Find the max compression of the spring

In summary, the conversation discusses solving for the velocity of a system using the equations for kinetic energy and potential energy. The initial energy of the system is 0, and the final energy is calculated using the final velocities and potential energies. There is a discrepancy in the final answer, but it is eventually resolved by omitting the negative sign and taking the square root. The conversation also touches on the positive nature of kinetic energy and the potential for gravitational potential energy to be zero or negative.
  • #1
isukatphysics69
453
8

Homework Statement



help.PNG

Homework Equations


ke = .5mv^2

The Attempt at a Solution


Ei= .5m1v21 + .5m2v22 + PEs(PEs initially 0)

Ef= .5m1v21 + .5m2v22 + PEs(PEs is now .5(16.2)(.2672)set them equal and then solve for velocity right? is this the right approach?
 

Attachments

  • help.PNG
    help.PNG
    10.1 KB · Views: 780
Physics news on Phys.org
  • #2
isukatphysics69 said:

Homework Statement



View attachment 225073

Homework Equations


ke = .5mv^2

The Attempt at a Solution


Ei= .5m1v21 + .5m2v22 + PEs(PEs initially 0)

Ef= .5m1v21 + .5m2v22 + PEs(PEs is now .5(16.2)(.2672)set them equal and then solve for velocity right? is this the right approach?
What about GPE?
 
  • Like
Likes isukatphysics69
  • #3
Why do you use two different velocities for two tied bodies?
And the system is set in motion by the gravity, and there is no GPE in the equations.
 
  • Like
Likes isukatphysics69
  • #4
omg thank you guys completely forgot gpe also yes i know the velocities are the same but just labeled like that thank you
@haruspex @Sergio Rodriguez
 
  • #5
ok what is going on here i took into account gpe and got 1.59m/s and incorrect
GPE = mgdcosθ
initially 0 because distance was 0 at initial state
The energy of the whole system is initially 0
 
  • #6
wait a second i will try something here i see something wrong. spe is negative, the spring wants to snap back i was using positive
 
  • #7
ok i need help
initial energy is 0 for this system, final energy is
EFINAL = .5m1v2FINAL + .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
So

0 = .5m1v2FINAL - .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
solve for vFINAL
why is this not correct?
 
  • #8
isukatphysics69 said:
ok i need help
initial energy is 0 for this system, final energy is
EFINAL = .5m1v2FINAL + .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
So

0 = .5m1v2FINAL - .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
solve for vFINAL
why is this not correct?
Why the sign change on the second term?
 
  • Like
Likes isukatphysics69
  • #9
haruspex said:
Why the sign change on the second term?
sorry that is not what i meant i meant

0 = .5m1v2FINAL + .5m2v2FINAL -.5(16.2)(.2672)+((.481)(9.8)(-.267))

i meant to put negative spring potential energy sorry for confusion
 
  • #10
ok i just got the answer by doing something i was telling myself not to do
so it says speed. I was getting a negative value and i was going to square root it but knew i didn't want to square root a negative. since it says speed i decided to omit the negative and got the correct answer of 1.37m/s
 
  • #11
0 = .5m1v2FINAL - .5m2v2FINAL - .5(16.2)(.2672)+((.481)(9.8)(.267))

and solving for v gave the correct answer

couple of things here, i thought that the ((.481)(9.8)(.267)) at the end of the equation should have been ((.481)(9.8)(-.267)) because it was moving downward
and when i was doing the algebra i got a negative value under the square root. i had to multiply it by negative 1 to get a positive and then took the square root and got the correct answer. My gut was telling me not to do that for hours and i have tried this way earlier and saw a negative under the square root and instantly gave up and tried a different approach.
 
  • #12
isukatphysics69 said:
i meant to put negative spring potential energy
Why? The KE and spring PE are both gains in energy. The only energy lost is the GPE.
isukatphysics69 said:
ok i just got the answer by doing something i was telling myself not to do
so it says speed. I was getting a negative value and i was going to square root it but knew i didn't want to square root a negative. since it says speed i decided to omit the negative and got the correct answer of 1.37m/s
Your reasoning was correct. v2 should not have turned out negative.
I am mystified as to how you got the right answer from that equation.
Go back to your post #7, correct the sign on the m2v2 term and try again.
 
  • Like
Likes isukatphysics69
  • #13
isukatphysics69 said:
0 = .5m1v2FINAL - .5m2v2FINAL - .5(16.2)(.2672)+((.481)(9.8)(.267))

and solving for v gave the correct answer

Yes, the KE is always positive.
The GPE could be zero at the lowest level and positive at the initial position or begin with zero and negative later. But the SPE is positive because the spring is stretching and gaining energy.
 
  • Like
Likes isukatphysics69

Related to Find the max compression of the spring

1. What is the maximum compression of a spring?

The maximum compression of a spring is the point at which the spring has been compressed as much as it can be without permanently deforming or breaking.

2. How do you calculate the maximum compression of a spring?

The maximum compression of a spring can be calculated using Hooke's Law, which states that the force applied to a spring is directly proportional to the spring's displacement from its equilibrium position. The formula is F = kx, where F is the force, k is the spring constant, and x is the displacement.

3. What factors affect the maximum compression of a spring?

The maximum compression of a spring is affected by its material, length, diameter, and the amount of force applied to it. A stiffer spring with a higher spring constant will have a lower maximum compression, while a more flexible spring with a lower spring constant will have a higher maximum compression.

4. Why is it important to know the maximum compression of a spring?

Knowing the maximum compression of a spring is important for designing and using springs in various applications. It ensures that the spring will not be over-compressed and damaged, and allows for proper selection of springs for different tasks.

5. How can the maximum compression of a spring be increased?

The maximum compression of a spring can be increased by using a spring with a higher spring constant, or by stacking multiple springs together. However, it is important to note that increasing the maximum compression can also lead to a decrease in the spring's durability and lifespan.

Similar threads

Spring Potential Energy: Horizontal Spring Problem
    • Introductory Physics Homework Help
Replies
6
Views
1K
How High Will the Ball Bearing Rise When Released from a Compressed Spring?
    • Introductory Physics Homework Help
Replies
5
Views
3K
Find elastic energy in the compressed spring.
    • Introductory Physics Homework Help
Replies
12
Views
1K
Projectile Motion/ Explosion Mid air
    • Introductory Physics Homework Help
Replies
6
Views
4K
How to find the max. compression of a spring given a block on an incline?
    • Introductory Physics Homework Help
Replies
17
Views
4K
Spring Constant when compressed
    • Introductory Physics Homework Help
Replies
1
Views
2K
Time to Max Compression of a Spring
    • Introductory Physics Homework Help
Replies
2
Views
2K
A block of mass m is dropped onto the top of a vertical spring....
    • Introductory Physics Homework Help
Replies
8
Views
4K
Distance of box pushed up incline by spring
    • Introductory Physics Homework Help
Replies
1
Views
2K
Pan suspended by a spring (Energy + SHM)
    • Introductory Physics Homework Help
Replies
20
Views
2K

Hot Threads

Recent Insights

Back
Top