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Physics 8 Work, Energy, and Power (26 of 37) A pulley with two weights

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  • čas přidán 7. 08. 2024
  • Visit ilectureonline.com for more math and science lectures!
    This six part physics lecture series will cover the concepts of Work and Energy.
    Problem Text:
    A pulley is suspended from the ceiling. A mass of 80kg is suspended on one side and a mass of 40kg on the other side. If the 80kg mass starts at a height of 2.0m, how fast will it be moving when it reaches the ground?

Komentáře • 115

  • @_wli
    @_wli Před 2 lety +13

    Thanks. Was just having a debate with my Tinder match on how to calucation acceleration inside a pulley system and this video helped me to clear things up. I think I will be getting a date out from this.

  • @xhantisinoxolo1867
    @xhantisinoxolo1867 Před 9 lety +21

    I hated ''PHYSICS'' but this guy makes it much more interesting

  • @lycan_system8427
    @lycan_system8427 Před 3 lety +3

    what if this was the same exact question but instead of how fast it goes , they ask that considering the big block goes down 2m from the initial and assuming that the smaller block does NOT hit the pulley, we are asked to find the height that the little block goes ( not velocity is given )

  • @TheIgnorant
    @TheIgnorant Před 10 lety +4

    My teacher didn't explain this concept as well as you did. Thank you!

  • @AuroraNora3
    @AuroraNora3 Před 5 měsíci

    I prefer this method with the Work-Energy Theorem:
    "The absolute vertical displacement 'h', and speed 'v'
    are the same for both.
    Starting from rest, assume G does all of the work;
    it works against the mass 'm', and with the mass 'M':
    W_net = ΔK
    (M - m) * g * h = 1/2 * (M + m) * v²
    v² = 2gh * (M-m) / (M+m)
    v = +sqrt(13 m²/s²)
    v = 3.6 m/s"

  • @Kurchack
    @Kurchack Před 9 lety +2

    Crystal clear. Thanks you.

  • @raflicky
    @raflicky Před 10 lety +1

    Thanks! It really helped

  • @msindisenindovela1020
    @msindisenindovela1020 Před 7 lety +1

    Very helpful, thanks very much

  • @DynestiGTI
    @DynestiGTI Před 5 lety +2

    Very helpful!

  • @fweezy31
    @fweezy31 Před 6 lety +6

    Not all heroes wear capes. Thank You!

  • @ibrahimnazemqader9153
    @ibrahimnazemqader9153 Před 5 lety +1

    another way:
    Initial potential energy:
    Pi=Fnet*h=(m1-m2)*g*h
    final potential energy (Pf) become zero, because h=0.
    then:
    W+Pi+KEi=Pf+KEf+heat lost
    0+(m1-m2)*g*h+0=0+(m1+m2)*v^2+0
    v=sqrt[(m1-m2)*g*h / (m1+m2)]

  • @sthelilesihle5102
    @sthelilesihle5102 Před 2 lety +1

    Really didn`t understand, but after the video all was cleared. Thank you!!

    • @MichelvanBiezen
      @MichelvanBiezen  Před 2 lety +2

      Glad this helped your understanding of the concepts. 🙂

  • @Mikebigmike94
    @Mikebigmike94 Před 2 lety +1

    ah, after a fair few mistakes i managed to solve it by solving for the tension force and finding out the work done by that and using that as the external work done, same answer, 3.6 m/s :) i like how the further you go in physics the more ways you learn how to solve similar problems. I am looking forward to classical mechanics

    • @MichelvanBiezen
      @MichelvanBiezen  Před 2 lety +1

      Well done! That is how you obtain a good understanding of the material. Keep it going! 🙂

  • @liambishop2365
    @liambishop2365 Před 6 lety +2

    good shit fam

  • @aaa34aaa
    @aaa34aaa Před 8 lety +1

    thank you sooooooooooo much!

  • @dhruvsharma8315
    @dhruvsharma8315 Před 9 lety +1

    U r supab sir!!!

  • @tsoojbaterdeneharvard3187

    thank you so much. God bless you🙏🙏🙏

  • @vonfiller5166
    @vonfiller5166 Před 4 lety

    You can also solve this using the force equation and finding the acceleration of the whole system! then by that you use the kinematics equation to find the velocity :))))

  • @lailai1395
    @lailai1395 Před 6 lety +1

    thank you so much

  • @khorkaixun2175
    @khorkaixun2175 Před rokem +1

    thank you!!!your explanation is so goood

  • @surenderchandra
    @surenderchandra Před 6 lety +1

    How to calculate " weight of 10 kg box when lifting at the height of 20 mtr through pulley "

    • @MichelvanBiezen
      @MichelvanBiezen  Před 6 lety +3

      With a simple pulley the force required to lift a weight at a constant speed equals the weight of the object. The height does not make a difference. You can also determine the weight by determining how much work it takes to lift it a certain height (like 20m). Then the work done = mgh.

  • @Peter_1986
    @Peter_1986 Před 9 lety

    I tried solving this problem by solving for the acceleration from Newton's second law: a = g*(m1 - m2)/(m1+m2), and then solved for the final velocity from a = v²/(2*s), that seemed to work as well.

    • @OMARYassin1
      @OMARYassin1 Před 8 lety

      +Laurelindo I did the same, but I ended up with a slightly different answer !!
      the final velocity with me is 2.55 m/s

    • @Peter_1986
      @Peter_1986 Před 8 lety +2

      +OMAR Yassin
      I'm not sure why you got that answer, but here is what I did:
      I started with the formula F = (m1 + m2)·a, where the force is made up by the weights - and since the smaller weight m2 is opposing the acceleration, this weight is given a negative sign.
      This gives you the formula m1·g - m2·g = (m1 + m2)·a.
      Now, use the fact that a = v² / (2·h) when the initial velocity is zero, and plug that into the previous equation;
      now you should get m1·g - m2·g = (m1 + m2)·v² / (2·h).
      Solving this for v will then give you v = sqrt(2·g·h·[m1 - m2] / [m1 + m2]).
      Now try plugging the values into that equation, and you should get 3.6 m/s.
      This looks really messy in this form, but hopefully it's clear what I am trying to say here. =P

  • @theazndonut1
    @theazndonut1 Před 4 lety

    if you didn't use the h=0 as reference point for weight B. Would the PE initial be (mgh)_1 - (mgh)_2? Because they are moving in a different direction?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 4 lety

      The PE always needs a reference point and does not depend on the velocity of the object, but only on the position relative to some height reference. The choice of the reference point is arbitrary.

  • @sunitasinha7785
    @sunitasinha7785 Před 7 lety +1

    can we apply law of conservation of energy for a single body separately in a system which has more than one body

  • @rashmisinha4859
    @rashmisinha4859 Před 7 lety

    very nyc explanation and one thing is that u have nyc personality very much personality

  • @RoIIingStoned
    @RoIIingStoned Před 7 lety +1

    Thanks!

  • @sabinayeasmin4116
    @sabinayeasmin4116 Před 5 lety

    I think it would be easy by $v^2=2ah$, where $a=(m1-m2)g/(m1+m2)$

  • @daveaberin4705
    @daveaberin4705 Před 7 lety +1

    Sir, what if the pulley has mass? Where should I put it in the equation?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 7 lety

      Look at the examples with moment of inertia that deal with pulleys that have mass in this playlist: PHYSICS MECHANICS 3: ROTATIONAL MOTION, MOMENT OF INERTIA, ANGULAR MOMENTUM, TORQUE czcams.com/play/PLX2gX-ftPVXX_8wOGXRHgCBC6Okijsag5.html

  • @nkosiz1593
    @nkosiz1593 Před 4 lety

    since the blocks are moving in different directions(one up and the other down) isn't necessary to specify the direction?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 4 lety +2

      It is better to indicated direction in terms of which way the pulley is turning.

  • @boboganbobogan9297
    @boboganbobogan9297 Před rokem +1

    if we knew the time it takes for 80 kg block to reach the ground, could we calculate velocity this way? acceleration could also be calculated and then we would use kinematics equations

    • @MichelvanBiezen
      @MichelvanBiezen  Před rokem +2

      I am assuming the you know the starting height, and that the block started from rest. If you know the time you can use h = (1/2) a t^2 where a is the acceleration. Then you can use v = a t to find the velocity when it hits the ground.

    • @boboganbobogan9297
      @boboganbobogan9297 Před rokem +1

      @@MichelvanBiezen even if we do not know the height, cannot we calculate acceleration by force analysis

    • @MichelvanBiezen
      @MichelvanBiezen  Před rokem +2

      Yes, it can be done using F = ma as well. We have a lot of examples of that in those playlists.

    • @preciousa.4384
      @preciousa.4384 Před rokem

      So i have a= m1g + m2g/mtotal = 3.21 but using vf2 =vo2+2as didnt give the answer

  • @sethother8012
    @sethother8012 Před 2 lety +1

    this helped me a lot

  • @drpeculiar1414
    @drpeculiar1414 Před 2 lety +2

    I don't understand why the kinetic energies of both the blocks were added. I've never seen it being done. Why is it do sir?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 2 lety +2

      Kinetic energy is NOT dependent on direction, (it is a scalar quantity), that is why we ADD the kinetic energy of each part of the total system.

    • @drpeculiar1414
      @drpeculiar1414 Před 2 lety +1

      @@MichelvanBiezen But in this case, since kinetic energy is conserved. Shouldn't kinetic energy be the same so there would be no need to add the other kinetic energy

    • @MichelvanBiezen
      @MichelvanBiezen  Před 2 lety +1

      Energy is conserved, but kinetic energy is changing

    • @drpeculiar1414
      @drpeculiar1414 Před 2 lety +1

      @@MichelvanBiezen I have to say this. I like you and your teaching style. You understand the things. Your answers are pregnant. I wish if u were my tutor

    • @MichelvanBiezen
      @MichelvanBiezen  Před 2 lety

      Thank you.

  • @aumsharma6970
    @aumsharma6970 Před 7 lety +1

    sir u rock

  • @jeramsantos3730
    @jeramsantos3730 Před 7 lety +1

    Sir, what if we use the location of the 80kg block as its reference rather than what you did. Will it still have a PE(initial)?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 7 lety

      You can pick the height reference point at any height and the answer will always be the same.

    • @jeramsantos3730
      @jeramsantos3730 Před 7 lety

      Michel van Biezen You said in 4:00 that we dont have PE(i) on block 40kg because we set its reference point in its position, but what if we did the same thing in the block 80kg does it mean we wont also have PE(i) on 80kg? But we will have PE(f) on both block? If we do that it will be 0 = PE(f) + KE(f)

    • @jeramsantos3730
      @jeramsantos3730 Před 7 lety

      Im confused sir. :( cause its giving me a squareroot of negative answer and that's imaginary. Right?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 7 lety +1

      Remember that it is only the DIFFERENCE in potential energy that matters. If you assume that neither had potential energy initially, then at the end the potential energy of the 80 kg mass will be - m1g delta h and for the 40 kg mass it will be + m2g delta h. Delta h is the same for both.

  • @DEXClanCompetitive
    @DEXClanCompetitive Před 4 lety

    How would you solve a problem like this where you are given only the masses and the initial velocities and asked how far up on of the masses moved?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 4 lety

      Use the energy equation, and the maximum height is obtained when the potential energy reaches its maximum value (which is when the kinetic energy is equal to zero).

  • @HimanshuKumar-lg4jm
    @HimanshuKumar-lg4jm Před 5 lety

    Thank you

  • @MAliK-ox2lz
    @MAliK-ox2lz Před 5 lety

    sir, shouldn't we write in the initial potential energy --> (m1-m2) *g*h?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 5 lety +1

      No, that would not be correct. Both object have positive potential energy if we take the ground as zero height.

  • @l3aIIin23
    @l3aIIin23 Před 10 lety

    thanks

  • @grahamflowers
    @grahamflowers Před 2 lety +1

    There is no kinetic energy in a moving mass there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers

  • @Ariellex0
    @Ariellex0 Před 8 lety

    How come you use the equation
    W + PE0 + KE0 = PEf + KEf + heat loss
    The equation that we are to use is very similar, K + U + Eth = Emech + Eth
    however the part which concerns me is where you have W, our formula was Eth which is the same as how you have heat loss. The Eth is heat loss from friction. So, in our formula we account for heat loss on both sides of the equation which is the W in terms of friction. Will this change the equation?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 8 lety +1

      +Arielle Arsenault It is just a method or technique used to solve these types of questions and different people prefer different methods. In the end they all help us get the correct answer.I found that the technique I teach here is easy to learn and solves any type of problem like this.In the end the energy sum on both sides of the equation must be equal.Left side: work put into the system from an outside source + initial KE + initial PE equals Right side: final KE + final PE + any heat lost due to friction etc.

    • @Ariellex0
      @Ariellex0 Před 8 lety

      No, I like your method. I'm just confused because the W on the left side of the equation is equal to F*d where are the Eth on the right side is equal to µFn*d. So I'm just not sure how those two would be considered equal to each other.

    • @MichelvanBiezen
      @MichelvanBiezen  Před 8 lety

      +Arielle Arsenault
      The W on the left represents word done on the system by an outside force that ADDS energy to the system, like someone pushing. The E lost on the right side is energy lost by the system by overcoming a retarding force such as friction or wind resistance etc.

    • @SMD1999
      @SMD1999 Před 8 lety

      +Arielle Arsenault shut up

  • @fizixx
    @fizixx Před 2 lety +1

    Great problem. Seems innocent enough, at first, but has a little more to it, which makes it more interesting.

    • @MichelvanBiezen
      @MichelvanBiezen  Před 2 lety +2

      It is good to start with a simple example like this.

    • @fizixx
      @fizixx Před 2 lety +1

      @@MichelvanBiezen :thumbs-up: definitely

  • @mallick9459
    @mallick9459 Před 5 lety +1

    But sir wont we add the potential energy which the 40 kg mass had because of it being at a certain height above the reference height

    • @MichelvanBiezen
      @MichelvanBiezen  Před 5 lety +2

      Each block can have its own reference height. (That is why the problem was written that way).

    • @mallick9459
      @mallick9459 Před 5 lety

      Sir thankyou, the video realy helped me in my JEE preparation god 😇🙏👼bless you

    • @alhadbhagwat6142
      @alhadbhagwat6142 Před 4 lety

      @@mallick9459 usko JEE pata hai kya bhai 🤣🤣

  • @IshitaKukreti
    @IshitaKukreti Před 9 lety

    sir why can't we simply take the 8 kg block as the system ?
    then we need not bother about the other block

    • @MichelvanBiezen
      @MichelvanBiezen  Před 9 lety +1

      Ishita Kukreti
      You can draw a free body diagram around each block, calculate the acceleration for each separately, (you will end up with two unknowns for each). Then you solve for the unknowns by solving the two equations simultaneously. That works well, but it is more work.

  • @user-bu8mg7uq3s
    @user-bu8mg7uq3s Před 3 lety

    thank you

  • @shivam-tiwari19
    @shivam-tiwari19 Před 4 lety

    Sir, why did you put the potential energy of the second block on the right side of the equation? Wouldn't it need to be the total potential energy equals the total kinetic energy when the blocks are moving?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 4 lety

      The energy equation places the total energy at some initial point = to the total energy at some final point. Therefore the energy on the right side is the PE and KE at the final point in time + any energy that may have been lost to friction.

    • @shivam-tiwari19
      @shivam-tiwari19 Před 4 lety

      Michel van Biezen Then what about the initial potential energy of the second block? You seemed to ignore it and only use the initial potential energy of the first block

    • @MichelvanBiezen
      @MichelvanBiezen  Před 4 lety

      It depends on where you place the reference point of the PE (which is arbitrary). If the height of the lowest block at the initial point in time is the zero reference height, then it has zero potential energy.

    • @shivam-tiwari19
      @shivam-tiwari19 Před 4 lety

      @@MichelvanBiezen Ok thank you so much! I understand it now

  • @aestheticnature4997
    @aestheticnature4997 Před 3 lety

    What if the pulley had friction and was exerting a constant force F against the direction of motion of mass?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 3 lety +1

      Then you would have to add the work done to overcome friction. (W = F x d)

    • @aestheticnature4997
      @aestheticnature4997 Před 3 lety

      @@MichelvanBiezen Thanks. Could u plz tell me what will be the value of "d" for this specific example? Will the "d" be equal to h (i.e 20) for this question

    • @MichelvanBiezen
      @MichelvanBiezen  Před 3 lety

      d is the distance traveled. It all depends on how the problem is defined. The tension in the rope produced a torque, turning the pulley. There are lots of examples in the playlists where the moment of inertial is taken into account.

  • @albertopoli8896
    @albertopoli8896 Před 5 lety

    Can I calculate a , with equation a=v^2/2x? And once I obtained a , can I calculate F =ma?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 5 lety

      Yes you can. Try it and see what happens.

    • @albertopoli8896
      @albertopoli8896 Před 5 lety +1

      Michel van Biezen F=389 Nw

    • @MichelvanBiezen
      @MichelvanBiezen  Před 5 lety +1

      Is that the tension in the string?

    • @albertopoli8896
      @albertopoli8896 Před 5 lety +1

      I think F=389 Nw = Fnet , otherwise If that was the tension in the string then Fnet =0......or not?

    • @MichelvanBiezen
      @MichelvanBiezen  Před 5 lety +3

      It would probably help if you watch this playlist: PHYSICS 4.8 FREE BODY DIAGRAMS

  • @totab6190
    @totab6190 Před 5 lety

    Thank u

  • @MultiShahem
    @MultiShahem Před 8 lety

    sir, i really misunderstand that how can we choose 2 different reference points in one problem ! -I'm talking about the high-

    • @MichelvanBiezen
      @MichelvanBiezen  Před 8 lety +4

      Choosing 2 reference points in a problem like this is OK ONLY IF you are calculating the DIFFERENCE in energy. If you are calculating the total energy, then you must use the same reference point.

    • @MultiShahem
      @MultiShahem Před 8 lety

      Understood. Thank you very much

    • @qedmath1729
      @qedmath1729 Před 7 měsíci

      just let the horizontal zero line be at the ground, h_1 be the initial height of block 1, and h_2 be the initial height of block 2. Initial potential energy will be m_1gh_1 + m_2gh_2, final potential energy will be m_2g(h_1 + h_2), equating these 2 as in the equation we see m_2gh_2 cancels, and therefore we can solve this problem independent of the height of block 2 from the ground.

  • @md.shahalamkhan335
    @md.shahalamkhan335 Před 6 lety +1

    Pls can u show the conservation of energy. U have shown final velocity. But I want to know that does it follow the law of conservation? That means Ef=Eo

    • @MichelvanBiezen
      @MichelvanBiezen  Před 6 lety +3

      That is exactly what the first equation shows, that Eo = Ef

  • @moazelsawaf2000
    @moazelsawaf2000 Před 5 lety

    Thank you sir ❤️
    Very helpful ❤️

  • @kumarsanman3
    @kumarsanman3 Před 7 lety +1

    HOW THE HEIGHT OF 40 KG IS 2 m

    • @MichelvanBiezen
      @MichelvanBiezen  Před 7 lety +1

      Regardless of where m2 starts in elevation, it will be 2 m higher when m1 reaches the ground. (We only care about the change in height, not the final height.

  • @itsjustferaas2680
    @itsjustferaas2680 Před 4 lety +1

    My exam is in less than 6 hours :')