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Test 3 Practice Test-sols.pdf

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Department
Physics
Course
PHY 303K
Professor
Rodenborn
Semester
Spring

Description
Version 001 – Test 3 Practice Test – rodenborn – (57770) 1 This print-out should have 20 questions. 002 10.0 points Multiple-choice questions may continue on Ablkidonhepofampas the next column or page – find all choices shown in the figure below. The block is re- before answering. leased from rest on the frictionless incline. When the block reaches the flat portion of Acceleration of an Oscillator the table it begins to feel a frictional force 001 10.0 points with the table characterized by a coe▯cient Amilamiaiasmied of kinetic frictiok µ . 0.85 m and period 2.3sec. What is the maximum acceleration? h 1. 1.00959 m/s2 µk Which of the following statements about 2. 3.17171 m/s2 the energy and work of the system could NOT be correct? 2 3. 6.34341 m/s correct 1. The sum of the potential energy and the 4. 0.369565 m/s2 kinetic energy of the block at any point along 2 the entire path of travel is conserved. correct 5. 14.5898 m/s 2. Along the incline, gravity does work on 6. 0.160681 m/s2 the block causing it to speed up. Explanation: 3. The block’s gravitational potential en- ergy loss is equal to kinetic energy gain as it Let : A =0 .85 m and descends. T =2 .3sec . 4. Along the incline, the potential energy of For a simple harmonic oscillator, the dis- the system decreases from its initial value. placement is ▯ ▯ 5. On the table, friction does work on the 2▯ block causing it to slow down. x = Acos T t + ▯ , 6. The normal force from the incline does no so the acceleration is work on the block. 2 ▯ ▯2 ▯ ▯ d x 2▯ 2▯ a = dt2 = ▯A T cos T t + ▯ . 7. When the block reaches the bottom of the incline, its kinetic energy reaches a maxi- Since ▯1 < cos▯< 1, the maximum acceler- mum. ation is 2 Explanation: A = 4▯ A max T2 The total mechanical energy of the block 2 (i.e.hemo fhepealngyadte = 4▯ (0.85 m) kinetic energy) is not conserved. Mechani- (2.3sec) cal energy is converted from potential energy = 6.34341 m/s . and eventually is dissipated by work done by friction. Concepts of Energy Hewitt CP9 15 E13 Version 001 – Test 3 Practice Test – rodenborn – (57770) 2 ▯ 003 10.0 points =2 .2 dna Which statement is wrong? 1. Adding the same amount of heat to two ▯ ▯ di▯erent objects will produce the same in- ▯T b T ▯bT =f8 .9 C ▯ 7.2 C crease in temperature. correct ▯ =91 .7 C, so 2. When the same amount of heat produces di▯erent changes in temperature in two sub- Q gained,waterQ lost,brass stances of the same mass, we say that they have di▯erent specific heat capacities. m w p,w ▯T w m c b p,b▯T b 3. Each substance has its own characteristic specific heat capacity. c = m w p,w ▯T w p,b mb▯T b 4. Temperature measures the average ki- ▯ netic energy of random motion, but not other (2.77 kg)(4186 J/kg · C) = (0.52 kg)(91.7 C) kinds of energy. ▯ ▯ (2.2 C) 5. Di▯erent substances have di▯erent ther- ▯ = 534.969 J/kg · C . mal properties due to di▯erences in the way energy is stored internally in the substances. Explanation: Power in a Car The same amount of heat does not necessar- 005 10.0 points ily produce the same increase in temperature When an automobile moves with constant in di▯erent objects. velocity, the power developed in a certain en- gine is 113 hp. Holt SF 10C 05 What total frictional force acts on the car 004 10.0 points at 35.4mi /h? 1 hp = 746 W and 1 m/s = Brass is an alloy made from copper and zinc. ▯ 2.2374 mi/h. A0 .52 kg brass sample at 98.9 Cirped into 2.77 kg of water at 5.0 C. Correct answer: 5327.92 N. If the equilibrium temperature is 7.2 C, Explanation: what is the specific heat capacity of brass? The specific heat of water is 4186 J/kg · C. ▯ Let : P =113hp and Correct answer: 534.969 J/kg · C. v =35 .4mi /h. Explanation: Let : m b0 .52 kg , T =98 .9 C, W Fd b P = = = Fv m w2 .77 kg , t t ▯ P 113 hp Tw=5 .0 C, F = v = 35.4mi /h = 5327.92 N . Tf=7 .2 C, and ▯ cp,w=4186J /kg · C. Falling Raindrops 006 10.0 points ▯T = T ▯ T =7 .2 C ▯ 5 C ▯ The force of air resistance on a raindrop is w f w Version 001 – Test 3 Practice Test – rodenborn – (57770) 3 5.3 ▯ 10▯5 Nw ntali ril path is n times as long, the work done will be n times as great. Clearly, the work by such velocity of 4.9m /s. 2 The acceleration of gravity is 9.8m /s . afewlotlysezooerald What is the mass of the raindrop? loop. Conservative forces do work that depends ▯6 Correct answer: 5.40816 ▯ 10 kg. only on the end-points of the integral (and is Explanation: completely independent of the path). When the raindrop is falling with constant velocity the sum of forces acting on it is zero. MandI: Strange Potential 008 10.0 points Fnet =0= mg ▯ F air Consider a uniform conservative force on some strange distant planet that is similar to the mg = F air F air gravitational force but pointing at an angle ▯ m = g to the vertical direction, 5.3 ▯ 10▯5 N = 9.8m /s2 F = ma sin▯ ˆı ▯ ma cos▯ ˆ▯. = 5.40816 ▯ 10 ▯6 kg y P(x,y) MandI: Conservative Force Concept 007 10.0 points Frictional forces notoriously fail to satisfy the x criterion for a conservative force; iet-,t O(0,0) tal work done by frictional forces does not necessarily equal zero when integrated over a ▯ loop, ▯ f · dr ▯=0 . f If you consider the planet that creates this force and an object of mass m as one system, Why don’t frictional forces satisfy the “con- what is the change in potential energy moving servative” criterion? the object from O(0,0) to P(x,y)? 1. Frictional forces are always proportional 1. ▯U =+ max sin▯ ▯ may tan▯ to the force of gravity. 2. None of these 2. ▯U =+ max tan▯ ▯ may tan▯ 3. ▯U =+ max cos▯ ▯ may sin▯ 3. Frictional forces can only do zero or neg- ative work. correct 4. ▯U = ▯max tan▯ + may tan▯ 4. Frictional forces always act to decrease 5. ▯U = ▯max sin▯ + may cos▯ correct the velocity v of an object. 6. ▯U = ▯max cos▯ + may tan▯ 5. Frictional forces are never constant, but are di▯erent at every point on an object’s 7. ▯U =+ max sin▯ ▯ may cos▯ path. Explanation: 8. ▯U = ▯max cos▯ + may sin▯ Frictional forces are always opposite to the displacement if there is one. Therefore if the 9. ▯U =+ max cos▯ ▯ may tan▯ Version 001 – Test 3 Practice Test – rodenborn – (57770) 4 10. ▯U = ▯max sin▯ + may tan▯ 4. PointS. correct Explanation: 5. The particle remains stationary at point A. Let : Fx=+ ma sin▯, Fy= ▯ma cos▯, Explanation: Since the total energy is constant, the max- Since the force is conservative, the workimum kinetic energy will occur when the po- required to move the object from O to P is tential energy is a minimum. Thus point S is independent of the path taken. Consider a the correct answer. straight horizontal path from O to (x,0), and then a straight vertical path from (x,0) to P. MandI: Potential Energy Graph The work is 010 10.0 points The following graph represents a hypothetical W =+ F xx+ F yy so potential energy curve for a particle of mass =+ max sin▯ ▯ ma y cos▯. m. U(r) The change is potential is the negative of the work done by the force or 3U 0 ▯U = ▯W 2U 0 = ▯max sin▯ + ma y cos▯. U 0 r O r Complicated Potential 02 0 2r0 009 10.0 points If the particle is released from rest at posi- Use the potential orpoitiss.op tion 0 ,whatwillbeitsspeed▯▯▯ at position shown below to answer the following questio2r0? Acsmfoi A and ▯ moves in the potential U(x).Suppose the 4U 1. ▯▯▯ = 0 correct mechanical energy of the system is conserved. m ▯ A U(x) U 0 T 2. ▯▯▯ = 2m x ▯ U 0 3. ▯▯▯ = ▯ 6m 2U 0 Z 4. ▯▯▯ = S m ▯ 5. ▯▯▯ = 8U 0 At which position(s) will the kinetic energy m ▯ of the particle have its maximum value? U 0 6. ▯▯▯ = 8m 1. PointT. ▯ U 0 7. ▯▯▯ = 2. PointZ. ▯ 4m U 0 8. ▯▯▯ = 3. PointsZ and S. m Version 001 – Test 3 Practice Test – rodenborn – (57770) 5 =2(6 .67259 ▯ 10▯11Nm /kg )2 ▯ 9. ▯▯▯ = 6U 0 (6.77 ▯ 104 kg) m ▯ 8 2 (2.99792 ▯ 10 m/s) Explanation: = 0.0100524 m . The total energy of the particle is con- served. So the change of the potential en- ergy is converted into the kinetic energy of Technically speaking, in a region where the particle, which gives gravity is extremely intense, Newton’s me- 1 2 chanics cannot be used. Rather, one needs to mv =3 U ▯ 0 0 apply the “general theory of relativity” devel- 2 ▯ 4U 0 oped by Albert Einstein. v = . It turns out that the theory of relativity m gives the same expression for this limiting MandI: Black Hole Radius radius, referred to as the “Schwarzschild ra- dius”. 011 10.0 points Ablackholeisanobjectsoheavythatneither matter nor even light can escape the influence Potential Graph 03 012 10.0 points of its gravitational field. Since no light can The graph is a potential energy diagram for escape from it, it appears black. Suppose a body with the same mass as the Earth 6 .77 ▯ animes satof 24 position x. 10 kg is packed into a small uniform sphere. Based on Newtonian mechanics, determine the radius 0 at which this body would appear to be a black hole. B Hint: The escape speed at the sur- face of the body must be equal to the x speed of light c =2 .99792 ▯ 10 m/s. The U universal gravitational constant is G = A 6.67259 ▯ 10▯11Nm /kg .2 x 0 x 2x 3x 4x 5x Correct answer: 0.0100524 m.
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