Physics 9 Conservation of Energy (8 of 11) Weights On A Table

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In this video I will show how to calculate the final velocity of two objects (one on a table top) around a pulley.
Next video in this series can be seen at:
• Physics 9 Conservati...

[@michelebianchi7861]

Your approach is always so organised and systematic yet you manage to clearly explain the underlying mechanisms ! Many thanks for the wealth of content that is your youtube channel.

[@Skittix447]

I'm binge watching most of your Physics-Mechanics videos and everything is making so much sense. I can look through my notes from class and know what's going on. I can't thank you enough

[@asantechiko3292]

I'm new and I'm already hooked, I'll be staying a little longer. Thank you sir,you explain so well everything seems so easy.

[@fas5]

Your videos are saving me, my good sir. Thank you, very much!

[@LikiLulgjuraj]

Absolutely love these videos, they help me get 96 points on my physics exam! Thank you very much Professor!!

[@MichelvanBiezen]

Great job! That is why we started the channel in the first place. So that students around the world can find the help they need.

[@UNEMPLOYEDSENSEI]

Professor you are such a big help! Please keep posting videos on physics problems because after this video the Energy unit is making loads of sense now!

[@paramnesia9110]

you can go further with this at the end - m1 has KE left over (98 J) with this you can calculate how far along the table m1 will continue to travel before stopping from friction and furthermore calculate the full distance m1 traveled. from the 98 J m1 has remaining it travels for 10 m after m2 impacts the ground + the 5 m m1 traveled prior therefore covering 15 m overall. my calculations might be wrong but it was a bonus i wanted to do.

[@MichelvanBiezen]

There are indeed more questions that can be attached to this type of problem.

[@theadel8591]

informative video, thanks. btw based on what I read, Ei=Ef alone does not hold true if there is a non-conservative force doing work on the system because the total energy of the system won't be conserved in this case. ΔEmechanical=W(applied)-w(resistive force) ✔

[@firomsamohammed3100]

man !!!! you really do make sense.what you did is just mashallah

[@MichelvanBiezen]

Glad you liked it. 🙂

[@AZ-zz4kn]

The explanation was clear and concise. Thank you very much

[@MichelvanBiezen]

Glad it was helpful!

[@MichelvanBiezen]

Gareth, Exactly as the problem explains. Use Newton's second law on the object at the table. The weight of the small mass is greater than the friction forces on the large mass.

[@garethm3171]

I agree that that m2g > Force friction by about 30N. But m1g > m2g. However, I'll have to set up an experiment to see if a 5kg mass can accelerate a 10kg mass across a table that has some friction of about 20N.

[@MichelvanBiezen]

Gareth M m1g (mu) < m2g

[@garethm3171]

Michel van Biezen That's correct, that's what I said above. As I said I'll get the experiment set up to see if a 5kg mass can accelerate a 10kg mass across a table with about 20N of friction.

[@schrod56]

Gareth M Allow me to explain it hopefully better, I appreciate it does look a bit counter intuitive. If the table was frictionless then I hope you can see that the 5kg mass would indeed accelerate the 10kg mass, indeed a small push would probably get the 10kg mass going. But we have a bit of friction from the table, about 20N, however the downward force (m2g) of the 5kg mass is still sufficient to overcome this friction. The (m2g) force is about 50N so about 20N of this is taken up overcoming friction from the table so there is still about 30N of force available from the 5kg mass to accelerate the 10kg mass. Don't forget the downward weight of the 10kg (m1g) is balanced out by the Normal force of the table ergo no acceleration in the vertical direction. Hope that explains it. :-)

[@MichelvanBiezen]

Schrodie 56 Coefficient of friction in this problem is 0.2

[@alexandrechartrand9866]

Thanks, it helps before an exam !

[@muhammadhamza4494]

I always love how you explain 😍

[@MichelvanBiezen]

Berk, There is no need to. Since the PE for m1 doesn't change, it will cancel out.

[@garethm3171]

Perhaps I'm missing something but how can a 5kg mass cause a 10kg mass with added friction from the table to accelerate?? Maybe if m1 was 5kg and m2 was 10kg then it could happen but not as in the diagram.

[@grimmjjac1190]

Thanks a lot sir!! this sure help me alot

[@aliabrahimi3711]

Professor Biezen here is my equation for finding final velocity for a system like this V=√ (m1*g* μ) Final velocity = Square root of mass on top of the table * Gravity * Coefficient of Friction This works only for this problem, I am not sure if it would for others

[@MichelvanBiezen]

Ali, I recommend that my students always start with basic principles like F= ma, etc. and then follow a general approach that solves all problems.

[@goodc.2853]

How would you solve this problem if M2 traveled a distance of, let's say, 5 meters but DOESN'T touch the ground? Could you still cancel out final potential energy?

[@jbonceu2457]

For some people, you could also assign friction as the work done by the system on the left side, but make it negative cause it's just like friction is doing negative work

[@MichelvanBiezen]

Indeed. That would be perfectly OK.

[@los.9757]

@@MichelvanBiezen so can I instead use PE + KE - (work done (W) as (Force of kinetic friction * Distance))=KE+PE for conservation of energy problems involving friction?

[@alexacaramel7]

I love your videos! very concise and straight forward!

[@MichelvanBiezen]

Thank you for the feedback. We started putting these videos up a year ago and we are trying to find out what people like. So this feedback is very valuable. Thanks,

[@enriquesoler3150]

I have a question, is the reason why there's no final potential energy is because there no increase in height and is just the same height?

[@MichelvanBiezen]

You can have a height reference point for each object in the system. The top block does not change height and we can call the low point for the hanging block zero height.

[@assaineindustries]

sir, can this question be solve by using newtons law to resolve the force and find the acceleration. Then use the second equation of kinematics to solve for the velocity.

[@MichelvanBiezen]

That is correct. That will work. 🙂

[@ahmedal-ebrashy3691]

Thank you very much. Though h=d is very intuitive, it still hard for me to proof that?

[@psilvakimo]

The rope is inflexible.

[@karlschmitkons5289]

Shouldn't there also be a factor on the "final" side of the equation to account for the rotational kinetic energy of the pulley?

[@MichelvanBiezen]

We are ignoring the moment of inertia of the pulley (assume a very light pulley) We cover that topic in a later chapter.

[@sunnyz3046]

If we assume the practical was in a perfect environment where there's no friction. Will it be just PE + KE = PE + KE? Because you can remove the friction away the equation and W is zero because it's stationary

[@MichelvanBiezen]

That is correct.

[@sunnyz3046]

@@MichelvanBiezen Okay thank you. But when I was doing my calculation, my PE1 doesn't equal with the KE2. Is that correct? The values that I have are the total mass of the system was 1.174 kg, carriage was weighed 1.124 kg , the load travelling downwards was 50 g, and dropped at a height of 86.5 cm. Because PE1 I got 0.423 j and KE2 was 9.963 j. This is assuming its a perfect environment.

[@aryansefidrokhsharahjin1000]

what should we do in a situation when the pulley is on an incline?

[@MichelvanBiezen]

Are you looking for examples like this? Physics - Mechanics: Applications of Newton's Second Law (3 of 20) kzbin.info/www/bejne/rYqUcp6tr7CjoJY

[@berk26092]

If there is a force pushing to the block, must I add W=(F)x(distance) to the equation? Thanks, sir.

[@exodia45]

what if Fnet=mt*a (m2*g-m1*g*mu)/(m2+m1)=a=1.96m/sec^2 Vf^2=Vo^2+2a(yf-yo) where Vo=0 , yo=0,yf=5 and a=1.96m/sec^2 so Vf=4.43m/sec i think that's another solution of it! @Michel van Biezen

[@MichelvanBiezen]

Yes indeed, you can also solve this problem using the equations of kinematics.

[@albertopoli8896]

If μ would be 0,5 then Vf=0 (if I did right calcule) : what does it mean? The Blocks would be in rest ( no moving)?

[@MichelvanBiezen]

that is correct

[@RaselAhmed-jc6iu]

How do we have kinetic final, isn't the block going to come to a stop? On some cases, the kinetic final is considered 0 because it comes to a stop. I can do these problems but I have the most trouble just deciding what energy to consider. Can you recommend something to get better at that? Thanks!

[@MichelvanBiezen]

Sometimes the question is very clear and sometimes we must make assumptions. The assumption here is to find the velocity of the block as it reaches the floor. After it reaches the floor the velocity if obviously zero, but we cannot use the conservation of energy equation when there is a collision. (For that we must use conservation of momentum).

[@officialspamaccount]

Hi sir, I had a quick question. Why did you not consider the rotational kinetic energy of the pulley?

[@MichelvanBiezen]

In this problem we assumed a "light" pulley. The concept of moment of inertia and rotational kinetic energy is covered in later chapters.

[@officialspamaccount]

@@MichelvanBiezen Thank you for your quick reply, I understand now. Also, thank you so much for making easy to understand online lectures they are really saving me!! 😊

[@elisears8501]

What if the given is the initial velocity and the time it takes for the block to land and you need to find the mass of the object ?

[@joeyborja423]

I am thinking it miight help if the distance from m1 and the pulley is given which should be > or = to 5m to properly equate with h.

[@MichelvanBiezen]

That makes no difference to the principle of the problem.

[@joeyborja423]

I was just thinking if the string is shorter mass m2 won't make it to the floor :) but yes principle and logic rule.

[@danielnalepa5724]

What if the table was at an incline?

[@MichelvanBiezen]

It will work exactly the same way. You do have to take in to account the change in height for both objects then.

[@TheDonofGaming]

hi sir I don't understand how the system is moving downwards when the mass1 is heavier then mass2 thus shouldint the mass attached to the pulley not budge

[@MichelvanBiezen]

It all depends on the net force. If the net force is greater than zero there must be an acceleration of the system. (Newton's second law).

[@binoanil6954]

Does m1 not have a potential energy?

[@MichelvanBiezen]

Yes, but since it doesn't change, we don't need to keep track of it.

[@tvlk444]

do it on inclined plane please

[@MichelvanBiezen]

We have a lot of examples on the inclines plane. You can find them easily from the HOME page of the channel.

[@sankalpjha2117]

Professor shouldn't we subtract friction as it is lost throughout the entire mechanism ??

[@sankalpjha2117]

Can we also apply Wnet =∆k.As far as the way you solved it, are you considering that the total initial energy= final energy + some energy lost.I am having problems in figuring out about friction if I am applying your method. Plzz help

[@MichelvanBiezen]

There are different methods. The one that I show requires the lost energy to be added to the right side or subtracted from the left side.

[@sankalpjha2117]

+Michel van Biezen why will it be added sir

[@MichelvanBiezen]

If you don't add it to the right side of the equation, the equation will not be correct. If you lost some energy due to friction that means that the PE and KE energy final is less than the PE and KE initial and you have to make up the difference.

[@sankalpjha2117]

+Michel van Biezen Thnx professor you are a sensation among my comrades.Professor I wanted to ask you will you make some videos on optics explaining the basics with some examples

[@erc333]

If the 10 kg block falls off the table before the 5kg is about to hit the ground then mustn't the PE of the 10kg change as well. Why can we just assume that the the string is long enough so that the 10kg block doesn't fall off the table? Also, why can you assume d and h are the same length ?

[@MichelvanBiezen]

You can assume any scenario and work the problem accordingly.

[@arijitroy9462]

Reuel Cuellar d=h as the string is inextensible.

[@TeenGohan9798]

let's say u have a rope 50 meters long. and u pull from one side by 1 meter. Won't it's other side have moved 1 meter towards the initial side?

[@fredydelamora9301]

My question is why aren't you multiplying the force of frictions and the distance by the angle in between them which should be cos180??shouldnt friction always be negative work?

[@RaselAhmed-jc6iu]

Fredy De La Mora it does but from the initial energy because that's your total and you can see it does get subtracted when he starts solving it. He set up the equation to make it look easy but you can move the energy lost on the other side with a negative sign and you should get the same answer. I know this is late but it's for people that might have the same question as you.

[@danielquinn2414]

@@RaselAhmed-jc6iu Can you take a look at this video? kzbin.info/www/bejne/o4ucfX2wjdunr9k&feature=emb_logo He sets up the equation as you described yet he multiplies it by cos180. Why does he do that there and not here?

[@coolwierdo]

Why did you say that the weight reaches the bottom for this problem (so that final PE is zero) but not for the last problem?

[@MichelvanBiezen]

Aida, It probably has to do with the conditions in the problem and what is asked. In this problem you are asked to find the final velocity when the block reaches the floor.

[@coolwierdo]

Michel van Biezen I never thanked you, so thanks!

[@iftikharakhan1256]

Hello. Why is the distance moved horizontally by m1 is equal to the vertical distance traveled by m2? Thank You in advance :)

[@MichelvanBiezen]

Iftikhar ahmed khan Both masses are connected to the same string, so they must move the same distance

[@iftikharakhan1256]

Your videos are really helpful. Thank You so much!

[@jennamonteleone5205]

How would you find acceleration for this example using the same method?

[@MichelvanBiezen]

Once you have the final velocity then you use: V^2 = Vo^2 + 2 a h and solve for a (h is the height)

[@jennamonteleone5205]

@@MichelvanBiezen And how would you solve for acceleration without a known height or velocity?

[@saharhussein04]

how is it that d is the same as h even though it's for friction. which means that it should be the distance on the table??

[@MichelvanBiezen]

the distance traveled by the hanging mass is the same as the distance traveled by the sliding mass.

[@saharhussein04]

@@MichelvanBiezen oh, now I feel stupid:). thank you for the reply. you're the best.

[@marinettecheng485]

What if there are three blocks? Is it still the same?

[@MichelvanBiezen]

We have examples with three blocks as well

[@sgwen10]

I thought we can't use conservation of energy due to the external forces such as friction, tension and gravity???

[@craiglistly8210]

there is an equation when none conservative forces(friction, air resistance etc) are doing work KE1+PE1+Work Other= KE2+PE2. PE is composed of gravitational potential and elastic potential and work other is the work done by non conservative forces. That equation is starting point for this problem but as we begin to work problem out some values are zero so the equation changes accordingly

[@MichelvanBiezen]

Gwen, The equation still works if you write it like this: W + PEo + KEo = PEf + KEf + Heat lost where the heat lost term is the amount of work done or energy lost to overcome friction, wind resistance, etc. NOTE that the heat lost term is POSITIVE on the right side of the equation. There are some videos that show this.

[@kyzel101]

how to solve for distance travelled by the 10 kg block?

[@MichelvanBiezen]

+kyzel101 Since they are connected, both blocks will travel the same distance. (5 m)

[@johnsumner1474]

WHAT ABOUT THE WORK DONE ON BLOCK M1 BY THE STRINGS TENSION?

[@MichelvanBiezen]

Since the string is internal to the system, it does not do any work on the system. If you draw a free body diagram around each block separately, then you would indeed need to include the force of the string on the block, but not the way the problem was solved.

[@johnsumner1474]

@@MichelvanBiezen Many thanks for clearing that up for me, and thank you for your excellent physics videos.

[@seungyeonpark3071]

If m1 is 30kg instead 10kg, what is v?

[@MichelvanBiezen]

If m1 was 30 kg, then the friction force would be too large and the weight of m2 would not be enough to get m1 to move.

[@seungyeonpark3071]

@@MichelvanBiezen I got it! thank you.

[@tsoojbaterdeneharvard3187]

But 5 kg mass cant move 10 kg mass.I think that final kinetic energy is zero ,Sir isnt it?

[@MichelvanBiezen]

The weight of the 5 kg mass can easily move the 10 kg mass.

[@jbonceu2457]

You seem to underestimate the power of gravity

[@berkk1993]

You forgot to add the P.E of m1 to the both sides of the equation.

[@psilvakimo]

m1 is where the datum is. So PE1 = 0.

[@leeovib9284]

What is E lost means 🙂

[@MichelvanBiezen]

E stand for energy. It is the energy lost due to the work performed to overcome friction.

[@leeovib9284]

Michel van Biezen thanks 🙏

[@ca8ine]

I HATE PHYSICS SO MUCH OMG I WISH THAT HELPED BUT I CANT FIGURE IT OUT AUAIAJKAHSHSJJS

[@elisears8501]

What if the given is the initial velocity and the time it takes for the block to land and you need to find the mass of the object ?

[@MichelvanBiezen]

+Eli Sears Use the equations of kinematics to find the acceleration. Then use F = ma to find the mass.