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# Spring potential energy example (mistake in math) | Physics | Khan Academy

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.

1. Khan Academy says:

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

2. anava84 says:

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

3. Julio Fernandez says:

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!

4. inzanozulu says:

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"

5. megaelliott says:

you didn't multiply 10 by 1/2.

6. megaelliott says:

I got X=4.43m

7. shortygtx says:

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.

8. Michael Kelly says:

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

9. skyfaze says:

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

10. Golden_Raven says:

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………….

11. Val Lumi says:

wow thanks 😛 very very helpful!!!

12. Devo says:

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

13. JonBlizzle2 says:

I lol'd at the 4kg ice cube xD

14. stratocaster1111 says:

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

aka the square root of 2g

15. Riley Lim says:

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

16. Alexiepower says:

haahah now i wont forget "mass/maths general hospital" 😀

17. Derek gorton says:

I think k 1/2=5

18. hahs4 says:

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

19. ksjin09 says:

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.

20. hihen says:

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.

21. Amir Mullick says:

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.

22. GOD_JDMasterFlex says:

Thanks

23. Maddie Meyers says:

10 x ½ = 5 (;

24. Frida Westerbergh says:

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

25. Steve Shaw says:

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

26. Angie Wiranata says:

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

27. Michael Persaud says:

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?

28. Venice3393 says:

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?

29. Vikram Singh says:

I think at the highest point of loop
Velocity should be square root of 5gr
So that ice cube completes the loop
:/

30. Alice Ma says:

is there an example of this with friction.?

31. Develhuntr75 says:

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

32. arnold lee says:

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

33. arnold lee says:

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

34. arnold lee says:

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

35. Happyman VDO says:

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.

36. Sriram B says:

5x^2

37. Jonathan Alveno says:

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.

38. Xela Namiti says:

x=4.43m

39. Faraz x says:

The stuttering is soo annoying

40. Omar Osama says:

loop d loop <3

41. Ian Lawson says:

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

42. Viktoriya T says:

It's actually 4.43 though right?

43. sumit kumar says:

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44. Obadah Adernali says:

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

45. Sohan Singh says:

Lol
46. Divad Ignawm says:
47. Pokemon Verse says:
48. Umar Hussain says: