Find the maximum elastic potential energy of the spring

AI Thread Summary
The discussion focuses on calculating the maximum elastic potential energy of a spring, maximum velocity of a mass, and maximum height traveled up a slope after the spring is released. The elastic potential energy is correctly calculated as 19.36 J using the formula E_p = 1/2 k x^2. The maximum velocity of the mass is derived as 4.4 m/s, based on energy conservation principles. The maximum height achieved by the mass on the slope is determined to be 0.97 m using the formula E_p = mgh. The term "maximum" is emphasized as the potential energy and other quantities vary, reaching their peak values at specific conditions.
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):
E_p spring = \frac{1}{2}kx^2
= \frac{1}{2} \cdot 800 \cdot 0.22^2
= 19.36J

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

Few notation questions:

What is the appropriate notation for E_p spring?
Is it ok to use "\cdot" in place of the regular multiplication sign whereever?
 
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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 \cdot 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. :)
 
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:

E_p = mgh
h = \frac{E_p}{mg}
= \frac{19}{2.0 \cdot 9.80}
= 0.97m

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?
 
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 1/2k x^2, it varies from zero to some maximum value (at x = x_{max}).
 
Oh, so they were asking for the maximum EP, etc. with the values they provided?
 
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