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Physics 18 Fall 2009 Midterm 1 Solutions For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck! 1. A typical laboratory centrifuge rotates at 4000 RPM. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. (a) What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation? (b) For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.0 ms (10−3 seconds) encounter with a hard floor? ———————————————————————————————————— Solution (a) Since it’s spinning, the test tube is subjected to a centripetal acceleration acent = v 2 /r, where v is the rotational velocity and r is the rotational distance. Now, we know the rotational frequency, f = 4000 rotations/min ÷ 60 sec/min ≈ 67 rotations per second. This means that the period is T = 1/f = 1/67 = 0.015 seconds. Now, the period is the time it takes to go around one orbit in the centrifuge, and is the total distance around (2πr) divided by how fast it’s going around (v). So, 2πr , T = v which means that v = 2πr/T . Thus, the acceleration is acent v2 (2π)2 = = r. r T2 Plugging in the numbers gives acent = 1790g! (2π)2 .0152 × 0.1 = 1.75 × 104 m/s2 , which is (b) If we drop a test tube, it experiences a big acceleration when it hits the ground. Since it goes from some velocity vdrop to zero over a time ∆t, then the acceleration is a = vdrop /∆t. What’s the initial speed? Since it’s dropped from a height −h, under the constant acceleration due to gravity then vf2 = vi2 − 2g(−h) = 0 + 2gh, since the tube is assumed to have dropped √ from√rest. The final velocity it gets √ 2gh is just vf = vdrop = 2gh. Thus, a = ∆t = 2×9.8×1 = 4400 m/s2 , which is 10−3 about 450g. So, the centrifuge subjects the tube to accelerations even greater than hitting the ground! 1 2. An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a 6.0 kg mass from the rope. The boot is suspended by the ropes and does not touch the bed. (a) Determine the amount of tension in the rope by using Newton’s laws to analyze the hanging mass. (b) The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope? (c) What is the net traction force pulling the leg? Hint: If the pulleys are frictionless, which we will assume, the tension in the rope is constant from one end to the other. ———————————————————————————————————— Solution (a) Since the tension is everywhere the same in the wire, then the tension is due to the hanging mass, only. We can draw the free-body diagram for the system. From the sum of the forces we see that if M is the mass of the hanging weight, X Fy = T − M g = 0, where the net force is zero since the system isn’t accelerating. So, T = M g. Plugging in the numbers gives T = M g = 6.0 × 9.8 = 58.8 N. 2 T FG (b) Now, in order for the traction force to be entirely horizontal, we need the tension forces to cancel with the weight of the boot. We can draw the force diagram for the boot in the figure to the right. In the vertical direction, the sum of the forces is (taking the weight of the boot FG = mg, where m is the mass of the boot) X Fy = T sin θ − T sin 15◦ − mg = 0. y T Ftraction θ 15 x T FG (foot) Solving for the angle gives −1 θ = sin T sin 15◦ + mg T , or since T = M g, −1 θ = sin m sin 15 + = sin−1 (0.26 + .66) = 67.7◦ . M ◦ (c) Now that we know that the vertical components of the forces balance, then the traction force is all horizontal. From the diagram, we see that (taking θ = 67.7◦ ) X Fx = T cos 67.7◦ + T cos 15◦ − Ftraction = 0. So, Ftraction = T (cos 67.7◦ + cos 15◦ ) = 58.8 (1.35) = 79 N. This is a force equivalent to hanging a bit less than four pounds directly from your foot, so it’s not huge. 3 3. To assess a patient who is suspected of having heart disease, the physician must examine the cardiac function when (a) the patient is at rest with a heart beating at a normal pace, and (b) when under stress, for example, after exercise. To simulate the conditions of stress, the patient exercises by walking a treadmill to increase heart beat and sustain high levels of cardiac stress. Suppose a 50 kg patient exercises on the treadmill, angled at 30◦ from the ground, exerting a constant force of 500 N up the slope of the treadmill while running a constant velocity of 4 m/s along the treadmill for 5 minutes. The coefficient of friction of the treadmill is µs = 0.45. Determine the work done by the patient. ———————————————————————————————————— Solution The work, W , done by the patient is given by y W = Fnet d cos θ, F where Fnet is the net force on the patient, N and d is the total distance that she runs. Because the net force is along the direction that she’s moving (she’s accelerating up along the treadmill), the angle is zero, and so cos θ = 1. Thus, we just find W = Fnet d. P So, in order to proceed, we need to know the total force acting on the patient. We can draw the force diagram seen to f the right, orienting our coordinate system along the treadmill to make life eas30 ier. Here F~N is the normal force, F~P is the constant force exerted by the patient, F~G = mg is her weight, and F~f is the fricG tional force. Let’s look at the forces along the x and y directions. Doing so gives P ◦ P Fx = −Ff − mg sin 30 +◦ FP = Fnet Fy = FN − mg cos 30 = 0. x F F F Solving the y components gives FN = mg cos 30◦ . Recalling that the frictional force Ff = µs FN gives Ff = µs mg cos 30◦ . Plugging this back into the x component expression we have the net force, Fnet = FP − mg (sin 30◦ + µs cos 30◦ ). Now, we need the distance. Since the patient ran at a speed v for a time t, the “distance” she ran along the treadmill is d = vt = 4 m/s × 5 min × 60 seconds = 1200 meters. Thus, ◦ W = Fnet d = [FP − mg (sin30◦ + µs cos 30 )] vt √ 3 1 = 500 − 50 × 9.8 × 2 + 0.45 × 2 × 1200 4 = 7.69 × 10 J. 4 4. Suppose you eat a Snickers bar that contains 280 Calories. (a) If one Calorie is 4187 joules, then how much energy is contained in the Snickers bar? (b) Digestion requires some energy, and so only about 85% of the food energy is directly available for other uses (like living). How much energy is this? (c) Suppose you want to work off the candy by climbing stairs. Because the efficiency of the muscles is relatively low, only about 20% of the food energy is available for conversion to mechanical energy. If you have a mass of 80 kg, and a single step is 15 centimeters tall, how many steps would you be able to climb from the energy in part (b) available to you? (d) How many steps would you need to climb in order to work off all the Calories from the candy? ———————————————————————————————————— Solution (a) Since one Calorie is 4187 joules, then the Snickers bar contains 280 × 4187 = 1.2 × 106 J. (b) Since only 85% is available for other uses, then we have available 0.85×1.2×106 = 1.02 × 106 J to us. (c) Due to the low efficiency, only 20% of the energy can be used for mechanical work. So, we have only 0.2 × 1.02 × 106 = 2.04 × 105 J that we can use from the candy bar. The energy that we gain in going up a height h is just mgh. This potential energy that we gain came at the expense of the food energy that we’re burning. So, Efood = mgh, which says that we can climb a height h= 2.04 × 105 Efood = = 260 meters. mg 80 × 9.8 If one step is 15 centimeters tall, then it would take 260/0.15 = 1733 stairs to completely burn off the energy available from the candy! Fortunately, other life processes are burning the Calories, too! (d) The total energy in the candy was found in part (a). To completely work off all this energy would require climbing to a height of h= Efood 1.2 × 106 = = 1531 meters, mg 80 × 9.8 which is 10,200 steps!! You would have to climb very nearly a mile up! 5 Extra Credit Question!! The following is worth 10 extra credit points! where and σ are constants, and r is the distance between the molecules. The potential energy is plotted in the figure to the right. The vertical axis is in units of , while the horizontal axis is in units of σ. Energy The potential energy between a pair of neutral atoms or molecules is very well-approximated by the Lennard-Jones Potential, given by the expression σ 12 σ 6 , P E(r) = 4 − r r Molcular Bond Energy 8 7 6 5 4 3 2 1 0 0.75 -1 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 -2 -3 -4 Distance (a) Why does the potential energy approach zero as the distance gets bigger? (b) At what separation distance, in terms of σ and , is the potential energy zero? (c) At approximately what distance is the system in equilibrium? What is the potential energy at that distance? (Express your answers in terms of σ and .) (d) How much energy would you need to add to the system at equilibrium in order to break the molecular bonds holding it together? Why? (e) How much energy is released in the breaking of those molecular bonds? Why? Note - no calculation is needed to answer these problems! ———————————————————————————————————— Solution (a) As the two molecules get further apart, the attractive force between them gets weaker and weaker. When the are very far apart, they hardly interact at all they are basically free molecules. The potential energy of a free particle is zero, since potential energy depends on the interaction between multiple particles. (b) We can just read the value off from the graph. We see that the potential energy crosses the x axis when x = 1, which means that r = σ. We can see this from the equation, too: setting r = σ gives P E(σ) = 0. 6 3.25 (c) The system is in equilibrium when the net force on it is zero. Since the force is the slope of the potential energy graph, this happens when the slope is zero. The potential energy graph has zero slope when it’s at it’s minimum point. Checking the graph, we see that this happens right around x ≈ 1.15, or r ≈ 1.15σ. We could d (P E(r)) = 0, which gives r = 21/6 σ ≈ 1.12σ, check the exact answer by finding dr and so we were close on our guess. The energy at this distance can just be read off the graph, giving y = −3, or P E = −3. (d) In order to break the molecular bonds apart, we’d need to raise the energy to zero. At equilibrium the energy is P E = −3, and so we’d need to add +3 units of energy. (e) There is no energy released in breaking these molecular bonds - we had to add the energy to break these bonds. Energy is never released in the breaking of bonds! One can obtain energy by breaking a less stable bond, then forming a more stable bond. The more stable bond has a more negative potential energy (a deeper potential “well”). The difference in energy between the initial and final states is released to the environment. This is where the energy comes from in the ATP reactions, and not by releasing energy from the breaking of bonds! 7