PHYS 111 University Physics I Problem Set 5 Winter Quarter, 007 SOLUTIOS Instructor Barry L. Zink Assistant Professor Office: Physics 404 Office Hours: (303) 871-305 M&F 11am-1pm barry.zink@du.edu Th -4pm http://portfolio.du.edu/bzink (or by appointment) Penny-Pedestrian Redux In an earlier Quiz, we calculated the velocity of a penny dropped from the 150 foot tall Empire State Building when it hits the sidewalk below. ow we consider the more realistic case by considering air resistance, or drag, on the penny. Assume you have a.50 gram penny, that tumbles as it falls. Assume a Coefficient of Drag, C 0.75 (0 points total). 1. Estimate the effective cross-sectional area of the tumbling penny using a simple average of minimum and maximum cross-sections. ASWER: The minimum cross-sectional area of the penny is edge-on and is simply A min d penny t penny, where D penny is the diameter and t penny is the thickness. The maximum crosssection is the circular cross-section when the penny is horizontal to the ground, A max π(d penny ) The average of these is A eff (A min + A max )/. Many methods to measure or estimate the diameter and thickness of a penny are acceptable. My methods gave A min 3.14 10 5 m, and A max.85 10 4 m, so that A eff 1.58 10 4 m.. What is the terminal velocity of the penny? ASWER: mg (.5 10 v t 3 kg)9.8 m/s CρA eff 0.75(1. kg/m 3.85 10 4 18.5 m/s m 3. How far does the penny fall before reaching terminal velocity? Here we want to see how far a truly free-falling penny would fall before reaching 18.5 m/s. Kinematics tells us: v gh h v g (18.5 m/s) 9.8 m/s 17.4 m 4. Are you afraid of the penny-wielding tourists atop the ESB? o, because getting hit by a penny falling from a great height is essentially no different than being hit by one dropped from 0 meters. It would probably hurt a bit, but not do any serious damage... Halliday, Resnick, and Walker (HRW) See following pages... 1
Chapter 5 1. (a) The slope of each graph gives the corresponding component of acceleration. Thus, we find a x 3.00 m/s and a y 5.00 m/s. The magnitude of the acceleration vector is therefore a (3.00 m/s ) + ( 5.00 m/s ) 5.83 m/s, and the force is obtained from this by multiplying with the mass (m.00 kg). The result is F ma 11.7. (b) The direction of the force is the same as that of the acceleration: θ tan 1 [( 5.00 m/s )/(3.00 m/s )] 59.0. 3. We resolve this horizontal force into appropriate components. (a) ewton s second law applied to the x-axis produces F cosθ mg sin θ ma. For a 0, this yields F 566. (b) Applying ewton s second law to the y axis (where there is no acceleration), we have F F sin θ mg cos θ 0 which yields the normal force F 1.13 10 3. 51. We apply ewton s second law first to the three blocks as a single system and then to the individual blocks. The +x direction is to the right in Fig. 5-49. (a) With m sys m 1 + m + m 3 67.0 kg, we apply Eq. 5- to the x motion of the system in which case, there is only one force T + T ɵ i. Therefore, 3 3 T3 msysa 65.0 (67.0 kg) a which yields a 0.970 m/s for the system (and for each of the blocks individually). 175
176 PS5 (b) Applying Eq. 5- to block 1, we find ( )( ) T1 m1a 1.0kg 0.970m/s 11.6. (c) In order to find T, we can either analyze the forces on block 3 or we can treat blocks 1 and as a system and examine its forces. We choose the latter. ( ) ( )( ) T m1 + m a 1.0 kg + 4.0 kg 0.970 m/s 34.9. 55. The free-body diagrams for m1 and m are shown in the figures below. The only forces on the blocks are the upward tension T and the downward gravitational forces F1 m1 g and F mg. Applying ewton s second law, we obtain: T m g m a 1 1 m g T m a which can be solved to yield a m m g 1 m + m1 Substituting the result back, we have T m m g 1 m1 + m (a) With m 1 1.3 kg and m.8 kg, the acceleration becomes a.80 kg + 1.30 kg (b) Similarly, the tension in the cord is.80 kg 1.30 kg (9.80 m/s ) 3.59 m/s. (1.30 kg)(.80 kg) (9.80 m/s ) 17.4. T 1.30 kg +.80 kg
177 73. Although the full specification of Fnet ma in this situation involves both x and y axes, only the x-application is needed to find what this particular problem asks for. We note that a y 0 so that there is no ambiguity denoting a x simply as a. We choose +x to the right and +y up. We also note that the x component of the rope s tension (acting on the crate) is F x Fcosθ (450 ) cos 38 355, and the resistive force (pointing in the x direction) has magnitude f 15. (a) ewton s second law leads to 355 15 Fx f ma a 310 kg 0.74m/s. (b) In this case, we use Eq. 5-1 to find the mass: m W/g 31.6 kg. ow, ewton s second law leads to 355 15 7.3 m/s. Tx f ma a 31.6 kg 74. Since the velocity of the particle does not change, it undergoes no acceleration and must therefore be subject to zero net force. Therefore, Thus, the third force F 3 is given by 3 1 Fnet F1 + F + F3 0. F F F i + 3j k 5i + 8j k 3i 11j + 4k. ( ) ( ) ( ) The specific value of the velocity is not used in the computation.
178 PS5 Chapter 6. To maintain the stone s motion, a horizontal force (in the +x direction) is needed that cancels the retarding effect due to kinetic friction. Applying ewton s second to the x and y axes, we obtain F fk ma F mg 0 respectively. The second equation yields the normal force F mg, so that (using Eq. 6- ) the kinetic friction becomes f k µ k mg. Thus, the first equation becomes F µ mg ma 0 k where we have set a 0 to be consistent with the idea that the horizontal velocity of the stone should remain constant. With m 0 kg and µ k 0.80, we find F 1.6 10. 1. (a) Using the result obtained in Sample Problem 6-, the maximum angle for which static friction applies is θ µ s 1 1 max tan tan 0.63 3. This is greater than the dip angle in the problem, so the block does not slide. (b) We analyze forces in a manner similar to that shown in Sample Problem 6-3, but with the addition of a downhill force F. F + mg sin θ fs, max ma 0 F mg cos θ 0. Along with Eq. 6-1 (f s, max µ s F ) we have enough information to solve for F. With θ 4 and m 1.8 10 7 kg, we find ( ) 7 F mg µ cosθ sinθ 3.0 10. s