Help Jane & John Pull a Rope: Solve Acceleration & Meetup

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SUMMARY

In the discussion, Jane and John, with masses of 51kg and 62kg respectively, are on a frictionless surface and connected by a rope. John exerts a force that accelerates Jane at 0.92 m/s², resulting in John's acceleration calculated as 0.756 m/s² using Newton's second law (f=ma). To determine where they meet, both must travel for the same time interval, utilizing the equation x = 1/2*a*t² to find their respective distances.

PREREQUISITES
  • Understanding of Newton's second law (f=ma)
  • Knowledge of kinematic equations, specifically x = 1/2*a*t²
  • Basic concepts of acceleration and mass
  • Familiarity with frictionless surfaces in physics
NEXT STEPS
  • Calculate the time taken for Jane and John to meet using their accelerations.
  • Explore the implications of frictionless surfaces in classical mechanics.
  • Investigate the effects of varying masses on acceleration in connected systems.
  • Learn about tension in ropes and its role in force transmission between objects.
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of acceleration and force interactions.

fitchguy316
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Here is a question:

Jane and John with masses of 51kg and 62kg respectively, stand on a frictionless surface 13m apart. John pulls on a rope that connects him to JAne, giving jane an acceleration of .92 m/s*s toward him. a. What is Johns acceleration?? b. IF the pulling force is applied constantly, where will Jane and John meet?

a. f=ma, manipulating this you get John's Acceleration at .756m/s*s

b. More towards John. I got 5.9m away from him but I believe this is wrong..Please help ASAP
 
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Both are starting from rest simultaneously.When they meet they must have traveled for the same time intrerval. Use the equation x = 1/2*a*t^2 for both and find x
 

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