27. Consider the block-spring-surface system in part (B) of Example 8.6. (a) Using an energy approach, find the position x of the block at which its speed is a maximum. Answer (b) In the What If? section of this example, we explored the effects of an increased friction force of 10.0 N. At what position of the block does its maximum speed occur in this situation? Answer Example 8.6 (B) Calculate the speed of the block as it passes through the equilibrium position if a constant friction force of 4.0 N retards its motion from the moment it is released. Solution Conceptualize The correct answer must be less than that found in part (A) because the friction force retards the motion. Categorize We identify the system as the block and the surface, a nonisolated system for energy because of the work done by the spring. There is a nonconservative force acting within the system: the friction between the block and the surface. Analyze Write the appropriate reduction of Equation 8.2 and substitute for the energy changes: AK + AEint = W, → (mv²³ — 0) + frd = Ws Solve for vţ: 2 V₁ = -(Ws - fkd) m Substitute for the work done by the spring, found in part (A): 2 1 Uf = kxmax - fkd) m2 Substitute numerical values: 2 vf = (1 000 N/m) (0.020 m)² - (4.0 N) (0.020 m)] = 0.39 m/s 1.6 kg 2 Finalize As expected, this value is less than the 0.50 m/s found in part (A). What If? What if the friction force were increased to 10.0 N? What is the block's speed at x = 0? Answer In this case, the value of fkd as the block moves to x = 0 is fkd (10.0 N) (0.020 m) = 0.20 J which is equal in magnitude to the kinetic energy at x = 0 for the frictionless case. (Verify it!). Therefore, all the kinetic energy has been transformed to internal energy by friction when the block arrives at x = 0, and its speed at this point is v = 0. In this situation as well as that in part (B), the speed of the block reaches a maximum at some position other than x = 0. Problem 27 asks you to locate these positions.
27. Consider the block-spring-surface system in part (B) of Example 8.6. (a) Using an energy approach, find the position x of the block at which its speed is a maximum. Answer (b) In the What If? section of this example, we explored the effects of an increased friction force of 10.0 N. At what position of the block does its maximum speed occur in this situation? Answer Example 8.6 (B) Calculate the speed of the block as it passes through the equilibrium position if a constant friction force of 4.0 N retards its motion from the moment it is released. Solution Conceptualize The correct answer must be less than that found in part (A) because the friction force retards the motion. Categorize We identify the system as the block and the surface, a nonisolated system for energy because of the work done by the spring. There is a nonconservative force acting within the system: the friction between the block and the surface. Analyze Write the appropriate reduction of Equation 8.2 and substitute for the energy changes: AK + AEint = W, → (mv²³ — 0) + frd = Ws Solve for vţ: 2 V₁ = -(Ws - fkd) m Substitute for the work done by the spring, found in part (A): 2 1 Uf = kxmax - fkd) m2 Substitute numerical values: 2 vf = (1 000 N/m) (0.020 m)² - (4.0 N) (0.020 m)] = 0.39 m/s 1.6 kg 2 Finalize As expected, this value is less than the 0.50 m/s found in part (A). What If? What if the friction force were increased to 10.0 N? What is the block's speed at x = 0? Answer In this case, the value of fkd as the block moves to x = 0 is fkd (10.0 N) (0.020 m) = 0.20 J which is equal in magnitude to the kinetic energy at x = 0 for the frictionless case. (Verify it!). Therefore, all the kinetic energy has been transformed to internal energy by friction when the block arrives at x = 0, and its speed at this point is v = 0. In this situation as well as that in part (B), the speed of the block reaches a maximum at some position other than x = 0. Problem 27 asks you to locate these positions.
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter8: Potential Energy And Conservation Of Energy
Section: Chapter Questions
Problem 72AP: A body of mass m and negligible size starts from rest and slides down the surface of a frictionless...
Related questions
Question
please help with 27 part A.
Transcribed Image Text:27. Consider the block-spring-surface system in part (B) of Example 8.6.
(a) Using an energy approach, find the position x of the block at which its speed is a maximum.
Answer
(b) In the What If? section of this example, we explored the effects of an increased friction force
of 10.0 N. At what position of the block does its maximum speed occur in this situation?
Answer
![Example 8.6 (B)
Calculate the speed of the block as it passes through the equilibrium position if a constant friction
force of 4.0 N retards its motion from the moment it is released.
Solution
Conceptualize The correct answer must be less than that found in part (A) because the friction
force retards the motion.
Categorize We identify the system as the block and the surface, a nonisolated system for energy
because of the work done by the spring. There is a nonconservative force acting within the system:
the friction between the block and the surface.
Analyze
Write the appropriate reduction of Equation
8.2 and substitute for the energy changes:
AK + AEint = W, → (mv²³ — 0) + frd = Ws
Solve for vţ:
2
V₁ =
-(Ws - fkd)
m
Substitute for the work done by the spring,
found in part (A):
2
1
Uf =
kxmax - fkd)
m2
Substitute numerical values:
2
vf =
(1 000 N/m) (0.020 m)² - (4.0 N) (0.020 m)] = 0.39 m/s
1.6 kg 2
Finalize As expected, this value is less than the 0.50 m/s found in part (A).
What If?
What if the friction force were increased to 10.0 N? What is the block's speed at x = 0?
Answer In this case, the value of fkd as the block moves to x = 0 is
fkd (10.0 N) (0.020 m) = 0.20 J
which is equal in magnitude to the kinetic energy at x = 0 for the frictionless case. (Verify it!).
Therefore, all the kinetic energy has been transformed to internal energy by friction when the block
arrives at x = 0, and its speed at this point is v = 0.
In this situation as well as that in part (B), the speed of the block reaches a maximum at some position
other than x = 0. Problem 27 asks you to locate these positions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee54b90d-e4d6-4f72-b48d-d71475e94862%2Fddf27213-f21e-46e0-a32c-2e50f7a42bdd%2Fu301cn9_processed.png&w=3840&q=75)
Transcribed Image Text:Example 8.6 (B)
Calculate the speed of the block as it passes through the equilibrium position if a constant friction
force of 4.0 N retards its motion from the moment it is released.
Solution
Conceptualize The correct answer must be less than that found in part (A) because the friction
force retards the motion.
Categorize We identify the system as the block and the surface, a nonisolated system for energy
because of the work done by the spring. There is a nonconservative force acting within the system:
the friction between the block and the surface.
Analyze
Write the appropriate reduction of Equation
8.2 and substitute for the energy changes:
AK + AEint = W, → (mv²³ — 0) + frd = Ws
Solve for vţ:
2
V₁ =
-(Ws - fkd)
m
Substitute for the work done by the spring,
found in part (A):
2
1
Uf =
kxmax - fkd)
m2
Substitute numerical values:
2
vf =
(1 000 N/m) (0.020 m)² - (4.0 N) (0.020 m)] = 0.39 m/s
1.6 kg 2
Finalize As expected, this value is less than the 0.50 m/s found in part (A).
What If?
What if the friction force were increased to 10.0 N? What is the block's speed at x = 0?
Answer In this case, the value of fkd as the block moves to x = 0 is
fkd (10.0 N) (0.020 m) = 0.20 J
which is equal in magnitude to the kinetic energy at x = 0 for the frictionless case. (Verify it!).
Therefore, all the kinetic energy has been transformed to internal energy by friction when the block
arrives at x = 0, and its speed at this point is v = 0.
In this situation as well as that in part (B), the speed of the block reaches a maximum at some position
other than x = 0. Problem 27 asks you to locate these positions.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 6 images
Recommended textbooks for you
University Physics Volume 1
Physics
ISBN:
9781938168277
Author:
William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:
OpenStax - Rice University
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
University Physics Volume 1
Physics
ISBN:
9781938168277
Author:
William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:
OpenStax - Rice University
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
Physics for Scientists and Engineers, Technology …
Physics
ISBN:
9781305116399
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Classical Dynamics of Particles and Systems
Physics
ISBN:
9780534408961
Author:
Stephen T. Thornton, Jerry B. Marion
Publisher:
Cengage Learning
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College