Welcome back. So let’s do a potential

energy problem with a compressed spring. So let’s make this an

interesting problem. Let’s say I have

a loop-d-loop. A loop-d-loop made out of ice. And I made it out of ice so that

we don’t have friction. Let me draw my loop-d-loop. There’s the loop, there’s

the d-loop. All right. And let’s say this loop-d-loop

has a radius of 1 meter. Let’s say this is– this right

here– is 1 meter. So of course the loop-d-loop

is 2 meters high. And let’s say I have a

spring here– it’s a compressed spring. Let’s say this is the wall. This is my spring, it’s

compressed, so it’s all tight like that. And let’s say its spring

constant, k, is, I don’t know, 10. Attached to that compressed

spring– so I have a block of ice, because I need ice on ice,

so I have no friction. This is my block of

ice, shining. And let’s say the block of ice

is, I don’t know, 4 kilograms. And we also know that we are

on Earth, and that’s important, because this problem

might have been different if we were

on another planet. And my question to you is how

much do we have to compress the spring– so, let’s say

that the spring’s natural state was here, right, if

we didn’t push on it. And now it’s here. So what is this distance? How much do I have to compress

this spring, in order for when I let go of the spring, the

block goes with enough speed and enough energy, that it’s

able to complete the loop-d-loop, and reach safely

to the other end? So, how do we do this problem? Well, in order– any loop-d-loop

problem, the hard part is completing the

high point of the loop-d-loop, right? The hard part is making sure

you have enough velocity at this point, so that you

don’t fall down. Your velocity has to offset the

downward acceleraton, in which case– and here, is going

to be the centripetal acceleration, right? So that’s one thing

to think about. And you might say, wow this is

complicated, I have a spring here, it’s going to accelerate

the block. And then the block’s going to

get here, and then it’s going to decelerate, decelerate. This is probably where it’s

going to be at its slowest, then it’s going to accelerate

back here. It’s a super complicated

problem. And in physics, whenever you

have a super complicated problem, it’s probably because

you are approaching it in a super complicated way,

but there might be a simple way to do it. And that’s using energy–

potential and kinetic energy. And what we learned when we

learned about potential and kinetic energy, is that the

total energy in the system doesn’t change. It just gets converted from

one form to another. So it goes from potential

energy to kinetic energy, or to heat. And we assume that

there’s no heat, because there’s no friction. So let’s do this problem. So what we want to know is, how

much do I have to compress this spring? So what I’m essentially saying

is, how much potential energy do I have to start off with–

with this compressed spring– in order to make it up here? So what’s the potential

energy? Let’s say I compress the

spring x meters. And in the last video, how

much potential energy would I then have? Well, we learned that the

potential energy of a compressed spring– and I’ll

call this the initial potential energy– the initial

potential energy, with an i– is equal to 1/2 kx squared. And we know what k is. I told you that the spring

constant for the spring is 10. So my initial potential energy

is going to be 1/2 times 10, times x squared. So what are all of the energy

components here? Well, obviously, at this point,

the block’s going to have to be moving, in order

to not fall down. So it’s going to have

some velocity, v. It’s going tangential

to the loop-d-loop. And it also is going to have

some potential energy still. And where is that potential

energy coming from? Well, it’s going to come because

it’s up in the air. It’s above the surface

of the loop-d-loop. So it’s going to have some

gravitational potential energy, right? So at this point, we’re going

to have some kinetic energy. We’ll call that– well, I’ll

just call that kinetic energy final– because this is while

we care about alpha, maybe here it might be the kinetic

energy final, but I’ll just define this as kinetic

energy final. And then plus the potential

energy final. And that of course, has to

add up to 10x squared. And this, of course, now, this

was kind of called the spring potential energy,

and now this is gravitational potential energy. So what’s the energy

at this point? Well, what’s kinetic energy? Kinetic energy final is going

to have to be 1/2 times the mass times the velocity

squared, right? And then what’s the potential

energy at this point? It’s gravitational potential

energy, so it’s the mass times gravity times this height. Right? So I’ll write that here. Potential energy final is going

to be mass times gravity times the height, which also

stands for Mass General Hospital, anyway. You can tell my wife’s

a doctor, so my brain just– anyway. So let’s figure out the kinetic

energy at this point. So what does the velocity

have to be? Well, we have to figure out

what the centripetal acceleration is, and then, given

that, we can figure out the velocity. Because we know that the

centripetal acceleration– and I’ll change colors for

variety– centripetal acceleration has to be the

velocity squared, over the radius, right? Or we could say– and what is

the centripetal acceleration at this point? Well it’s just the acceleration

of gravity, 9.8 meters per second squared. So 9.8 meters per second

squared is equal to v squared over r. And what’s the radius

of this loop-d-loop? Well it’s 1. So v squared over r

is just going to be equal to v squared. So v squared equals 9.8– we

could take the square root, or we could just substitute the

9.8 straight into this equation, right? So the kinetic energy final is

going to be equal to 1/2 times the mass times 4 times

v squared times 9.8. And that equals– let’s just use

g for 9.8, because I think that might keep it

interesting. So this is just g, right? So it’s 2 times g. So the kinetic energy final

is equal to 2g– and g is normally kilogram meters per

second squared, but now it’s energy, right? So it’s going to be in joules. But it’s 2g, right? And what is the potential

energy at this point? Well, it’s the mass, which is

4, times g times the height, which is 2. So it’s equal to 8g. Right. So what’s the total energy

at this point? The kinetic energy is 2g, the

potential energy is 8g, so the total energy at this

point is 10g. 10g total energy. So if the total energy at this

point is 10g, and we didn’t lose any energy to friction

and heat, and all of that. So then the total energy

at this point has also got to equal 10g. And at this point we have no

kinetic energy, because this block hasn’t started

moving yet. So all the energy is

a potential energy. So this also has to equal 10g. And this g, I keep saying,

is just 9.8. I just wanted to do that just

so you see that it’s a multiple of 9.8, just for

you to think about. So what do we have here? [? I’ll do ?] these numbers worked out well. So let’s divide both

sides by 10. You get x squared is equal

to g, which is 9.8. So the x is going to be equal to

the square root of g, which is going to be equal to what? Let’s see– if I take 9.8, take

the square root of it, it’s like 3.13. So x is 3.13. So we just did a fairly– what

seemed to be a difficult problem, but it wasn’t so bad. We just said that, well the

energy in the beginning has to be the energy at any point in

this, assuming that none of the energy is lost to heat. And so we just figured out

that if we compress this spring, with the spring

constant of 10. If we compress it 3.3 meters–

3.13 meters– we will have created enough potential

energy– and in this case, the potential energy is 10 times

9.8, so roughly 98 joules. 98 joules of potential energy

to carry this object all the way with enough velocity at the

top of the loop-d-loop to complete it, and then come

back down safely. And so if we wanted to think

about it, what’s the kinetic energy at this point? Well we figured out it

was 2 times g, so it’s like 19.6 joules. Right. And then at this point,

it is 98 joules. Right? Did I do that right? Well, anyway I’m running out

of time, so I hope I did do that last part right. But I’ll see you in

the next video.

You're right. It should be 5. I need to be more careful

I got X=4.47m at initial PE=5x^2 and gravity=10m/s^2

im in college right now and (modesty aside) im fairly good in phisics, i did well on my first mid term but on my second mid term (tommorow) i wasnt feeling real sure, ur videos cleared everything up in a why neither my book nor cramster could! thank you so much! your a brilliant teacher and awsome in phisics!

The only thing he didn't tell the units of was K, notice it makes very little difference (none, actually). Also notice that v^2 is measured in m^2/s^2, whereas mas is measured in kg, so in the end, his unit does equal a Joule.

Besides, if you are -at all- familiar in the material, you should know that units and whatnot are mostly a formality.

"Don't be stingy with your units"

you didn't multiply 10 by 1/2.

I got X=4.43m

Thank you sooo much!!!! I can only dream of my university professor giving me such a clear and understandable explanation. I must say you're truly gifted as a teacher.

Why not pop in one of those little bubble notes when the mistake is made, just in case people are questioning their maths…

well its not that big of a math mistakes u just do 10/5 = 2 so sqrt(2*9.8)

There is an error in the calculations from the very start, P(initial) = 1/2 K*x^2

P(initial)= 1/2 *10 N* x^2

P(initial)= 5 X^2 not 10 X^2………….

AND THANK YOU FOR UPLOADING APPRECIATE IT VERY MUCH!!!

wow thanks ðŸ˜› very very helpful!!!

the kinetic energy just needs to be >0 for it to go around the loop….. it does not need to be 2g.

I lol'd at the 4kg ice cube xD

@RayMeel it's (2g)^1/2 = 4.43m

aka the square root of 2g

mgh….haha mass/maths general hospital HAHAHA :))

haahah now i wont forget "mass/maths general hospital" ðŸ˜€

I think k 1/2=5

thank you sal, you really saved my ass (again), thanks

I like that you referred to MGH as mass general hospital.. are you from Boston?? haha.

and THANKS for this!! It's helped me so much in preparing for my physics exam tomorrow.

Thank you so much. It seemed like an intimidating problem at first, but you made it easier to understand. You're a lot better than my professor.

Khan, is there any way I can get in touch with you? I have posted on the FACEBOOK khanacademy page, messaged the khanacademy website, and much stuff. I really wish to discuss my own set of lessons for exam prep with you and be a part of khanacademy because I share the same virtues and love for knowledge and teaching as I believe you do. Please help me reach out to you.

Thanks

10 x Â½ = 5 (;

Doesn't make sense… Shouldn't the cube be able to finish the loop as long as v > 0?

There is Another error. You haven't considered the height of the cube, around 17cm for a 4kg block of ice. So the center of gravity of the cube would only have to rise 200-17 =183cm, not 2m

when you input k=10, shouldn't it be 5x^2 instead of 10x^2?

So to summarize,

Your looking for the distance x, the spring has to be from equilibrium?

You have to find the sum of the potential AND kinetic energy at the top of the loop then take that sum and plug it into the potential energy of the springs initial formula?

Please reply, thank you ðŸ™‚

To be precise, I think that this example is missing a normal force going with the centripetal acceleration when the ice cube is at the top. So n+mg=mAr, right?

I think at the highest point of loop

Velocity should be square root of 5gr

So that ice cube completes the loop

:/

is there an example of this with friction.?

watch the first 30 secs of this at .5 speed lol

so disappointed that u made an error..shame on you..i failed my exam.

so disappointed that u made an error..shame on you..i failed my exam.

so disappointed that u made an error..shame on you..i failed my exam.i got a 69

the kinetic energy needed is actually not that small because the centripetal force include the normal force plus the gravitational force when the ice block is going up the circle.

5x^2

Just set K = 20(unit= N/m) instead of 10 and the unit Sal uses for g is 9.8 kg x m^2/s^2. Then, follow along homies.

x=4.43m

The stuttering is soo annoying

loop d loop <3

Was anyone able to find the amount of time for which the force was applied?

It's actually 4.43 though right?

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here too you can fine the velocity of the cube after completing the circular turn by consirding that the PE at the end is 0 , then just say that the whole energy that we have is equal to the KE , then you can calculate the speed from it >> like KE = 1/2 * m * V^2 = 98J

Lol

Answer should be 4.4

another mistake. The potential energy isn't exactly at 2r because of the center of mass of that block will be below that loop when it reaches the top.

Doesn't it depend on the length of loop d loop?

Why is the acceleration 9.81 ? 9.81 is for free fall.