Find the maximum elastic potential energy of the spring

Is that what the question was asking for?Yes, exactly! The question is asking for the maximum values of each quantity, given the specific conditions and values provided.
  • #1
TheShapeOfTime
A 2.0 kg mass is pressed against a spring (k = 800N/m) such that the spring has been compressed 0.22 m. The spring is released and the mass moves along a horizontal frictionless surface and up a frictionless slope. Calculate:
a) the maximum elastic potential energy of the spring
b) the maximum velocity of the mass
c) the maximum vertical height the mass will travel up the slope

This is what I've done so far (I'm not sure if it's correct):
[tex]E_p spring = \frac{1}{2}kx^2[/tex]
[tex]= \frac{1}{2} \cdot 800 \cdot 0.22^2[/tex]
[tex]= 19.36J[/tex]

[tex]E_p spring = E_k mass[/tex]
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]
[tex]v = 4.4m/s^2[/tex]

Few notation questions:

What is the appropriate notation for [itex]E_p spring[/itex]?
Is it ok to use "[itex]\cdot[/itex]" in place of the regular multiplication sign whereever?
 
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  • #2
What you've done so far looks correct to me as well, as long as you understand why they're using the adjective "maximum" to describe the quantities you're calculating. :)
The [tex]\cdot[/tex] is fine to use between numbers in R, as the multiplication operator satisfies the definition of the dot product on R. You can use Us to represent potential energy of the spring if you want to use a common potential energy symbol, but as long as you define your notation and its not unnecessarily convoluted, it's fine to use. :)
 
  • #3
Thanks for checking over my work! I hadn't completed (c) because I wasn't sure how to do it, but I found out today it was just me forgetting about one formula I had:

[tex]E_p = mgh[/tex]
[tex]h = \frac{E_p}{mg}[/tex]
[tex] = \frac{19}{2.0 \cdot 9.80}[/tex]
[tex] = 0.97m[/tex]

Could you elaborate a bit on the adjective "maximum" and why it's used? Why might there be lesser values than the ones I calculated with these formula's?
 
  • #4
TheShapeOfTime said:
Could you elaborate a bit on the adjective "maximum" and why it's used? Why might there be lesser values than the ones I calculated with these formula's?
The quantities in question (spring PE, speed of mass, height up the slope) are not constants. For example, since spring PE is [itex]1/2k x^2[/itex], it varies from zero to some maximum value (at [itex]x = x_{max}[/itex]).
 
  • #5
Oh, so they were asking for the maximum EP, etc. with the values they provided?
 

1. What is elastic potential energy?

Elastic potential energy is the energy stored in a stretched or compressed elastic material, such as a spring. It is the potential energy that is stored due to the deformation of the material.

2. How is the elastic potential energy of a spring calculated?

The elastic potential energy of a spring can be calculated using the formula E = 1/2kx^2, where E is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

3. What is the maximum elastic potential energy of a spring?

The maximum elastic potential energy of a spring is the highest amount of energy that the spring can store when it is stretched or compressed to its maximum limit, also known as its elastic limit.

4. How does the mass of an object affect the maximum elastic potential energy of a spring?

The mass of an object does not directly affect the maximum elastic potential energy of a spring. However, the displacement of the spring will be greater with a heavier object, resulting in a higher elastic potential energy.

5. How can the maximum elastic potential energy of a spring be increased?

The maximum elastic potential energy of a spring can be increased by increasing the spring constant, which can be achieved by using a stiffer material or by making the spring longer. It can also be increased by increasing the displacement of the spring.

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