Hooke's Law and Elastic Potential Energy
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- čas přidán 12. 07. 2024
- This video provides a basic introduction into Hooke's law. It explains how to calculate the elastic potential energy and how to determine the amount of work required to stretch a spring.
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Provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the load or applied force. (Hooke's law)
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x and delta x have been used interchangeably in the video, which makes it confusing. If x is assumed to be the distance between ends of spring, then F=-k*(delta x), W=-1/2 k*(delta x)^2.
Yes very confusing part because x is usually the strain caused by the stress so how they can be used interchangeably. If x is total length then using ∆x makes sense
I love how he says length. “Lenf”
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can the same theory be applied if we press the spring ?? How ever as everyone else say you are truly a life saver
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17:57 dont we define the change in X as Xf - Xi ? when we make (Xf - Xi)^2 , doesnt it have to become Xf^2 +Xi^2 -2Xf.Xi ?
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I have one small question: Why doesn't work done equal to the negative change of potential energy, which is represented by W=-deltaPE. I remembered that this is what my teacher taught me.
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I'm a bit confused abt the last part involving the Work done by the restoring force, I mean initially when the applied force is stretching or compressing the spring to a certain, I have the belief that only the applied force is stretching the spring, not the restoring force, I hqve the believe the restoring only acts when the spring is returning to its natural position , and then the displacement (opposite to the stretched displacement) and the restoring force are in the same direction, so I'm confused how it's work done by the restoring force is negative, to me it feels its should be +ve.
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for 4th question just do fp=kx
then w/2 = fs where f=fp and s = x
so fp= kx == 250x0.75 == 187.5
187.5 x 0.75 = 140.625 / 2
therefore work required = 70.3125 J or 70.3 J
If k is a constant how it's value can change from spring to spring
Every time he says hookes law I hear it as cole slaw
Hooke slaw 🥰
at 16:17 I can't understand how when the acceleration is 0, the work done is equal to the potential energy change, can someone please explain ?
It has nothing to do with acceleration, there are numerous ways to calculate the force exerted that don’t include acceleration. If work is equal to force times distance, then the force is equal to the work applied divided by the distance.
@@yuvalkoren4070 Thankyou !
How will you find x if the only present is the mass?
Springs in parallel and series plz
23:29
Molto Gracie
wow😻
at 16:15 why isnt the force negative? isnt it 1/2-KX^2?
Elastic Potential Energy (PE) = 0.5kx^2. The unit will be N/m • m^2 = Nm = joules. Energy is measured in joules. Note: 1 joule = 1Nm.
However, the restoring force, F is measured in N.
its not in the same direction with force thats why it is positive
On #4 can someone explain why we multiply it by 0.5 and what does the triangle have to do with out problem in the first place?
The 0.5 is just the ½ in decimal. It makes it easier to multiply and we need the triangle because the force is varying.
So for problem nr 7 (which we see a glimpse of it at the end) i got:
a) 490 N/m;
b) 6.272 J;
c) -12.544 J;
d) 40 cm.
How far was i?
c should be 11.392 i think because we use the formula Ug=mgh
how did u solve d?
i hope u still remmber after 5 months
@@sarahnaji7563 lol
16:51 doesn't work equals the change in kinetic energy and not the potential energy ?
So it would be
W = PE = 1/2 F x
There is a mistake on question 6; you multiplied 1/2k by the difference of the squares of x rather than by the square of the difference of x.
Cheers
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At minute 22:58 why is ΔX² = (Xբ² - Xᵢ²). I don't really understand the logic behind it, it seems more plausible for it to be (Xբ - Xᵢ)². Can someone please explain?
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16:52 what if acceleration is not 0,
I wonder if f net is always 0 in this law?
if the acceleration is not 0, does the total work equal KE + PE?
if it is equal how do we find KE?
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How does the gravitational wave emitting from a planet differ from a rubbee band being stretched out maybe the gravitational field is stretching out some other force than elastic energy stretches out any ideas to prove or disprove this? I need all possible awnsers please no insults those are a waste of time im trying to learn and grow not hear little 2 year olds throw insults
Question 6 is incorrect 45cm is 0.45m not 0.045m. Therefore the answer is -14.6832J
4.5 cm, not 45.
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@vivansh و ما لغت الاستعمار هذه؟
Isn't work = - change in potential energy?
nvm he explains that in the end of the video lol
How do you expand Deltax^2 as x_f^2-x_i^2? You are missing a term, aren't you?
I thought w= change in KE
what is "m"
????
Am I the only one who thinks he kinda sounds like Ben Feldman?
for question's 1 and 2, why is the formula: F=kx? Shouldn't it be this: F=-kx, that's what my textbook is telling me?
To stretch a spring it’s positive and to compress it, it’s negative
Yo did not explain what k and the other formulas meant you were just saying them but not telling us what it really meant
Well its explained here...
czcams.com/video/HALbtyDUjp0/video.html
I heard a sumsang phone in the background 😅.
i didn't understand question number 4
Hhhhh I ans the (Q) that been end this video 🤣🤣