COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 57QAP
To determine

(a)

Speed of the object, when it is 8.00 cm from equilibrium

Expert Solution
Check Mark

Answer to Problem 57QAP

Speed of the object, when it is 8.00 cm from equilibrium= 0.42 ms-1

Explanation of Solution

Given info:

  Object mass= 5.00 kgspring constant = 250 Nm-1spring is compressed 10 cm

Formula used:

  E=12kx2E=energyk=spring constantx=distance

  E=12mv2E=kinetic energym=massv=velocity

Calculation:

  Potential energy of the spring when it is compressed 10.0 cm,E=12kx2E=12×250 Nm-1×(0.1 m)2E=1.25 JWhen released, potential energy is converted to kinetic energy.

  As the total energy is constant, when it is at 8 cm,Ek=Ep10-Ep8Ek=kinetic energyEp8=potential energy at 8 cmEp10=potential energy at 10 cmEk=1.25 J-12*250 Nm-1*(0.08 m)2Ek=0.45 J

  E=12mv20.45 J=12×5 kg×v20.45 J=12×5 kg×v2v=0.42 ms-1

Conclusion:

Speed of the object, when it is 8.00 cm from equilibrium= 0.42 ms-1

To determine

(b)

Speed of the object, when it is 5.00 cm from equilibrium

Expert Solution
Check Mark

Answer to Problem 57QAP

Speed of the object, when it is 5.00 cm from equilibrium= 0.61 ms-1

Explanation of Solution

Given info:

  Object mass= 5.00 kgspring constant = 250 Nm-1spring is compressed 10 cm

Formula used:

  E=12kx2E=energyk=spring constantx=distance

  E=12mv2E=kinetic energym=massv=velocity

Calculation:

  Potential energy of the spring when it is compressed 10.0 cm,E=12kx2E=12×250 Nm-1×(0.1 m)2E=1.25 JWhen released, potential energy is converted to kinetic energy.As the total energy is constant, when it is at 5 cm,Ek=Ep10-Ep5Ek=kinetic energyEp5=potential energy at 5 cmEp10=potential energy at 10 cmEk=1.25 J-12×250 Nm-1×(0.05 m)2Ek=0.94 J

  E=12mv2Ek=0.94 J=12×5 kg×v20.94 J=12×5 kg×v2v=0.61 ms-1

Conclusion:

Speed of the object, when it is 5.00 cm from equilibrium= 0.61 ms-1

To determine

(c)

Speed of the object, when it is at equilibrium

Expert Solution
Check Mark

Answer to Problem 57QAP

Speed of the object, when it is at equilibrium= 0.71 ms-1

Explanation of Solution

Given info:

  Object mass= 5.00 kgspring constant = 250 Nm-1spring is compressed 10 cm

Formula used:

  E=12kx2E=energyk=spring constantx=distance

  E=12mv2E=kinetic energym=massv=velocity

Calculation:

  Potential energy of the spring when it is compressed 10.0 cm,E=12kx2E=12×250 Nm-1×(0.1 m)2E=1.25 JWhen released, at equilibrium all the potential energy is converted to kinetic energy.

  E=12mv2Ek=1.25 J=12×5 kg×v20.94 J=12×5 kg×v2v=0.71 ms-1

Conclusion:

Speed of the object, when it is at equilibrium= 0.71 ms-1

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 6 Solutions

COLLEGE PHYSICS

Ch. 6 - Prob. 11QAPCh. 6 - Prob. 12QAPCh. 6 - Prob. 13QAPCh. 6 - Prob. 14QAPCh. 6 - Prob. 15QAPCh. 6 - Prob. 16QAPCh. 6 - Prob. 17QAPCh. 6 - Prob. 18QAPCh. 6 - Prob. 19QAPCh. 6 - Prob. 20QAPCh. 6 - Prob. 21QAPCh. 6 - Prob. 22QAPCh. 6 - Prob. 23QAPCh. 6 - Prob. 24QAPCh. 6 - Prob. 25QAPCh. 6 - Prob. 26QAPCh. 6 - Prob. 27QAPCh. 6 - Prob. 28QAPCh. 6 - Prob. 29QAPCh. 6 - Prob. 30QAPCh. 6 - Prob. 31QAPCh. 6 - Prob. 32QAPCh. 6 - Prob. 33QAPCh. 6 - Prob. 34QAPCh. 6 - Prob. 35QAPCh. 6 - Prob. 36QAPCh. 6 - Prob. 37QAPCh. 6 - Prob. 38QAPCh. 6 - Prob. 39QAPCh. 6 - Prob. 40QAPCh. 6 - Prob. 41QAPCh. 6 - Prob. 42QAPCh. 6 - Prob. 43QAPCh. 6 - Prob. 44QAPCh. 6 - Prob. 45QAPCh. 6 - Prob. 46QAPCh. 6 - Prob. 47QAPCh. 6 - Prob. 48QAPCh. 6 - Prob. 49QAPCh. 6 - Prob. 50QAPCh. 6 - Prob. 51QAPCh. 6 - Prob. 52QAPCh. 6 - Prob. 53QAPCh. 6 - Prob. 54QAPCh. 6 - Prob. 55QAPCh. 6 - Prob. 56QAPCh. 6 - Prob. 57QAPCh. 6 - Prob. 58QAPCh. 6 - Prob. 59QAPCh. 6 - Prob. 60QAPCh. 6 - Prob. 61QAPCh. 6 - Prob. 62QAPCh. 6 - Prob. 63QAPCh. 6 - Prob. 64QAPCh. 6 - Prob. 65QAPCh. 6 - Prob. 66QAPCh. 6 - Prob. 67QAPCh. 6 - Prob. 68QAPCh. 6 - Prob. 69QAPCh. 6 - Prob. 70QAPCh. 6 - Prob. 71QAPCh. 6 - Prob. 72QAPCh. 6 - Prob. 73QAPCh. 6 - Prob. 74QAPCh. 6 - Prob. 75QAPCh. 6 - Prob. 76QAPCh. 6 - Prob. 77QAPCh. 6 - Prob. 78QAPCh. 6 - Prob. 79QAPCh. 6 - Prob. 80QAPCh. 6 - Prob. 81QAPCh. 6 - Prob. 82QAPCh. 6 - Prob. 83QAPCh. 6 - Prob. 84QAPCh. 6 - Prob. 85QAPCh. 6 - Prob. 86QAPCh. 6 - Prob. 87QAPCh. 6 - Prob. 88QAPCh. 6 - Prob. 89QAPCh. 6 - Prob. 90QAPCh. 6 - Prob. 91QAPCh. 6 - Prob. 92QAPCh. 6 - Prob. 93QAPCh. 6 - Prob. 94QAPCh. 6 - Prob. 95QAPCh. 6 - Prob. 96QAPCh. 6 - Prob. 97QAPCh. 6 - Prob. 98QAPCh. 6 - Prob. 99QAPCh. 6 - Prob. 100QAPCh. 6 - Prob. 101QAPCh. 6 - Prob. 102QAPCh. 6 - Prob. 103QAPCh. 6 - Prob. 104QAPCh. 6 - Prob. 105QAPCh. 6 - Prob. 106QAPCh. 6 - Prob. 107QAP
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Kinetic Energy and Potential Energy; Author: Professor Dave explains;https://www.youtube.com/watch?v=g7u6pIfUVy4;License: Standard YouTube License, CC-BY