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ashna

  • 2 years ago

Two identical 5kg blocks are moving with same speed of 2m/s towards each other along a frictionless horizontal surface . The blocks collide , stick together and comes to rest . Considering the two blocks as a system , calculate the work done by (1) external forces (2) internal forces

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  1. ashna
    • 2 years ago
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    @akash123

  2. akash123
    • 2 years ago
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    work done by all the forces= change in the kinetic energy

  3. ashna
    • 2 years ago
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    okay

  4. akash123
    • 2 years ago
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    so what are the external forces exerting on these two blocks?

  5. ashna
    • 2 years ago
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    i did'n get the question , what external and what internal ?

  6. ashna
    • 2 years ago
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    gravity ?

  7. akash123
    • 2 years ago
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    there is no friction...so only external forces are gravity and normal reaction

  8. ashna
    • 2 years ago
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    okay

  9. akash123
    • 2 years ago
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    internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force

  10. ashna
    • 2 years ago
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    okay

  11. akash123
    • 2 years ago
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    it's fine?

  12. akash123
    • 2 years ago
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    ok

  13. akash123
    • 2 years ago
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    now will the external forces( gravity and normal reaction) do work?

  14. ashna
    • 2 years ago
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    no ?

  15. akash123
    • 2 years ago
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    |dw:1349445206541:dw|

  16. akash123
    • 2 years ago
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    yes...they wont do any work because they are perpendicular to d displacement...so no work done

  17. ashna
    • 2 years ago
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    okay understood

  18. akash123
    • 2 years ago
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    so from work energy theorem work done= change in the kinetic energy --> work done by external forces+ work done by the internal forces= change in the kinetic energy of the system

  19. akash123
    • 2 years ago
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    so u can calculate the work done by the internal forces

  20. akash123
    • 2 years ago
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    u know the work done by the external forces(=0) and the change in the KE of the system

  21. ashna
    • 2 years ago
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    okay

  22. akash123
    • 2 years ago
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    u'll calculate the change in the KE of the system?

  23. akash123
    • 2 years ago
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    final KE=0 initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2

  24. ashna
    • 2 years ago
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    okay

  25. akash123
    • 2 years ago
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    m1= m2=5kg and v1=v2=2m/s

  26. ashna
    • 2 years ago
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    20 J ?

  27. akash123
    • 2 years ago
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    yes

  28. akash123
    • 2 years ago
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    so work done by internal forces= (final KE- initial KE)= -20 J

  29. ashna
    • 2 years ago
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    okay

  30. akash123
    • 2 years ago
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    u got the meaning of -ve sign? I mean...why work done is -ve?

  31. ashna
    • 2 years ago
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    yes 0-20 ?

  32. akash123
    • 2 years ago
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    -ve work done because they r pulling down the energy of the system to zero

  33. ashna
    • 2 years ago
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    okay .. i understand ! thank you :)

  34. akash123
    • 2 years ago
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    :)

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