R = Kx^2 / mg <- anybody reconise this?

  • Thread starter StheevilH
  • Start date
Thanks for the help =]In summary, the conversation discusses a formula for a spring experiment, R = Kx^2 / mg, and how to prove it using conservation of energy and projectile motion equations. It is clarified that the spring itself is the projectile in this experiment. The formula for conservation of energy is mentioned as PE + KE = constant, and the formula for KE is derived as v^2 = Kx^2 / m. The concept of symmetric parabolas in projectile motion is also mentioned. It is noted that the directions should not be ignored in calculations and the sin2 factor should be included.
  • #1
StheevilH
13
0
R = Kx^2 / mg <-- anybody reconise this?

Well..

I had experiment about springs...

I got this R = K x^2 / m g.. formula to explain...

where

R = range (displacement of x)
K = constant of spring (elastic)
x = extension of spring
m = mass
g = gravity

Can anybody prove this formula for me??

It says use conservation of energy and 4 formulae of projectile motion..

I'd really appreciate if anybody can prove it...


PS it is fired at angle of 45 degrees
 
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  • #2
Welcome to PF!

Hi StheevilH! Welcome to PF! :smile:
StheevilH said:
I had experiment about springs...

I got this R = K x^2 / m g.. formula to explain...

It says use conservation of energy and 4 formulae of projectile motion..

PS it is fired at angle of 45 degrees

uh-uh … you have to do the work! :smile:

First write out the equation of conservation of energy, as it applies to the spring and the projectile.

Then write out as many equations for projectile motion as you can think of.

What do you get? :smile:
 
  • #3


tiny-tim said:
Hi StheevilH! Welcome to PF! :smile:


uh-uh … you have to do the work! :smile:

First write out the equation of conservation of energy, as it applies to the spring and the projectile.

Then write out as many equations for projectile motion as you can think of.

What do you get? :smile:

Eh... write out as many equations for projectile motion?

I learned only 4 types of them though... and none of them matchs..

Before that...what is the equation of conservation of energy??

I don't get the whole thing..to begin with...so..would you be able to be more detail??

please?? =]
 
Last edited:
  • #4


Wait.. don't worry about the above one I wrote...

I found out potential energy of the spring is 1/2 * K * x^2..

but this doesn't match with any projectile formula I got..

is x = Vx?

or V itself? not in component form?

Gah...I'm not good at this...

Waiting for a reply...
 
  • #5
StheevilH said:
I found out potential energy of the spring is 1/2 * K * x^2..

but this doesn't match with any projectile formula I got..

Hi StheevilH! :smile:

(btw, is the spring launching a separate projectile, or is the spring itself a projectile? :confused:)

No, your PE won't fit into a projectile formula … it'll fit into a conservation of energy formula! :wink:

And conservation of energy is PE + KE = constant …

so what formula do you have for KE? :smile:
 
  • #6


tiny-tim said:
Hi StheevilH! :smile:

(btw, is the spring launching a separate projectile, or is the spring itself a projectile? :confused:)

No, your PE won't fit into a projectile formula … it'll fit into a conservation of energy formula! :wink:

And conservation of energy is PE + KE = constant …

so what formula do you have for KE? :smile:


Ah...I worked out...I think...

lol..yeah I thought PE was meant to be in projectile formula..

Eh...from what I got...is KE = PE...

1/2 * K * x^2 = 1/2 * m * v^2

Hence, v^2 = K * x^2 / m ..

Using range formula; R = Vo^2 * Sin 2 (*) / g..

R = K * x^2 / m * g...Sin 2 (45) = 1...so can be ignored...

though...

I don't get how initial velocity and the final velocity can be same?

Can you tell me how they works?

Spring itself is the projectile motion...yeah...I think...lol

Thanks if you tried!

Oh..don't forget to reply my last question =]
 
  • #7
StheevilH said:
Using range formula; R = Vo^2 * Sin 2 (*) / g..

R = K * x^2 / m * g...Sin 2 (45) = 1...so can be ignored...

though...

I don't get how initial velocity and the final velocity can be same?

Conservation of energy again … initial and final velocity are at the same height … so PE is the same … so KE must be the same! :biggrin:

Alternatively, just look at the curve the projectile follows … it's a parabola, and it's symmetric … so speed and angle on the way up will always be equal and opposite to speed and angle at the same height on the way down. :smile:
Oh..don't forget to reply my last question

Sorry … what question? :confused:
 
  • #8


tiny-tim said:
Conservation of energy again … initial and final velocity are at the same height … so PE is the same … so KE must be the same! :biggrin:

Alternatively, just look at the curve the projectile follows … it's a parabola, and it's symmetric … so speed and angle on the way up will always be equal and opposite to speed and angle at the same height on the way down. :smile:


Sorry … what question? :confused:

Oh..lol question...

That was pointing at the Vo being equal to V...

So... we can ignore about the directions in the calculations??

The magnitudes are same...so we can just put them as same?

and...did I actually get it that way.. @_@
 
  • #9
StheevilH said:
So... we can ignore about the directions in the calculations??

No … don't forget the sin2 factor … at 45º, it just happens to be 1, but you would usually have to include it. :smile:
 
  • #10


tiny-tim said:
No … don't forget the sin2 factor … at 45º, it just happens to be 1, but you would usually have to include it. :smile:

Roger, copy that

(Y)
 

1. What does the equation R = Kx^2 / mg represent?

The equation R = Kx^2 / mg represents the relationship between force, spring constant, distance, mass, and gravity in a spring system. It is commonly known as Hooke's Law and is used to calculate the force exerted by a spring when it is stretched or compressed.

2. What does each variable in the equation stand for?

The variable R represents the force exerted by the spring, K represents the spring constant, x represents the distance the spring is stretched or compressed, m represents the mass attached to the spring, and g represents the acceleration due to gravity.

3. How is this equation used in science?

This equation is commonly used in physics and engineering to calculate the force exerted by a spring in various applications such as in mechanical systems, elastic materials, and simple machines. It is also used to study the behavior of springs and their relationship with other variables.

4. Can this equation be used for non-spring systems?

While this equation is specifically derived for spring systems, it can also be applied to other systems that exhibit similar behavior, such as elastic materials or objects with certain degrees of flexibility. However, the values for the variables may differ depending on the specific system being studied.

5. How does this equation relate to other scientific laws and principles?

This equation is closely related to other scientific laws and principles, such as Newton's Second Law of Motion and the Law of Conservation of Energy. It also has connections to the concepts of potential energy and work. Understanding this equation can help in understanding these other laws and principles in physics and engineering.

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