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Work Done on Two Blocks

  1. Jun 17, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    A mass m1 = 5.1 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 5 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.83 m.

    workontwoblocks.png

    2. Relevant equations


    3. The attempt at a solution

    I don't need help with a specific answer, instead I need help understanding something.

    The work done by gravity on the system is 40.71 J. (W = M2gh, W = 5*9.81*0.83 = 40.71)
    The work done by the tension of the string on M1 is 20.57 J.
    Why is the work done by tension on M1 half of the work done by gravity? I only used the mass of M2 to find the work done on the system by gravity, so shouldn't the work done by the tension be more since M1 has more mass?

    (On the other hand, gravity is the one pulling M2 in the first place, which is supplying the tension on M1. So confused...)
     
  2. jcsd
  3. Jun 17, 2015 #2

    Nathanael

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    Be more than what? More than half? (It is more than half, not by much, but m1 is not larger by much, either.)
     
  4. Jun 17, 2015 #3

    SammyS

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    The tension also does work on m2 . (This is a negative quantity.)
     
  5. Jun 17, 2015 #4

    Drakkith

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    More than the work done by gravity.
    But of course that's impossible, as gravity is doing the work to begin with...

    Am I correct in assuming that if M2 wasn't bound to M1, the work done by gravity would accelerate M2 to a higher velocity than when M1 and M2 are connected, but the work done would still equal 40.71 J?

    I'm sorry if this is confusing. It's a multi-step problem that I've gotten almost every single step wrong the first time through.
     
  6. Jun 17, 2015 #5

    SammyS

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    The tension is an internal force as regards the entire system. Therefore, it does no net Work to the system as a whole.
     
  7. Jun 17, 2015 #6

    Drakkith

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    Okay, so gravity is pulling down on M2. M2 wants to accelerate under the force of gravity, but the tension in the rope only allows it to accelerate if M1 also accelerates. The force from M2is transferred through the string to M1 which accelerates under the force from tension. The acceleration of both blocks is the same and is equal to (M2g)/(M1+M2), or 4.86 m/s2.

    The work done on each block is equal to their final kinetic energy and added together equals 40.71 J, which is the total work done on the system.
     
  8. Jun 17, 2015 #7

    Nathanael

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    The tension does negative work on m2 (it slows it down) and positive work on m1 (speeds it up). You don't need to consider it to solve any problems because (as Sammy just said) it is zero for the entire system (it does the same amount of positive work on m1 as it does negative work on m2).

    If (out of curiosity) you want to find the work done by tension (call it WT) this is how I would find it:
    (KE)1=WT
    (KE)2=Wg-WT
    (Wg is the work done by gravity and (KE)n is the kinetic energy of block 1 and 2 respectively)

    First note about this system of equations: If you look at the total energy, (KE)1+(KE)2, it is equal to the work done by gravity (tension does no work on the system).

    Second note: if m1=m2 then (KE)1=(KE)2 (because they have the same speed) which leads to WT=0.5Wg

    (If the masses are the same the work done by tension is half the work done by gravity.)


    Since they have the same speed, the ratio of kinetic energies will be the ratio of their masses. Thus you have [itex]\frac{m_2}{m_1}=\frac{W_g-W_T}{W_T}[/itex]

    In this case, this equation leads to [itex]W_T=\frac{W_g}{\frac{5}{5.1}+1}=\frac{W_g}{1.98}=[/itex] slightly more than half the work done by gravity.



    Again, none of this is important information. The fact that the work done by tension essentially makes it ignorable. I just wrote this to explain your observation that the work done by tension is (slightly more than) half.
     
  9. Jun 17, 2015 #8

    Nathanael

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    Yes I'd say this is a satisfactory understanding. Calling tension "the force from M2" is a little iffy to me but you have the right idea.

    (You said in your OP M2 was "supplying the tension" which is also a little iffy to me.)
     
  10. Jun 17, 2015 #9

    Drakkith

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    I'm sorry, I don't know how else to describe it. How would you?
     
  11. Jun 17, 2015 #10

    Nathanael

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    Yeah you have the right idea it doesn't really matter.

    Instead of saying "The force from M2 is transferred through the string to M1" I would say something like "The tension is the same throughout the rope, so it is the same on M_1 and M_2"

    Sorry I probably shouldn't have complained because I don't really have a good way to say it either :oldsmile:
     
  12. Jun 17, 2015 #11

    SammyS

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    That looks very satisfactory to me.
     
  13. Jun 17, 2015 #12

    Drakkith

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    Thanks guys!
     
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