College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
Bartleby Related Questions Icon

Related questions

Q: Sketch a phenomenon or an application that can be observed from the concept in the image
A:
Q: A special spring has relation between applied force F and the stretch x as: F = 60x² 18x, where x is…
A:
Q: See image. Thanks!! :)
A:
Q: A block of mass 5.5 kg is suspended from two vertical ropes attached to the ceiling. What is the…
A:
Q: A car turns from a road into a parking spot. The car's initial velocity was 4.0m/s [E45N]. The cars…
A:
Q: What would be considered true for the given two vectors attached in the image?a.) Result of scalar…
A: Step 1: Step 2: Step 3: Step 4:
Q: 11.8. The threshold frequency for a certain metal is 3.3x 104 Hz. If light of frequency 8.2 x 10* Hz…
A:
Q: FG) = BinCt) + sin(at) 14 is appbed to the Ryatem, Aoume egqual to twicó the velocuty, and find the…
A: Spring Constant, k = 10 lb/ft Mass of the object, m = 32lb = 32lb/32lb = 1 slug of mass…
Q: What is the step by step procedure to solve any mesh analysis problem
A: Mesh analysis problem
Q: Example 10 A 4.00 kg mass is attached to a vertical rod by the means of two 1.25 m strings which are…
A:
Q: 5. Examine the four following graphs of frictional force vs. normal force. II. III. H I. (N 4) IV.
A: Frictional force which acts opposite to the direction of motion. Frictional force depends on normal…
Q: 2. Prove o, , 01,, 2 0,5 h (i,) where o represents the uncertainty., Ix represents the x-axis…
A:
Q: 8. Given ä = 31 + 4j – 3k, b= –4ỉ – 2j + 6k and i = [10,t,–15] , find the value of t such that a) b…
A: The given vectors are: a→=3i^+4j^-3k^b→=-4i^-2j^+6k^c→=10i^+tj^-15k^
Q: 12. A steel ball is launched at 37 degrees above horizontal from the top of a platform that is 55…
A: Given, The initial speed of the ball, u = 20.0 ms height of the platform, h = 55 m
Q: Moses wanders through the wilderness, starting at the position 10 m [E]. (1) On the v-t graph,…
A:
Q: Two crates on inclined planes are connected by a wire and a frictionless pulley as shown on the…
A:
Q: An object is at a distance of 60 cm from a concave mirror of 20 cm focal length, find the image…
A: An object is at a distance of 60 cm from a concave mirror of 20 cm focal length, find the image…
Q: Coherent light of frequency 6.40 x 101¹4 Hz passes through two thin slits and falls on a screen 82.0…
A:
Q: A 2kg ball is rotated in a vertical direction. The ball is attached to a light string of length 3m…
A: Given data:  Mass of ball (m) = 2 kg String length= Radius of circular path (r) =3m Speed (V) = 12…
Q: The initial velocity and acceleration of four moving objects at a given instant in time are given in…
A:
Q: 2 Exercise: Go through the algebra to show that the above leads to
A:
Q: Draw a diagram explaining how a person underwater views a clock on a wall outside of the swimming
A: concept used: Refraction of light
Q: The virtual image produced by a convex mirror is one-third the size of the object. A)If the object…
A:
Q: The image produced by a concave mirror is located 33 cm in front of the mirror. The focal length of…
A:
Q: Calculate the image distance dị. di = ст Calculate the image height hị. hi cm
A: In first part, we need to select the correct ray diagram.  In first diagram, image height and object…
Q: The following diagram represents this situation. The mass on the table is 2.8 kg. The hanging mass…
A: solution is given as
Q: 1. Band theory, used to investigate the physics of semiconductors, involves an understanding of…
A:
Q: 3. The throe forcas shown in Figure 5 act on an object. The objact is in aquilibrium. Calculate the…
A:
Q: An object of height 10 cm, and an image of it was formed using a concave mirror with magnification M…
A: Given,object height ho =10 cm Magnification M = 5 Let hi be the height of image.  WWe know that…
Question
The second image has background info used to help solve the practice problem.
130 CHAPTER 5 Applications of Newton's Laws EXAMPLE 5.7 An air track in a physics lab Now we will work a problem involving two objects tied together by a string. Figure 5.9a shows a glider with mass m, that moves on a level, frictionless air track in a physics lab. It is connected by a string passing over a small frictionless pulley to a hanging weight with total mass m₂. The string is light and flexible, and it doesn't stretch. Find the acceleration of each object and the tension in the string. ►FIGURE 5.9 (a) Apparatus SOLUTION SET UP The two objects have different motions, so we need to draw a separate free-body diagram and coordinate system for each, as shown in Figure 5.9b and c. We are free to use different coordinate axes for the two objects; in this case, it's convenient to take the +x direction to the right for the glider and the +y direction downward for the hanging weight. Then the glider has only an x component of acceleration (i.e., ax = a₁, a₁ = 0), and the weight has only a y component (i.e., a2x = 0, a2y = a₂). There is no friction in the pulley, and we consider the string to be massless, so the tension T' in the string is the same throughout; it applies a force with magnitude 7 to each object. The weights are mig and m2g. SOLVE We apply Newton's second law, in component form, to each object in turn. For the glider on the track, Newton's second law gives ΣF₁ = T = max EF, = n + (-m₁g) = m₁a₁y = 0, and for the hanging weight, EF, = m28 + (-T) = m₂a2y- Now, because the string doesn't stretch, the two objects must move equal distances in equal times, and their speeds at any instant must be equal. When the speeds change, they change by equal amounts in a given time, so the accelerations of the two bodies must have the same magnitude a. With our choice of coordinate systems, if a₁x is positive, azy will also be positive, and if a₁, is negative, so will be azy. We can express this relationship as alx = a2y = a. (The directions of the two accelerations are different, of course.) The two Newton's second law equations are then T = mya, m28 + (-T) = m₂a. We'd like to get separate expressions for T and a in terms of the masses and g. m₂ m₁9 (b) Free-body diagram for glider DEXO (n F X m₂9 Y (c) Free-body diagram for weight An easy way to get an expression for a is to replace -T in the second equation by -m₁a and rearrange the result; when we do so, we obtain m2g + (-ma) = m₂a, m2g = (m₁ + m₂)a, and, finally, m1m2 T = 8 = m₁ + m₂ m₂ 8. a = m₁ + m₂ Video Tutor Solution Then, to get an expression for T, we substitute this expression for a back into the first equation. The result is m1 m₁ + m₂ m₂8. 200 REFLECT It's always a good idea to check general symbolic results such as this for particular cases where we can guess what the answer ought to be. For example, if the mass of the glider is zero (m₁ = 0), we expect that the hanging weight (mass m₂) will fall freely with accelera- tion g and there will be no tension in the string. When we substitute m₁ = 0 into the preceding expressions for T and a, they do give T = 0 and a = g, as expected. Also, if m₂ = 0, there is nothing to create ten- sion in the string or accelerate either mass. For this case, the equations give T = 0 and a = 0. Thus, in these two special cases, the results agree with our intuitive expectations. We also note that, in general, the tension T is not equal to the weight m2g of the hanging mass m2, but is less by a factor of m₁/(m₁ + m₂). If I were equal to m2g, then m₂ would be in equilibrium, but it isn't. Practice Problem: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F, in terms of the two masses in the problem. Answer m28. 18d) 2005 hile wary
expand button
Transcribed Image Text:130 CHAPTER 5 Applications of Newton's Laws EXAMPLE 5.7 An air track in a physics lab Now we will work a problem involving two objects tied together by a string. Figure 5.9a shows a glider with mass m, that moves on a level, frictionless air track in a physics lab. It is connected by a string passing over a small frictionless pulley to a hanging weight with total mass m₂. The string is light and flexible, and it doesn't stretch. Find the acceleration of each object and the tension in the string. ►FIGURE 5.9 (a) Apparatus SOLUTION SET UP The two objects have different motions, so we need to draw a separate free-body diagram and coordinate system for each, as shown in Figure 5.9b and c. We are free to use different coordinate axes for the two objects; in this case, it's convenient to take the +x direction to the right for the glider and the +y direction downward for the hanging weight. Then the glider has only an x component of acceleration (i.e., ax = a₁, a₁ = 0), and the weight has only a y component (i.e., a2x = 0, a2y = a₂). There is no friction in the pulley, and we consider the string to be massless, so the tension T' in the string is the same throughout; it applies a force with magnitude 7 to each object. The weights are mig and m2g. SOLVE We apply Newton's second law, in component form, to each object in turn. For the glider on the track, Newton's second law gives ΣF₁ = T = max EF, = n + (-m₁g) = m₁a₁y = 0, and for the hanging weight, EF, = m28 + (-T) = m₂a2y- Now, because the string doesn't stretch, the two objects must move equal distances in equal times, and their speeds at any instant must be equal. When the speeds change, they change by equal amounts in a given time, so the accelerations of the two bodies must have the same magnitude a. With our choice of coordinate systems, if a₁x is positive, azy will also be positive, and if a₁, is negative, so will be azy. We can express this relationship as alx = a2y = a. (The directions of the two accelerations are different, of course.) The two Newton's second law equations are then T = mya, m28 + (-T) = m₂a. We'd like to get separate expressions for T and a in terms of the masses and g. m₂ m₁9 (b) Free-body diagram for glider DEXO (n F X m₂9 Y (c) Free-body diagram for weight An easy way to get an expression for a is to replace -T in the second equation by -m₁a and rearrange the result; when we do so, we obtain m2g + (-ma) = m₂a, m2g = (m₁ + m₂)a, and, finally, m1m2 T = 8 = m₁ + m₂ m₂ 8. a = m₁ + m₂ Video Tutor Solution Then, to get an expression for T, we substitute this expression for a back into the first equation. The result is m1 m₁ + m₂ m₂8. 200 REFLECT It's always a good idea to check general symbolic results such as this for particular cases where we can guess what the answer ought to be. For example, if the mass of the glider is zero (m₁ = 0), we expect that the hanging weight (mass m₂) will fall freely with accelera- tion g and there will be no tension in the string. When we substitute m₁ = 0 into the preceding expressions for T and a, they do give T = 0 and a = g, as expected. Also, if m₂ = 0, there is nothing to create ten- sion in the string or accelerate either mass. For this case, the equations give T = 0 and a = 0. Thus, in these two special cases, the results agree with our intuitive expectations. We also note that, in general, the tension T is not equal to the weight m2g of the hanging mass m2, but is less by a factor of m₁/(m₁ + m₂). If I were equal to m2g, then m₂ would be in equilibrium, but it isn't. Practice Problem: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F, in terms of the two masses in the problem. Answer m28. 18d) 2005 hile wary
Page 129 Practice Problem 5.7: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F in terms of the two a a masses in the problem. Answer: m₂g.
expand button
Transcribed Image Text:Page 129 Practice Problem 5.7: To set up for another run, you pull the glider back up the track at a constant speed with a horizontal force F. Determine the magnitude of F in terms of the two a a masses in the problem. Answer: m₂g.
Expert Solution
Check Mark
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
bartleby
This is a popular solution
See solutionCheck out a sample Q&A here
Step 1
VIEW
Step 2
VIEW
Step 3
VIEW
Step 4
VIEW
bartleby

Trending nowThis is a popular solution!

bartleby

Step by stepSolved in 4 steps with 2 images

See solution
Check out a sample Q&A here
Blurred answer
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
SEE MORE TEXTBOOKS