Grade 11 Physics Problem About Work and Energy

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Homework Help Overview

The discussion revolves around a Grade 11 physics problem related to work and energy, specifically involving elastic potential energy and kinetic energy. The scenario describes an object with a given mass and initial velocity colliding with a spring, prompting questions about how to determine the spring's compression distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between kinetic energy and elastic potential energy, questioning how to apply relevant equations. Some express uncertainty about the initial conditions and the implications of the object's velocity before and after impacting the spring.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the initial energy and the role of gravitational potential energy. Some participants have suggested specific equations and methods for solving the problem, while others are seeking clarification on the setup and assumptions involved.

Contextual Notes

There is a noted lack of consensus on the interpretation of the problem's wording, particularly regarding the object's velocity and the implications for the spring's behavior. Participants mention the absence of an answer key and express confusion about the calculations involved.

Felicity26
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1. The energy from elastic potential is 1/2 kx^2. K is a spring constant and x is the displacement distance of the spring from its equilibrium position. How far will a spring of spring constant k=25 N/m compress if an obejct of mass 4 kg has an initial velocity of 20 m/s runs into the spring without slowing down or speeding up?



2. 1/2 kx^2,



3. Basically, all I could come up with was putting the variables into the equation. To be honest, I don't know where to start solving this question or what equations I should use other than the six kinematic equations. The best I can think of after this is converting the mass into Newtons. Actually, what I wanted to know was how do I start solving this question? Sorry:(
 
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Felicity26 said:
1. The energy from elastic potential is 1/2 kx^2. K is a spring constant and x is the displacement distance of the spring from its equilibrium position. How far will a spring of spring constant k=25 N/m compress if an obejct of mass 4 kg has an initial velocity of 20 m/s runs into the spring without slowing down or speeding up?



2. 1/2 kx^2,



3. Basically, all I could come up with was putting the variables into the equation. To be honest, I don't know where to start solving this question or what equations I should use other than the six kinematic equations. The best I can think of after this is converting the mass into Newtons. Actually, what I wanted to know was how do I start solving this question? Sorry:(

Conservation of energy. What is the initial energy?
 
I would take that 20m/s and find the KE(.5mv^2). then that is converted into "spring energy" solve from there.
 
pat666 said:
I would take that 20m/s and find the KE(.5mv^2). then that is converted into "spring energy" solve from there.

Right. Initially the only energy is the kinetic energy of the object. When it compresses the spring to the maximum distance its final speed will be v = 0 (no kinetic energy). So the only energy at the maximum distance compressed is the potential energy in the spring.
 
hey, just wondering is the answer 17.9m. also the question is not written well. when you say that it doesn't slow down i assumed that it meant that the final velocity is also 20m/s(y/n) if it doesn't slow down at all the spring will be destroyed and carried away with the projectile.
 
pat666 said:
hey, just wondering is the answer 17.9m. also the question is not written well. when you say that it doesn't slow down i assumed that it meant that the final velocity is also 20m/s(y/n) if it doesn't slow down at all the spring will be destroyed and carried away with the projectile.

It says the INITIAL VELOCITY = 20m/s and it doesn't speed up or slow down. All that means is that it stays at v = 20m/s until it hits the spring. No friction is slowing it down and no other force is speeding it up. Once it hits the spring it does start slowing down as the kinetic energy is converted into potential energy of the spring.
 
yep that's what i thought, do you have an answer for the question?
 
pat666 said:
yep that's what i thought, do you have an answer for the question?

That is not the answer I got. Show your work.
 
Sorry, I don't know if the answer is 17.9m; I wasn't given an answer key. I put the numbers in and got 800 J. I was wondering if the equation Eg=mgh would be any help to figuring out the distance?
 
  • #10
Felicity26 said:
Sorry, I don't know if the answer is 17.9m; I wasn't given an answer key. I put the numbers in and got 800 J. I was wondering if the equation Eg=mgh would be any help to figuring out the distance?

Your still not showing your work and thus I cannot help you. Why would gravitational potential energy have to do with this system when it is all on a flat table (presumably)? The height does not change in the problem.
 
  • #11
yeah mgh is static(no movement in the j direction)... 1/2mv^2=1/2kx^2 mv^2=kx^2
4*20^2=25x^2 x=8m not sure how i got 17 or 19 or whatever..
 
  • #12
pat666 said:
yeah mgh is static(no movement in the j direction)... 1/2mv^2=1/2kx^2 mv^2=kx^2
4*20^2=25x^2 x=8m not sure how i got 17 or 19 or whatever..

I got 8m as well.
 
  • #13
I thought there was some kind of relation between height and distance. Right now, all I have is Us=1/2 mv^2, which would be 1/2 (25)(x)^2. Since I'm trying to find x, I'd have to find Us first. But I also have a kinetic energy of 800 J. So I'm guessing I'll have to find an equation that would find Us that includes the kinetic energy? Am I on the right track?
 
  • #14
Felicity26 said:
I thought there was some kind of relation between height and distance. Right now, all I have is Us=1/2 mv^2, which would be 1/2 (25)(x)^2. Since I'm trying to find x, I'd have to find Us first. But I also have a kinetic energy of 800 J. So I'm guessing I'll have to find an equation that would find Us that includes the kinetic energy? Am I on the right track?

No 8m is the answer. Height and distance of what? Do you understand what is happening in this problem? All of the kinetic energy is transferred into elastic potential energy of the spring. At that point the spring is the most compressed and that compression distance is x = 8m. what is Us?
 
  • #15
Sorry, I didn't refresh the page when I submitted the reply. I think I get it now. But, I was wondering, if the equation is 1/2mv^2, why is the half missing? (Oh, I was trying to look up the notation for potential energy and I got U and s as the sub for the spring, but I'm guessing I looked up the wrong thing)
 
  • #16
When I say the missing 1/2, I mean when you put then numbers into solve.
 
  • #17
[tex]E_i = K_i + Ui = K_i + U_gravity + U_spring = (1/2)M(V_i)^2 + Mgh + 0[/tex]
[tex]E_f = K_f + U_f = K_f + U_gravity + U_spring = (1/2)M(V_f)^2 + Mgh + (1/2)kx^2[/tex]

[tex]E_i = E_f[/tex]and [tex]V_f = 0[/tex]

[tex](1/2)M(V_i)^2 + Mgh = (1/2)kx^2 + Mgh[/tex]

Mgh's cancel out. We could have ignored gravitational potential energy from the beginning since we know it did not change during the problem.

[tex](1/2)M(V_i)^2 = (1/2)kx^2[/tex]

Solve for x.
 
  • #18
do the math there is a half on both sides of the equation so you can cancel them, if you prefer leave them it will give you the same answer. hope i was helpful.
 
  • #19
Thank you!
 

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