Conservation of Energy: Spring pushing a block up an incline

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SUMMARY

The discussion centers on a physics problem involving a 250-gram block on a 10-degree frictionless incline, launched by a spring with a constant of 1.2 N/cm and an initial compression of 6 cm. Participants emphasize the importance of considering all forms of mechanical energy, including kinetic energy (KE), gravitational potential energy (PE), and spring potential energy, to solve for the block's velocity at various points and its maximum height on the incline. Key equations include KE = 1/2mv² and PE = mgh, with a focus on accurately applying these concepts to derive the correct solutions.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) equations
  • Familiarity with spring constant and spring potential energy
  • Basic knowledge of trigonometry, particularly cosine functions
  • Ability to draw and interpret free body diagrams
NEXT STEPS
  • Study the work-energy principle in the context of spring systems
  • Learn how to calculate spring potential energy using the formula PE_spring = 1/2kx²
  • Explore the application of trigonometric functions in physics problems involving inclines
  • Practice solving similar problems involving conservation of energy in mechanical systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their teaching strategies for energy conservation concepts.

CrazyCatMom

Homework Statement


"Consider a 250 gram block on a 10 degree frictionless incline and in contact with a spring of constant 1.2 N/cm. If the block is launched from rest by the spring with an initial compression of 6cm, how fast is the block moving at the point of release from the spring? How fast is the block moving at 20cm away from the release? How far up the incline will the block slide before stopping?

Homework Equations


KE = 1/2mv^2
PE = mgh

The Attempt at a Solution


First I converted everything to kg and m to make the answer easier. Then I took the equation for KE and rearranged it to be v^2= 2k/m, but the answer I got wasn't correct. I feel like I am almost there, or at least maybe on the right track, but I'm not sure. I've been working on this for a while and could really use the help. Thanks in advance!
 
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CrazyCatMom said:
rearranged it to be v^2= 2k/m
Do you mean you divided k by the mass? What dimensions or units would that yield?
Also, you do not seem to have considered the incline.
 
I think I'm just kind of grasping at straws and hoping something will come of it. I really don't know how to approach the problem.
 
CrazyCatMom said:
I think I'm just kind of grasping at straws and hoping something will come of it. I really don't know how to approach the problem.
Please post the details of the attempt you described.
 
Don't look for a formula for v. First, draw a diagram. In the diagram, show all the four different positions mentioned in the statement of the problem. At each position, write down the total mechanical energy. Then you will yourself see how to solve the problem.
 
CrazyCatMom said:

Homework Statement


"Consider a 250 gram block on a 10 degree frictionless incline and in contact with a spring of constant 1.2 N/cm. If the block is launched from rest by the spring with an initial compression of 6cm, how fast is the block moving at the point of release from the spring? How fast is the block moving at 20cm away from the release? How far up the incline will the block slide before stopping?

Homework Equations


KE = 1/2mv^2
PE = mgh
You ignore the spring energy.
CrazyCatMom said:

The Attempt at a Solution


First I converted everything to kg and m to make the answer easier. Then I took the equation for KE and rearranged it to be v^2= 2k/m, but the answer I got wasn't correct.
The initial spring energy is converted to kinetic energy and potential energy. What is the initial energy of the spring if it is compressed by 6 cm?
 
haruspex said:
Please post the details of the attempt you described.

I took the equation KE = (1/2)mv^2
Since I know I need to find the velocity, I divided both sides by (1/2)m and got the formula v^2=2k/m
I know the velocity is going to be in cm/s, so those were the units I was aiming towards.
 
The initial spring energy is converted to kinetic energy and potential energy. What is the initial energy of the spring if it is compressed by 6 cm?[/QUOTE]

If, I use the W=fd formula, would it mean I use 1.2 N/cm for the Force and then 6cm for the Distance?
So, W=(1.2)(6)
7.2 J?
 
CrazyCatMom said:
got the formula v^2=2k/m
Where k is what? According to your description of what you did, that k would be the KE, but it would be more usual to use k for the spring constant.
And, as I wrote, you are neglecting the slope. As Chandra Prayaga instructed, consider all the mechanical energy when the system is initially at rest, and again when the spring is fully decompressed.
 
  • #10
As ehild pointed out, you are completely overlooking the spring potential energy in your attempts. You just used the first formula in front of you without looking at all the aspects. You also seem to be unclear about what the symbols mean. I sincerely suggest that you draw a careful diagram. You have three different forms of mechanical energy in your problem. The kinetic energy, the spring potential energy, and the gravitational potential energy. In the equations that you wrote, I don't see an expression for the spring potential energy. The sum of all three forms of energy must remain constant, in the absence of friction. Really that is the clue. There is not a single formula that you can manipulate, plug numerical values in, and get the answer.
 
  • #11
Chandra Prayaga said:
you are completely overlooking the spring potential energy
Not if that is what CCM means by "k".
 
  • #12
haruspex said:
Not if that is what CCM means by "k".

In all honesty, I don't know what it is. I thought it was just a place holder for the compression of the spring.
 
  • #13
CrazyCatMom said:
In all honesty, I don't know what it is. I thought it was just a place holder for the compression of the spring.
So where did you get this k from? It does not appear in the relevant equations you listed.
 
  • #14
If that is so, then CCM is overlooking the gravitational potential energy. In either case, the best way to learn how to solve the problem is by drawing a careful diagram, labeling each position and writing down the energy terms (all three of them) at each position. The solution will then fall out very simply.
CrazyCatMom, in your last post, you were mentioning W = f.d. You cannot use that formula in the case of a spring force. Also, the value of k = 1.2 N/cm is not a force. It is the spring constant. In the case of a spring, the simple formula of W = f.d will not work because the spring force is not a constant. You should look up the expression for the work done by a spring.
 
  • #15
Chandra Prayaga said:
If that is so, then CCM is overlooking the gravitational potential energy. In either case, the best way to learn how to solve the problem is by drawing a careful diagram, labeling each position and writing down the energy terms (all three of them) at each position. The solution will then fall out very simply.
CrazyCatMom, in your last post, you were mentioning W = f.d. You cannot use that formula in the case of a spring force. Also, the value of k = 1.2 N/cm is not a force. It is the spring constant. In the case of a spring, the simple formula of W = f.d will not work because the spring force is not a constant. You should look up the expression for the work done by a spring.

I drew out a free body diagram, so I know I have to use cos(10) somewhere in the problem, but I'm not sure where.
I guess this just isn't clicking for me.
 
  • #16
CrazyCatMom said:
I drew out a free body diagram, so I know I have to use cos(10) somewhere in the problem, but I'm not sure where.
I guess this just isn't clicking for me.
Take the height as zero at the initial position. Please post attempted answers to each of the following six questions.

In the initial position, what is the spring PE, what is the gravitational PE, and what the KE?
When the spring reaches zero compression, what is the spring PE and what is the GPE? If the speed is now v, what is the KE?
 

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