Gravitational Potential Energy - General Physics - Past Paper, Exams of Physics

Download Gravitational Potential Energy - General Physics - Past Paper and more Exams Physics in PDF only on Docsity! Physics 2101, Final Exam, Spring 2006 May 11, 2006 Name : Section: (Circle one) 1 (Rupnik, MWF 7:40am) 2 (Rupnik, MWF 9:40am) 3 (Rupnik, MWF 11:40am) 4 (Kirk, MWF 2:40pm) 5 (Kirk, TTh 10:40am) 6 (González, TTh 1:40pm) • Please be sure to write your name and circle your section above. • For the problems, you must show all your work. Explain your thinking clearly. Lonely right answers will not receive full credit, lonely wrong answers will receive no credit. • For the questions, no work needs to be shown (there is no partial credit). • Please carry units through your calculations when needed, lack of units will result in a loss of points. • You may use scientific or graphing calculators, but you must derive your answer and explain your work. • Feel free to detach, use and keep the formula sheet. No other reference material is allowed during the exam. • GOOD LUCK! 1 Problem 1 (11 points) A 180 g block is dropped from rest onto a relaxed vertical spring that has a spring constant of k = 300 N/m. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping. ppppp pppp ppppp pppp ppppp ? 1 (a) (3 pts) Calculate the work done on the block by the gravi- tational force from the time the block falls on the spring until the time the block momentarily stops. The work done by the gravitational force Fg = mg force is Wg =∫ ∆x 0 F · ds = mg∆x. Also, the work done by the gravitational force is minus the change in gravitational potential energy. The change in gravitational energy is negative, since the mass moves down a distance ∆x: ∆Ug = −mg∆x = −0.18kg × 9.8m/s2 × 0.12m = −0.21J, and thus Wg = +0.21J . (b) (4 pts) Calculate work done on the block by the spring force from the time the block falls on the spring until the time the block momentarily stops. The work done by the spring force Fs = −kx force is Ws = ∫ |Deltax 0 F· ds = −(1/2)k(∆x)2. The work done by the spring is minus the change in spring poten- tial energy. Initially, the spring is not compressed or stretchesd and the potential energy is zero; when the block momentarily stops, the spring potential energy is (1/2)k(∆x)2. The change in potential energy is pos- itive: ∆Us = (1/2)k(∆x)2, and the work done by the spring (which is negative) is Ws = −(1/2)k(∆x)2 = −0.5× 300N/m× (0.12m)2 = −2.16J. (c) (4 pts) Calculate the height relative to the fully compressed spring from which the block was initially dropped. The kinetic energy of the block is zero at the time where it momentarily stops, and it is also zero at the time when it is released from rest at a distance H from the lowest point. Thus, the change in potential energy is zero: 0 = ∆K + ∆U = 0 + ∆U = ∆Ug + ∆Us = −mgy + (1/2)k(∆x)2 = 0 y = k(∆x)2 2mg = 300N/m× (0.12m)2 2× 0.18kg × 9.8m/s2 = 1.22m 2 Problem 3 (11 points) A spherical shell of mass 4.0 kg rolls without slipping down an incline with an inclination angle of θ = 35◦, starting from rest. At the bottom of the incline the center of mass of the sphere has a translational speed of vcom,f = 6.0 m/s. The rotational inertia for a spherical shell about any diameter is Icom = 2 3 MR2. &% '$r &% '$r         *  dmax θ vcom,f             (a) (5 pts) Calculate the kinetic energy of the shell at the bottom of the incline. The kinetic energy of a rolling body has a translational part (1/2)Mv2com and a rotational part (1/2)Icomω 2. The angular velocity is related to the velocity of the center of ass, ω = vcom/R, so K = 1 2 Mv2com + 1 2 Icomω 2 = 1 2 Mv2com + 1 2 2 3 MR2 (vcom R )2 = 5 6 Mv2com Kf = 5 6 × 4.0kg × (6.0m/s)2 = 120J (b) (6 pts) How far, dmax, along the incline, did the shell travel? The mechanical energy is conserved in the rolling of the shell down the incline, so the (negative) change in gravitational potential energy plus the (positive) change in kinetic energy is equal to zero. The change in gravitational potential energy is ∆Ug = −Mgy = −Mgdmax sin θ, and the change in kinetic energy is ∆K = Kf − 0 = Kf , so ∆K + ∆U = Kf −Mgdmax sin θ = 0 ⇒ dmax = Kf Mg sin θ = 120J 4.0kg × 9.8m/s2 × sin(35o) = 5.3m 5 Problem 4 (11 points) A sphere of mass M is held stationary on the surface of a frictionless inclined plane by means of a cable, as illustrated in the figure below. The base angle of the inclined plane is θ, and the angle between cable and the surface of the inclined plane is φ. p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p θ φ r Cable


R R (a) (3 pts) Draw a free body diagram for the sphere, showing all external forces. Label each force. (You may draw directly on the diagram above). (b) (4 pts) Calculate the tension in the cable, expressing your answer in terms of M, g, θ, φ and numerical constants, as necessary. The sum of the forces along the incline add up to zero in equilibrium: 0 = F‖ = +T cos φ−Mg sin θ = 0 ⇒ T = Mg sin θ cos φ (c) (4 pts) Calculate the normal force exerted on the sphere by the surface of the inclined plane. Express your answer in terms of M, g, θ, φ and numerical constants, as necessary. The sum of the forces perpendicular to the incline add up to zero in equilibrium: 0 = F⊥ = +FN − Tsinφ−Mg cos θ ⇒ FN = T sinφ + Mg cos θ = Mg sin θ cos φ sinφ + Mg cos θ = Mg cos(φ− θ) cos φ 6 Question 2 (5 points) The figure shows a square of edge length L formed by four spheres of equal mass m. A fifth mass, mA, is at the center of the square. (a) (1pt) What is the direction of the gravitational force on the mass mA at the center of the square produced by the pair of masses m1,m4? m1 and m4 are at the same distance from mA, so the forces have the same magnitude; since the forces will have opposite directions, they will cancel each other. ı̂ −ı̂ ̂ −̂ the magnitude is zero (b) (2 pt) What is the direction of the gravitational force on the mass mA at the center of the square produced by the pair of masses m1,m3? m1 and m3 are at the same distance from mA, so the forces have the same magnitude; the force F1 will point towards the upper left corner, and the force F3 towards the lower left corner. The vertical components will cancel, and the resulting horizontal component will point in the negative x direction. ı̂ −ı̂ ̂ −̂ the magnitude is zero (c) (2 pts) What is the magnitude of the work done by an external agent to bring mass mA to the center of the square, with all the other four masses in place? The magnitude work done by an external agent to bring the mass mA to the center of the square is equal to the mass mA time the magnitude of the potential energy at the center of the square, produced by the other masses. The potential of any one of the masses at the center is Ui = −Gm/r = −Gm/(L/ √ 2) = − √ 2Gm/L. The magnitude of the total potential energy at the center is |U | = 4 √ 2Gm/L and the work done to bring mass mA is W = 4 √ 2GmmA/L. 0 2GmmAL √ 2GmmAL 4 √ 2GmmAL 4G mmA L 7 Problem 5 (11 pts) A block of copper whose mass mc is 80 g is heated in a laboratory oven to a temperature of Ti, c = 300 ◦C and put into the thermally isolated container together with a piece of ice of mass mice = 250 g and temperature Ti, ice = −10◦C. Eventually the two substances come to thermal equilibrium at final temperature Tf . The specific heat of copper is 386 J/kg·K. Copper Ice Insulation

Ti, c Ti, ice (a) (3 pts) How much energy would be released if the copper block cools down to 0◦C? Qc = mccc∆Tc = 0.08kg × 386J/kgoC× (0oC− 300oC) = −9, 264J The negative sign indicates that the copper releases heat when cooling down. (b) (4 pts) How much energy would the ice absorb if it would warm up and melt at 0◦C? Qice = micecice∆Tice + miceLmelt Qice = 0.25kg × 2220J/kgoC× 10oC + 0.25kg × 3.33× 105J/kg = 5, 550J + 83, 250J = 88, 800J (c) (4 pts) What is Tf , the final equilibrium temperature? Explain. Since the heat released by copper cooling down to oC, 9,264 J, is enough to warm up the ice to melting temperature (which needs 5,550 J), but not to melt it all (a process that needs 83,250 J), the final temperature will be zero degrees Celsius. The mass of ice melted will be mice,melted =(9,264 J - 5,550 J)/3.33× 105 J/kg = 1.1× 10−2kg = 11 g, a very small fraction of the initial mass. 10 Problem 6 (11 points) A bubble of air with initial volume V0 is located at the bottom of a lake of depth D. The temperature at the bottom of the lake is Tb, and the temperature at the surface of the lake is Ts, with both temperatures measured in Kelvin. The bubble rises to the surface. During its ascent, the temperature of the bubble is equal to the temperature of the water that surrounds it. Treat the air in the bubble as an ideal gas. (a) (3 pts) Find the pressure at the bottom of the lake. Express your answers in terms of atmospheric pressure pa, density of water ρ, acceleration of gravity g and D. From the equation or pressure in fluids, we know that pb = pa + ρgD (b) (3 pts) Find the number of molecules in the bubble, expressing your answer in terms of atmospheric pressure pa, density of water ρ, acceleration of gravity g, D, V0, Tb, Boltzmann constant k and numerical factors, as necessary. At the bottom of the lake, the pressure on the bubble is pb, the volume is V0, and the temperature is Tb. We can use the law for ideal gases: pbV0 = NkTb ⇒ N = pbV0 kTb = (pa + ρgD) V0 kTb (c) (5 pts) Calculate the volume Vf of the bubble just as it reaches the surface of the lake, ex- pressing your answer in terms of atmospheric pressure pa, density of water ρ, acceleration of gravity g, D, V0, Tb, Ts, Boltzmann constant k and numerical factors, as necessary. We use again the law for ideal gases, this time at the surface of the lake: paVf = NkTs ⇒ Vf = NkTs pa = (pa + ρgD) V0 kTb NkTs pa Vf = ( 1 + ρgD pa ) Ts Tb V0 11 Question 5 (5 points) In the pV diagram shown below, the gas does 10 J of work when taken along the isotherm ab and 6 J when taken along the adiabat bc. (a) (3 pts) What is the change in the internal energy of the gas when it is taken along the straight path from a to c? If we consider a reversible cycle abca, we know the net change in internal energy is zero, so 0 = ∆Eab + ∆Ebc + ∆Eca, and thus ∆Eac = −∆Eca = ∆Eab + ∆Ebc. Since the process ab is along an isotherm, ∆Eab = 0. Since the process bc is adiabatic, we know Qbc = 0 = ∆Ebc + Wbc and then ∆Ebc = −Wbc = −6 J. (a) −4 J (b) +4 J (c) +6 J (d) −6 J (e) 16 J (f) none of the above (b) (2 pts) The work delivered by the gas to the environment when taken along the straight line from a to c is (a) exactly 16 J (b) greater than 16 J (c) less than 16 J (d) negative in sign (meaning work is performed on the gas, not by the gas) (e) none of the above The work is the are under the curve in the process; thus the work done under the curve ac is smaller than the work done under the curves abc, which is equal to 10J+6J=16J. 12