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Physics: Post your doubts here!

Discussion in 'International A And AS Level' started by XPFMember, Sep 29, 2011.

  1. Hamnah Zahoor

    Hamnah Zahoor

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    Clark20 You know how to make the equations right?
     
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  3. Clark20

    Clark20

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    Can you please explain more about how we can think of the tension forces 'T' acting on the masses, for better understanding? Like, how are the tension forces even exerted on the masses?
    Also, is T in both masses equal in magnitude? If yes, why?[/QUOTE]
     
    Last edited: Feb 12, 2019
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  4. Clark20

    Clark20

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    Yes
     
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  5. Hamnah Zahoor

    Hamnah Zahoor

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    Whenever an object is moving downward we take the weight of the object to be greater than the tension in the string, and when the weight is moving upward the tension is greater than the weight we use this concept while writing the equations.
    Yes, T is equal in both cases for both objects are attached to the same string and the string used is usually a light inextensible string which exerts same forces on both masses attached.
     
  6. Clark20

    Clark20

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    Isnt mass M pulling mass m and hence exerting the tension force on it? Or is it just the string which is exerting forces of tension on both the masses?

    Like, are both the masses completely independent of each other?
     
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  7. Hamnah Zahoor

    Hamnah Zahoor

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    Yes mass ' m ' is pulling mass M which is exerted in the form of tension.
    the mass which is heavier exerts the force on the other when it is released.
     
  8. Clark20

    Clark20

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    Then how is tension exerted on the heavier mass?
     
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  9. Hamnah Zahoor

    Hamnah Zahoor

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    At the start you had both masses stationary then you released the system i.e whatever support was holding both masses was removed resulting in heavier mass going down with lighter mass going upwards simultaneously.
     
  10. Clark20

    Clark20

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    Yes, I get that. I understood that the heavier mass is exerting force of tension on the light mass, so it's known where tension acting on the lighter mass comes from. What I dont understand is why is there tension acting on the heavier mass? What exerts it on the heavier mass?
     
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  11. Hamnah Zahoor

    Hamnah Zahoor

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    Why are you making the question so complicated?

    I don't know if I am right here
    The string due to inertia will resist this change in motion and will impose an opposite force to stop the it.
    Thus, the tension in the string acting on the heavier mass.
     
  12. Clark20

    Clark20

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    I'm just trying to be able to write the heavier mass' equation (Mg-T=Ma) with proper understanding of why even there exists force of tension on it.

    From my knowledge of force of tension, it is an upward force exerted by a string/rope on an object when its pulled upwards by it. Hence, it made more sense why the lighter mass experiences force of tension as it actually is being pulled upwards by the string connected to the heavier mass. Heavier mass isnt being shown as being pulled by the string, but the lighter mass opposes it from going down freely and probably that way it's exerting force of tension on it.
     
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  13. Hamnah Zahoor

    Hamnah Zahoor

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    Agreed
     
  14. Clark20

    Clark20

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    Need help with this question.
     

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  15. Hamnah Zahoor

    Hamnah Zahoor

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    T - m1g = m1(a)
    T = m1(a) + m1g -------(1)

    m2g - T = m2(a)
    T = m2g -m2(a) --------(2)

    placing both equations equal to each other
    m1a + m1g = m2g - m2a
    m1a + m2a = m2g - m1g
    (m1 + m2 )a= (m2-m1) g
    a = (m2-m1)g/(m1+m2)

    As the question asks about the rate provided by the energy thus,
    P = Fv
    P = (ma) v

    As in the whole system the elevator is going upwards against gravity thus acceleration is negative
    P = m(-a)v
    P = (m1 + m2)(-(m2-m1)g/m1+m2)v
    cancelling out the common terms
    P = -(m2-m1)gv
    P = (m1-m2)gv ---------- option D
     
  16. Hamnah Zahoor

    Hamnah Zahoor

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  17. Clark20

    Clark20

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    What power are we asked to calculate in the question?

    Also, can you please explain how you used the formula P=fv? Why did you substitute those values?
     
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  18. Hamnah Zahoor

    Hamnah Zahoor

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    Power = energy/time
    Power is the rate at which energy is transferred
    thus, we use the power formula here.
     
  19. Clark20

    Clark20

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    We are told to calculate the power of the system/masses.
    The formula P=FV is used. What do you substitute for m in (ma×v)?
     
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  20. Hamnah Zahoor

    Hamnah Zahoor

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    (m1 + m2) ------- total mass
     
  21. Clark20

    Clark20

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    Thank you.
     
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