Nice, I was able to do this one by myself from watching the previous 30+ videos in this playlist. When I started yesterday, there was no way I could complete these types of questions... I'm finally starting to grasp these spring and energy concepts, Thank You again good sir :)
@@MichelvanBiezen I do. Before this time I was really insecure and unsure about studying civil engineering cus I didn't even know why I was solving most questions . I was only good at solving them but didn't know how I'm gonna apply it to the real job. But everything is clear now thanks .
Thank you Mr. Biezen. In the same scenario, how to find the position of ball (after it contacts the spring) with respect to time. I have a similar problem in my work and your help will be much appreciated.
For that you will have to look at the simple harmonic motion videos where we use trig functions to define position vs time. See playlist: PHYSICS 16 SIMPLE HARMONIC MOTION AND PENDULUM
Honestly, my professors had taken the exact same example for the term paper!!!!!!!!!!!!!!! Thank you very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very much sir!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You're welcome very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very much!!!!!
The maximum speed is calculated when all the of the potential energy is converted to kinetic energy. (At the moment the object touches the spring) mgh = (1/2) mv^2 or v = sqrt( 2 g h)
Sir, what is the point of equilibrium .... where the Velocity is zero and the whole system is in equilibrium and the weight of the object is balanced out by the force thats produced by the spring
The "equilibrium point" is the point where all the forces are balanced. If the object is motionless at the equilibrium point, it will not move. If the object is moving at the equilibrium point, it will oscillate about the equilibrium point.
You can choose your reference zero height anywhere you like and you will get the same answer. Turning the positive direction to downward could be confusing.
@@MichelvanBiezenBut it should still lead to the correct result, right? I thought so and tried to write down the energy conservation equation: -m*g*h = m*g*x + 0.5*k*x^2. If I was to find the h, for example, this gives wrong result.
why don't consider the initial point (h=0) when the spring on the equilibruim position means the height of the object is just h so the equation will be mgh=1/2kx^2
You can do it any way you like as long as you consider the total height from the starting point to the point where the spring is fully compressed and that potential energy loss is then equal to the potential energy gain from the compressed spring.
We want the height to be relative to the object's low point in order to correctly calculate the change in potential energy. The reference height to the ground is not relevant here in this problem.
The first term on the left (W) represent work put into the system during the event. (This does not include the work done by gravity, since that is accounted for by the potential energy and kinetic energy terms).
It depends on the reference point (where you consider the potential energy to be zero). That point can be arbitrarily places anywhere you like. We placed it at the lowest point of the compression of the spring.
If a mass is hung and the spring is initially stretched and another mass is dropped on the spring, how to find the distance it could be further stretched?
First find the spring constant using the first mass. Then the first mass will now be at the "new" equilibrium point. Now you add the second mass and F = kx or mg = kx so x = mg/k
I have an doubt.. If x=1.5m that mean at that time the spring is giving kx force to the block . Thais 750 N also block is giving this much at that specific instant . But how can block of 5kg under only influence of g=10m/s2 give more than 50N ? ... Plz help me
Since at that moment, the force of the spring exerted on the block is greater than the weight of the block, the spring will then push the block back in the air. It was the potential energy of the block that was converted to the potential energy of the compressed spring.
Sir kindly solve this problem A block of 1.2 kg is hung from a spring and after elongation of 0.08m it comes to equilibrium. (intially block was at rest) Then, some extra mass is added with the 1.2kg mass and so the spring elongates extra 0.16m. Now find the amount of extra mass.
Is the object at rest? Or is it dropped on the spring and from how high? If the object is at rest, then the reaction force equals the weight of the object, and the spring constant does not play a role.
Aha, that is the question. Since the spring equation is: F = - kx x = - F/k thus the spring is compress by : mg/k = (3)(9.8)/20000 Note that you have to convert to N/m so 20 N/mm = 20000 N/m
@@MichelvanBiezen I see a paradox here: you have written that x = mg/k and I don't see any flaw in your procedure. On the other hand, if we solve the same problem using energy conservation, we will write: mgx=kx²/2 → mg=kx/2 → x = 2mg/k. So, by considring the forces, we get x = mg/k but by considering energy we get x = 2mg/k. I have been thinking about this for hours and I can't understand how is that possible!
SIR KINDLY ANS PLEASE Suppose a block of mass m hangs from a spring of spring constant k. To calculate its extension, do we have to use energy conservation or balance the forces as both gives different answers?
Such an underrated video. What a life saver it is. Thank you so much man.
Glad you liked it!
Nice, I was able to do this one by myself from watching the previous 30+ videos in this playlist. When I started yesterday, there was no way I could complete these types of questions... I'm finally starting to grasp these spring and energy concepts, Thank You again good sir :)
Yes, after you see a number of them, you can see that pattern.
Best hand writing ever! Makes it very clear
Thank you! 😊
Although this video explains only one question, it helped me to understand lots of other questions.....thank you prof and yes it is very underrated
Great. It is good when you see the pattern on how to apply the energy equation. 🙂
Sir all your vedios are very useful and makes it easier to understand
Videos
You're the first to make me under why how physics is related to normal life.
Yes, that is a big aspect of physics - what does it mean in real life. I am glad you are enjoying the videos.
@@MichelvanBiezen I do. Before this time I was really insecure and unsure about studying civil engineering cus I didn't even know why I was solving most questions . I was only good at solving them but didn't know how I'm gonna apply it to the real job. But everything is clear now thanks .
now put a mass below the spring too. and solve for minimum height required to bounce the whole system off the ground
Damn intelligence sir ,you are absolutely a genius
Thanks!
You are amazing, sending thanks from Ithaca, NY
Thank you from LA California.
Say hi to professor delchamps for me.
Thank you Mr. Biezen. In the same scenario, how to find the position of ball (after it contacts the spring) with respect to time. I have a similar problem in my work and your help will be much appreciated.
For that you will have to look at the simple harmonic motion videos where we use trig functions to define position vs time. See playlist: PHYSICS 16 SIMPLE HARMONIC MOTION AND PENDULUM
Honestly, my professors had taken the exact same example for the term paper!!!!!!!!!!!!!!!
Thank you very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very much sir!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You're welcome very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very much!!!!!
This must’ve taken for ever to write lol
@@unknown-mn9wo nah.I am on PC. So just copy paste.....
You sir, are a god send
Thank you. Glad you found our videos and you found them helpful. 🙂
How would you calculate the max speed of the object?
The maximum speed is calculated when all the of the potential energy is converted to kinetic energy. (At the moment the object touches the spring) mgh = (1/2) mv^2 or v = sqrt( 2 g h)
If the initial PE takes into consideration of unknown height x (PEo = mg(h+x)), doesn't this imply that the block has somehow gained potential energy?
We want to compute the change in PE from where it starts to where it ends.
Thank you, this is very helpful.
Glad you found our videos. 🙂
Any jee aspirant watching this video ⤵️⤵️⤵️
Me
Sir, what is the point of equilibrium .... where the Velocity is zero and the whole system is in equilibrium and the weight of the object is balanced out by the force thats produced by the spring
The "equilibrium point" is the point where all the forces are balanced. If the object is motionless at the equilibrium point, it will not move. If the object is moving at the equilibrium point, it will oscillate about the equilibrium point.
Thank you once again! A lovely lesson.
Glad you found the video useful. 🙂
Sir, why don't we use mg=kx to solve this problem as the block is in equilibrium with the spring when the spring is compressed right??
When the block reaches its lowest point, the block-spring system is not in equilibrium. Moments later the block will be moving upwards again.
Thank you sir
What if we choose the initial spring position as a reference line and downward direction positive?
You can choose your reference zero height anywhere you like and you will get the same answer. Turning the positive direction to downward could be confusing.
@@MichelvanBiezenBut it should still lead to the correct result, right? I thought so and tried to write down the energy conservation equation: -m*g*h = m*g*x + 0.5*k*x^2. If I was to find the h, for example, this gives wrong result.
sir, what would be the bounce back height of block!!after hitting the spring
Assuming that no energy is lost, the block will bounce back to the same height.
when does the block reaches its maximum kinetic energy and why ?
The kinetic energy of any object = m v ^2, which means that the maximum kinetic energy is reached at the moment that its maximum velocity is reached.
why don't consider the initial point (h=0) when the spring on the equilibruim position means the height of the object is just h so the equation will be mgh=1/2kx^2
You can do it any way you like as long as you consider the total height from the starting point to the point where the spring is fully compressed and that potential energy loss is then equal to the potential energy gain from the compressed spring.
@@MichelvanBiezen can you please show how the equation would be in this case?
That is the way it was done in the video. (total change in height = h + x)
Sir why we took mg(h+x) rather then mgh as we have taken the starting point of spring as baseline, plz help me
You can choose the h = 0 point anywhere you like, it doesn't matter. You will get the same answer
@@MichelvanBiezen ok thankyou sir
Thank you Sir.
You are most welcome
What if there is a 10kg mass fixed to the top of the spring. How would you adapt the equation?
Then you would have to use the conservation of momentum to handle the collision between the 2 masses.
Sir, thank you very much
Most welcome
is it a simple harmonic motion ?
Not in this case. Conservation of energy works much better here.
why is the height h+x but not the total height from the ground to the object?
We want the height to be relative to the object's low point in order to correctly calculate the change in potential energy. The reference height to the ground is not relevant here in this problem.
life saver.
Why the initial kinetic energy is 0?
The object will fall with a velocity right??
At the moment the object is dropped it will not be moving and therefore the initial kinetic energy is zero.
@MichelvanBiezen yes. But if it falls with a velocity, i mean if it is thrown instead of dropping then won't it have a kinetic energy?
But sir how the final kinetic energy become zero, by conserved energy at top P.E =mgh and k.E =0 and at bottom the P.E=0 and K.E= 1/2 mv^2
The energy is stored in the spring (PE) and thus all the energy now becomes potential energy stored in the spring.
Best lecture sir.... 👍
You're welcome!
Very nice.....fun problem!
Glad you liked it. 🙂
@@MichelvanBiezen :)
@@MichelvanBiezen :thumbs-up:
Thanks a lot
Glad is was helpful
You are my angel❤️
It is 0.5 kx2 or - 0.5 kx2
thanks
Welcome
Sir, why is the work equals to zero if the system undergoes the force of its weight ?
The first term on the left (W) represent work put into the system during the event. (This does not include the work done by gravity, since that is accounted for by the potential energy and kinetic energy terms).
@@MichelvanBiezen Thanks a lot sir .
why isn't the potential energy of the spring negative
It depends on the reference point (where you consider the potential energy to be zero). That point can be arbitrarily places anywhere you like. We placed it at the lowest point of the compression of the spring.
What if we were looking for h and had the x value?
You would work it out exactly the same way, but solving for h instead of x.
@@MichelvanBiezen thank you. I have an initial velocity as well. Would that be KE intial = 1/2Kmv^2?
The initial energy = KEo + PEo = (1/2)mVo^2 + (1/2)kXo^2
@@MichelvanBiezen Great! That's what I did!
thank you so much!!!
You're welcome! Glad you found our videos. 🙂
If a mass is hung and the spring is initially stretched and another mass is dropped on the spring, how to find the distance it could be further stretched?
First find the spring constant using the first mass. Then the first mass will now be at the "new" equilibrium point. Now you add the second mass and F = kx or mg = kx so x = mg/k
@@MichelvanBiezen Thank you sir. You are a lifeline
bring something for jee plese
We have many videos on the JEE main and JEE advanced tests, and will continue to add to them.
@@MichelvanBiezen i mean plese upload chapterwise question bank for jee .. for jee
We are currently working on paper 1 and paper 2 of the JEE advanced 2022 (physics)
What if the spring is initially compressed state
What would compress it? (It would be an unlikely scenario).
sir what if there is friction opposing the motion downwards in scenarios like an elevator?
Then you have to add energy lost due to friction (a positive value) on the right side of the equation
Thanks sir
Legend
Thank you. Glad you find our videos helpful.
Thanks a lot ❤️🔥
I have an doubt..
If x=1.5m that mean at that time the spring is giving kx force to the block . Thais 750 N also block is giving this much at that specific instant . But how can block of 5kg under only influence of g=10m/s2 give more than 50N ? ...
Plz help me
Since at that moment, the force of the spring exerted on the block is greater than the weight of the block, the spring will then push the block back in the air. It was the potential energy of the block that was converted to the potential energy of the compressed spring.
Thanks sir!
So nice of you
@@MichelvanBiezen 😃
Sir kindly solve this problem
A block of 1.2 kg is hung from a spring and after elongation of 0.08m it comes to equilibrium. (intially block was at rest)
Then, some extra mass is added with the 1.2kg mass and so the spring elongates extra 0.16m.
Now find the amount of extra mass.
Tx
Thank you sir
Mass of 3kg object is on the spring.What is the reaction force from the ground?k=20n/mm
Is the object at rest? Or is it dropped on the spring and from how high? If the object is at rest, then the reaction force equals the weight of the object, and the spring constant does not play a role.
@@MichelvanBiezen Thank you so much.😍😍😍
The object is at rest. How to find the extension of the spring?
Aha, that is the question. Since the spring equation is: F = - kx x = - F/k thus the spring is compress by : mg/k = (3)(9.8)/20000 Note that you have to convert to N/m so 20 N/mm = 20000 N/m
@@MichelvanBiezen I see a paradox here: you have written that x = mg/k and I don't see any flaw in your procedure. On the other hand, if we solve the same problem using energy conservation, we will write: mgx=kx²/2 → mg=kx/2 → x = 2mg/k.
So, by considring the forces, we get x = mg/k but by considering energy we get x = 2mg/k.
I have been thinking about this for hours and I can't understand how is that possible!
SIR KINDLY ANS PLEASE
Suppose a block of mass m hangs from a spring of spring constant k. To calculate its extension, do we have to use energy conservation or balance the forces as both gives different answers?
kindly make some example videos on it.
Understand how**
And it does help in the understanding.
From india preparing for jee
Thank you sir ♥️
Most welcome 🙂
thanks
You're welcome!