Conservation of Energy: Spring pushing a block up an incline

In summary, the problem involves a 250 gram block on a 10 degree frictionless incline and in contact with a spring of constant 1.2 N/cm. The block is launched from rest by the spring with an initial compression of 6cm. To find the velocity of the block at the point of release and 20cm away from the release, as well as the distance up the incline the block will slide before stopping, the initial energy of the spring must be considered along with the kinetic and potential energy. The formula v^2=2k/m is used, but the spring potential energy is being overlooked. Drawing a careful diagram and labeling each position while taking into account all three forms of energy will lead to a
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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.
 
  • #8
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?
 
  • #9
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?
 

What is the basic concept of conservation of energy?

The basic concept of conservation of energy is that energy cannot be created or destroyed, but it can be transferred from one form to another.

How does a spring pushing a block up an incline demonstrate conservation of energy?

When a spring pushes a block up an incline, the potential energy of the spring is converted into kinetic energy of the block as it moves up the incline. At the top of the incline, the kinetic energy is converted back into potential energy. This demonstrates the conservation of energy, as the total energy remains constant throughout the process.

What are the different forms of energy involved in this scenario?

In this scenario, the forms of energy involved are potential energy (stored in the spring and the block's position), kinetic energy (due to the motion of the block), and thermal energy (due to friction between the block and the incline).

What factors affect the conservation of energy in this scenario?

The conservation of energy in this scenario can be affected by the amount of force applied by the spring, the mass of the block, the angle of the incline, and the presence of any external factors such as friction.

How is the conservation of energy related to the law of conservation of mass?

The law of conservation of mass states that matter cannot be created or destroyed, only transformed. This is closely related to the conservation of energy, as energy and mass are interchangeable and cannot be created or destroyed. Therefore, the total amount of energy and mass in a closed system remains constant.

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