Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ashna

  • 3 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

  • This Question is Closed
  1. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @akash123

  2. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    work done by all the forces= change in the kinetic energy

  3. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  4. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so what are the external forces exerting on these two blocks?

  5. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i did'n get the question , what external and what internal ?

  6. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    gravity ?

  7. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    there is no friction...so only external forces are gravity and normal reaction

  8. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  9. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    internal forces..when they'll collide with each other then they will exert forces on each other..that's the internal force

  10. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  11. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it's fine?

  12. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  13. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now will the external forces( gravity and normal reaction) do work?

  14. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no ?

  15. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1349445206541:dw|

  16. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes...they wont do any work because they are perpendicular to d displacement...so no work done

  17. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay understood

  18. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so u can calculate the work done by the internal forces

  20. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    u know the work done by the external forces(=0) and the change in the KE of the system

  21. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  22. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    u'll calculate the change in the KE of the system?

  23. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    final KE=0 initial KE= 1/2 m1 v1^2 + 1/2 m2 v2^2

  24. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  25. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    m1= m2=5kg and v1=v2=2m/s

  26. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    20 J ?

  27. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

  28. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so work done by internal forces= (final KE- initial KE)= -20 J

  29. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  30. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    u got the meaning of -ve sign? I mean...why work done is -ve?

  31. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes 0-20 ?

  32. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    -ve work done because they r pulling down the energy of the system to zero

  33. ashna
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay .. i understand ! thank you :)

  34. akash123
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    :)

  35. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy