Center of Mass Identical Size Object Connected by Spring

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SUMMARY

The discussion centers on calculating the final location of the center of mass for a system of two identical 0.19 kg blocks connected by a spring, subjected to an 8 N force. The initial position of block 2 is at x2 = 0.05 m, while the final positions are x1 = 0.01 m for block 1 and x3 = 0.12 m for block 2. The incorrect calculation of the center of mass by using the formula (x3 - x1)/2 = 0.055 m is highlighted, indicating a misunderstanding of the center of mass formula AM xcm = (m1x1 + m2x2)/(m1+m2) and the role of the spring's force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of center of mass calculations
  • Familiarity with spring mechanics and Hooke's Law
  • Basic principles of forces and accelerations in a two-body system
NEXT STEPS
  • Study the calculation of center of mass in multi-body systems
  • Learn about the dynamics of spring systems and their equations of motion
  • Explore the implications of applying external forces on connected objects
  • Investigate the relationship between acceleration and force in a two-block system
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to center of mass and spring dynamics.

Crazynutjob
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Homework Statement


Two identical 0.19 kg blocks (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper diagram, where x2 = 0.05 m. Then a constant force of 8 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower diagram, where x1 = 0.01 m and x3 = 0.12 m. At this final time, the system is moving to the right and also vibrating, and the spring is stretched.

http://www.webassign.net/mi3/09-056-two_blocks_and_spring.jpg
http://www.webassign.net/mi3/09-056-two_blocks_and_spring.jpg

Homework Equations



What is the final location of the center of mass of the real system?
xCM,final =



POINT PARTICLE SYSTEM
What is the initial location of the point particle system?




The Attempt at a Solution



I tried (x3 -x1)/2 = (.12- .01)/2 = 0.055 m
this is incorrect according to web assign

what am i doing wrong.
 
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Crazynutjob said:

Homework Statement


Two identical 0.19 kg blocks (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper diagram, where x2 = 0.05 m. Then a constant force of 8 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower diagram, where x1 = 0.01 m and x3 = 0.12 m. At this final time, the system is moving to the right and also vibrating, and the spring is stretched.

http://www.webassign.net/mi3/09-056-two_blocks_and_spring.jpg
http://www.webassign.net/mi3/09-056-two_blocks_and_spring.jpg


The Attempt at a Solution



I tried (x3 -x1)/2 = (.12- .01)/2 = 0.055 m
this is incorrect according to web assign

what am i doing wrong.
Perhaps you could explain your reasoning. I am not sure that the problem is fully explained. Is the "final" position supposed to represent the maximum stretch of the spring?

What is the force on block 1 (hint: you have to assume a spring constant k)? How does the acceleration of block 1 affect the acceleration of Block 2?

AM
 
xcm = (m1x1 + m2x2)/(m1+m2)
 

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