = 50 m/s. Now divide v by the time interval to get the average acceleration a = v/ t = 50/5 = 10 m/s 2. Finally, multiply the mass time the average acceleration to get the average force F = ma = 2 10 = 20 N. 7. A merry-go-round has a radius R = 2 m and a moment of inertia I = 250 kg-m 2. It is rotating about its center at 10 revolutions per minute. A child of mass 25 kg hops onto edge of the merry-go-round. What is the new rotational speed of the merry-go-round with the child at the outer edge (answer in revolutions per minute)? a) 10 b) 9 c) 8 d) * 7 e) none of these Solve by using conservation of angular momentum L 1 = I 1 ω 1 = I 2 ω 2 = L 2. We are given I 1 and ω 1 and we are asked to find ω 2. So we have to compute I 2. We take the child as a point particle for which I C = mr 2 = 25(2) 2 = 100 kg-m 2. Add the child s moment of inertia to the original moment of inertia to get the final moment of inertia: I 2 = I 1 + I C = = 350 kg-m 2. Now compute ω 2 = (I 1 /I 2 )ω 1 = (250/350)10 = 7.1 RPM. Normally we quote ω in rad/s, but there s not point in converting from RPM if the answer is in RPM. 8. A spaceship of mass m circles a planet of mass M at a distance R from the center of the planet. How much energy is required to transfer the spaceship to a new circular orbit at a distance of 3R? a) GmM/(2R) b) * GmM/(3R) c) GmM/(4R) d) GmM/(6R) e) none of these The total energy in orbit of radius R is given by E = GMm/2R where M is the mass of the planet (or Sun) and m is the mass of the orbiting object, and R is the distance between the centers of the masses. So E = E 3R E R = GMm/2R1/3 1 = +GMm/3R. 9. The work done in the expansion of an ideal gas from an initial state P 1, V 1 to a final state P 2, V 2 is a) (P 2 P 1 )(V 2 V 1 ) b) * the area under the P V curve c) independent of the path taken d) the slope of the P vs. V curve e) none of these. 10. A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s along the direction r = î + ĵ. What is its angular momentum (in kg m2 /s) relative to the (x, y) point (0,5)? a) 2 ˆk b) 1 ˆk c) 3 ˆk d) d ˆk e) * none of these The angular momentum of a particle is given by l = r m v. So here we have l = 5 ĵ (2)(3)( î + ĵ) = 30 ˆk kg-m 2 /s. 2

3 11. A satellite is in a circular orbit about the earth at an altitude where air resistance is insignificant. Ignoring any other astronomical bodies, which of the following is true? a) * There is only one force acting on the satellite. b) There are two forces acting on the satellite with a resultant of zero. c) There are two forces acting on the satellite with a non-zero resultant. d) There are three forces acting on the satellite whose resultant may be zero. e) None of the above statements is true. 12. Two stars of masses M and 6M are separated by a distance D. How from from star M should a third mass be placed in order that it have no net gravitational force acting on it? a) 0.41D b) 0.33D c) 0.37D d) * 0.29D e) none of these Call the third mass m and its distance from the M mass is x. So its distance from the 6M mass is D x. The m mass must be between the the star masses so that their gravity forces cancel out. This means F M = G Mm = G 6Mm x 2 (D x) 2 We obtain a quadratic equation for which the solutions are 6x 2 = D 2 2Dx + x 2 = 5x 2 + 2Dx D 2 = 0 d = 2D ± 24D 2 10 Only the positive solution makes sense: x =.29D. =.20D ±.49D 13. Two vectors have magnitudes 10 and 11, and their scalar product is 100. What is the vector sum of these two vectors? a) 6.6 b) * 4.6 c) 8.3 d) 9.8 e) none of these We take the scalar product of the sum of the two vectors as: From what is given So the magnitude of the sum is 21 = 4.6. ( A + B) ( A + B) = A 2 + B A B ( A + B) ( A + B) = (100) = 21 3

4 14. A rock is thrown downward from an unknown height h with an initial speed of 10 m/s. It strikes the ground 3.0 seconds later. What is the value of h (answer in meters)? a) 44 b) 14 c) * 74 d) 30 e) none of these Use one of the one dimensional kinematic equations of motion, taking ground level as y = 0. y(t) = y 0 + v 0 t gt 2 /2 In this case y 0 = h and v 0 = 10 m/s, where down is the negative direction. So at t = 3 seconds we have y(t = 3) = 0 = h 10(3) 4.9(9) = h = = 74.1 m. 15. A person whose weight is 800 N is riding in an elevator which is accelerating downward at the rate of 1.5 m/s 2. What is the magnitude of the force of the elevator floor on the person (answer in N)? a) * 680 b) 800 c) 920 d) 120 e) none of these The force of the floor of the elevator would be the weight of the person if the elevator were not accelerating. Since the elevator is accelerating downward by a rate of 1.5 m/s 2, then the force is reduced correspondingly. The person s mass = 800/9.8 = 81.6 kg, so the force is F = m(g 1.5) = 81.6(8.3) = 678 N 16. A 50 kg child is riding in a Ferris wheel which has a radius of 10 m and travels in a vertical circle. The Ferris wheel completes one revolution every 10 seconds. What is the magnitude of the force on the child exerted by the seat in which the child is sitting when the Ferris wheel is at the top if its motion (answer in N, ignore any forces exerted by a seatbelt or a restraining bar)? a) * 290 b) 490 c) 690 d) 200 e) none of these The centripetal force on the child is mv 2 /r = /10 = 200 N. The weight of the child is 490 N, of which 198 N goes into the centripetal force. That leaves 290 N for the force of the seat on the child. 17. A person lifted a 2.0 kg object from the bottom of a well at a constant speed of 2.0 m/s for 5.0 s. How much work was done (answer in Joules)? a) 220 b) * 200 c) 240 d) 270 e) none of these W = F d = wvt = = 200 Joules 4

5 18. A 1.2 kg mass is projected from ground level at some unknown angle with speed of 30 m/s. The mass is seen to just clear a 16 m high fence before falling back to the ground. What was the kinetic energy of the mass when it cleared the fence (ignore air friction and use conservation of energy, answer in J)? a) * 352 b) 188 c) 0.0 d) insufficient information provided e) none of these The initial energy is all kinetic = 0.5 mv 2 = 540 Joules. At the 16 m maximum height there is gravitational potential energy mgh = 188 Joules. So the difference is the remaining kinetic energy. 19. A 2.0 kg ball moving with a velocity of 3 î 4 ĵ bounces off the floor such that its new velocity is 3 î + 4 ĵ. What is the impulse exerted on the ball by the floor ( ĵis in the vertical direction, answer in Newton-seconds)? a) * +16 ĵ b) 16 ĵ c) +12 î d) 12 î e) none of these J F t = p = m(3 î + 4 ĵ) m(3 î 4 ĵ) = 2 8 ĵ = +16 ĵ 20. A horizontal disk with a 10 cm radius rotates about a vertical axis through its center. The disk starts from rest at t = 0 seconds and has a constant angular acceleration of 2.1 rad/s 2. At what time t will the centripetal (a c or a r ) and the tangential (a t ) components of the total acceleration be equal in magnitude (answer in seconds)? a) 0.55 b) 0.63 c) * 0.69 d) 0.59 e) none of these a c = v2 r = rα2 t 2 = a r = αr = t = 1/α = 0.69 s 21. A 3.0 kg object moving at 8.0 m/s in the positive x direction has a one dimensional elastic collision with an object of mass M initially at rest. After the collision the mass M has a velocity of 6.0 m/s in the positive x direction. What is the value of M in kilograms? (HINT: there are two unknowns M and the final speed of the 3.0 kg mass. Therefore you need to use two conservation equations.) a) 7.5 b) 6.0 c) 4.2 d) 3.0 e) * 5.0 The final speed of the 3.0 kg mass is symbolized as v. Then momentum conservation gives: 3 8 = 3v + M 6 = v = 2M 8 Kinetic energy conservation (elastic collision) gives 1 2 3(8)2 = 1 2 3(v) M(6)2 = 64 = v M = (2M 8) M = 4M 2 32M M = 4M M = M 2 + 5M = M = 5 kg 5

6 22. A cylinder of ideal gas is compressed to 1/16 of its initial volume while its pressure is increased 50 times. If the initial temperature was 20 o C, what is the final temperature (answer in degrees Celsius)? a) * 643 b) 916 c) d) 6.40 e) 293 Use the Ideal Gas Law P 1 V 1 = P 2V 2 = 50P 1(V 1 /16) T 1 T 2 T 2 T 2 = T 1 = (293) = 916 K = 643o C 23. Two particles each of mass 0.20 kg are place on a meter stick of negligible mass. One particle is placed at the 40 cm mark, and the other particle is placed at the 100 cm mark. The meter stick is free to rotate about an axis through its 0.0 cm mark, and is released from rest in the horizontal position. What is the initial angular acceleration of this system about the pivot point on the 0.0 cm end? (answer in rad/s 2, ignore the moment of inertia of the meter stick) a) * 11.8 b) 5.9 c) 8.4 d) 5.4 e) 1.20 Each mass exerts a torque about the pivot point, and the total torque is equal to the total moment of inertia times the angular acceleration: α = (r 1 + r 2 )g (r r 2 2) τ = τ 1 + τ 2 = Iα = (I 1 + I 2 )α mg(r 1 + r 2 ) = m(r r 2 2)α = ( )9.8 (0.40) 2 + (1.00) 2 ) = 11.8 rad/s2 24. A 100 kg (220 lb) student regrets having just eaten a 200 Calorie (=200,000 calorie) jelly doughnut. To burn it off, he decides to climb the steps of a tall building. How high up (in meters) would he have to climb in order to expend the energy contained in the doughnut? a)273 b) 623 c) 418 d) 185 e) * 854 mgh = h = 837, 200 = h = 854 meters which is more than twice the height of the Sears Tower, the tallest building in the world. 6

7 1. A TVA electric power plant has an overall efficiency of 15%. The plant produces 150 MW of electric power to Nashville by burning coal to produce steam at 190 C. The steam is condensed into water at 25 C in cooling towers in contact with some water from the Cumberland river whose temperature is 20 C. a) How much coal does the plant burn in one day (24 hours) if 1 kg of coal produces 33 x 10 6 J? The power plant must use up thermal power at the rate of 150 MW/0.15 = 1,000 MW because of the efficiency factor. In 24 hours then the plant will use 1, = Joules Dividing by Joules/kg, one will obtain kg of coal per day, corresponding to about 1300 tons of coal per day being burned. b) How much does the fuel cost TVA per year if the delivery price is $8 per thousand kg of coal? Just multiply out the numbers: Cost = = $ per year. c) How much river water in m 3 /sec, starting out at 20 C, must flow through the cooling towers if the water reaches 25 after it exits the towers? The river water must be absorbing the 85% of the thermal power which is wasted. The wasted power is 850 MW or J/second. So in one second a given amount of water m w must be raised from 20 to 25 degrees C with Joules of heat (= Q C ) Q C = m w c w T = m w where we use the specific heat of water as 1 cal/(g-c deg) or J/(g-C deg). This yields m w = grams of water in one second. Since the density of water is 1,000 kg per cubic meter m w = 40.6 m 3 in each second. 7

8 2. A mass of 1.30 kg is placed at the end of a light rod m long having negligible mass. This system rotates in a horizontal circle about the other end of the rod at 5010 revolutions/minute. a) What is the moment of inertia of this system about the axis of rotation? For discrete masses I = i m i r 2 i = (1.30)(.780) 2 =.79 kg-m 2 b) Air resistance exerts a force of Newtons on the mass, opposite to the direction of motion at all times. How much torque must be applied to the system in order to keep it rotating at constant angular velocity? The applied torque must be equal to the torque from the air resistance τ air = F r = = N-m 3. A 5.00 kg block is pulled along a horizontal frictionless table as shown in the figure. The force F is 12.0 N acting at 25 o above the horizontal. a) What is the acceleration of the block? The horizontal component of the force accelerates the block while the vertical component of the force just decreases the normal force of the table. So. F h = F cos(25) = = ma = 5.0 a = = 2.2 m/s 2 b) Suppose the force F is increased slowly, but still acts at 25 o above the horizontal. What is the maximum value of the magnitude of F consistent with the block remaining on the table? The maximum force would be such that the vertical component of F would be equal to the weight force: F v = F sin(25) = w = F = sin 25 = 116 n c) What is the acceleration of the block for the force F given in part b)? Repeat the solution for part a) except use the 116 N value from part b). This will give a = 21.0 m/s 2. 8

9 4. An 8.0 cm disk rotates at a constant rate of 1200 revolutions/minute about its axis. a) What is the angular velocity of the disk? Convert RPM to radians/second ω = π 60 = 126 rad/s b) What is the linear speed of a point 3 cm from the center of the disk? v = rω = = 3.77 m/s c) What are the radial and tangential accelerations of a point on the rim of the disk? a r = v2 r = ( )2.08 = 1, 270 m/s 2 a t = rα = 0 Since ω is constant, then α = dω dt = 0. d) What is the total distance that a point on the rim moves in 2 seconds? θ(t) = θ 0 + ω 0 t at2 = 126(2) = 252 radians 9

10 5. The coefficient of kinetic friction between the 3.0 kg object and the surface of the table is This mass is connected by a light string passing over a pulley and on to a 5.0 kg mass hanging down. What is the speed of the 5.0 kg mass when it has fallen a vertical distance of 1.5 meters. (Neglect the effect of the pulley.) We must obtain the acceleration of the system after which we can obtain the speed requested. First put in the free body forces for the masses and apply Newton s Second Law: For the 5 kg mass m 1 hanging down: where T is the tension in the string. For the 3 kg mass m 2 on the table: F 1 = m 1 a = m 1 g T F 2 = T f k = T µ k m 2 g = m 2 a Adding these two equations together, in order to cancel out the T tension, we obtain m 1 g µ k m 2 g = (m 1 + m 2 )a a = m 1 µ k m 2 g m 1 + m 2 a = 5 (3)(.4) (9.8) = 4.66 m/s 2 8 Now compute the speed from one of the three one-dimensional kinematic equations of motion In this case v 0 = 0 and so v 2 = v as v 2 = = 14.0 = v = 14.0 = 3.74 m/s 10

11 6. A 40 kg mass initially at rest is pulled a distance of 5 meters along a rough (µ k = 0.10) horizontal floor with a constant force of 130 N at an angle of 15 o above the horizontal. a) What is the acceleration, if any, of the mass? b) What is the total work done by the applied force? c) What is the (negative) work done by the frictional force? d) What is the kinetic energy of the mass at the end of the 5 meter distance? e) What is the final speed of the mass? a) The forces acting on the 40 kg mass are the applied 130 N (F ) force, the force of gravity (mg), the normal force of the floor (N), and the force of kinetic friction f k = µ k N. The mass is in equilibrium in the vertical direction (since it is not moving in the vertical direction), but it may have an acceleration in the horizontal direction. vertical direction f y = 0 = N + F sin 15 o mg = N = mg F sin 15 o = 358 Newtons horizontal direction f x = F cos 15 o f k = F cos 15 o µ k N = 89.8 Newtons = a x = f x m = 89.8 = 2.24 m/s2 40 b) The total work done by the F force is the force times the displacement times the cosine of the angle between the displacement and the force. Equivalently, it is the product of the component of the force in the direction of the displacement times the displacement. W F = F cos (15 o )s = 130 cos(15 o )5 = 628 J c) The (negative) work done by friction is the friction force times the displacement times the cosine of that angle which is always 180 o which accounts for the negative sign W f = fs = µ k Ns = (0.10)(358)(5) = 179 J d) The net work done on the mass goes into the kinetic energy of the mass. K = W F + W f = = 449 J e) The final speed of the mass is just obtained from the kinetic energy equation K = 1 2 mv2 = v = 2K 898 m = 40 = 4.74 m/s One can verify this result with the result obtained from the acceleration in part a) v 2 = v as = 0 + 2(2.24)(5) = = v = 4.74 m/s 11

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