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isukatphysics69
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What about GPE?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?
Why the sign change on the second term?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?
sorry that is not what i meant i meantharuspex said:Why the sign change on the second term?
Why? The KE and spring PE are both gains in energy. The only energy lost is the GPE.isukatphysics69 said:i meant to put negative spring potential energy
Your reasoning was correct. v2 should not have turned out negative.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
isukatphysics69 said:0 = .5m1v2FINAL - .5m2v2FINAL - .5(16.2)(.2672)+((.481)(9.8)(.267))
and solving for v gave the correct answer
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.
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.
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.
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.
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.