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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
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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?
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.
Good luck with your date.
I hated ''PHYSICS'' but this guy makes it much more interesting
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 )
My teacher didn't explain this concept as well as you did. Thank you!
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"
Crystal clear. Thanks you.
Thanks! It really helped
Very helpful, thanks very much
Very helpful!
Not all heroes wear capes. Thank You!
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)]
Really didn`t understand, but after the video all was cleared. Thank you!!
Glad this helped your understanding of the concepts. 🙂
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
Well done! That is how you obtain a good understanding of the material. Keep it going! 🙂
good shit fam
thank you sooooooooooo much!
U r supab sir!!!
thank you so much. God bless you🙏🙏🙏
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 :))))
thank you so much
thank you!!!your explanation is so goood
You are welcome! Glad you found our videos. 🙂
How to calculate " weight of 10 kg box when lifting at the height of 20 mtr through pulley "
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.
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.
+Laurelindo I did the same, but I ended up with a slightly different answer !!
the final velocity with me is 2.55 m/s
+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
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?
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.
can we apply law of conservation of energy for a single body separately in a system which has more than one body
Sunita Sinha yes you can...
very nyc explanation and one thing is that u have nyc personality very much personality
Thanks!
I think it would be easy by $v^2=2ah$, where $a=(m1-m2)g/(m1+m2)$
Sir, what if the pulley has mass? Where should I put it in the equation?
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
since the blocks are moving in different directions(one up and the other down) isn't necessary to specify the direction?
It is better to indicated direction in terms of which way the pulley is turning.
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
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.
@@MichelvanBiezen even if we do not know the height, cannot we calculate acceleration by force analysis
Yes, it can be done using F = ma as well. We have a lot of examples of that in those playlists.
So i have a= m1g + m2g/mtotal = 3.21 but using vf2 =vo2+2as didnt give the answer
this helped me a lot
Glad it was helpful.
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?
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.
@@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
Energy is conserved, but kinetic energy is changing
@@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
Thank you.
sir u rock
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)?
You can pick the height reference point at any height and the answer will always be the same.
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)
Im confused sir. :( cause its giving me a squareroot of negative answer and that's imaginary. Right?
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.
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?
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).
Thank you
sir, shouldn't we write in the initial potential energy --> (m1-m2) *g*h?
No, that would not be correct. Both object have positive potential energy if we take the ground as zero height.
thanks
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
KE = (1/2) mv^2
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?
+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.
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.
+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.
+Arielle Arsenault shut up
Great problem. Seems innocent enough, at first, but has a little more to it, which makes it more interesting.
It is good to start with a simple example like this.
@@MichelvanBiezen :thumbs-up: definitely
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
Each block can have its own reference height. (That is why the problem was written that way).
Sir thankyou, the video realy helped me in my JEE preparation god 😇🙏👼bless you
@@mallick9459 usko JEE pata hai kya bhai 🤣🤣
sir why can't we simply take the 8 kg block as the system ?
then we need not bother about the other block
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.
thank you
You're welcome
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?
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.
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
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.
@@MichelvanBiezen Ok thank you so much! I understand it now
What if the pulley had friction and was exerting a constant force F against the direction of motion of mass?
Then you would have to add the work done to overcome friction. (W = F x d)
@@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
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.
Can I calculate a , with equation a=v^2/2x? And once I obtained a , can I calculate F =ma?
Yes you can. Try it and see what happens.
Michel van Biezen F=389 Nw
Is that the tension in the string?
I think F=389 Nw = Fnet , otherwise If that was the tension in the string then Fnet =0......or not?
It would probably help if you watch this playlist: PHYSICS 4.8 FREE BODY DIAGRAMS
Thank u
sir, i really misunderstand that how can we choose 2 different reference points in one problem ! -I'm talking about the high-
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.
Understood. Thank you very much
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.
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
That is exactly what the first equation shows, that Eo = Ef
Thank you sir ❤️
Very helpful ❤️
HOW THE HEIGHT OF 40 KG IS 2 m
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.
My exam is in less than 6 hours :')
Good luck on your exam