Use the work-energy theorem to derive an expression for v^2

AI Thread Summary
The discussion centers on using the work-energy theorem to derive an expression for v^2 in a physics problem involving a spring and friction. Participants highlight confusion regarding the variables k (spring constant) and x (position), clarifying that x should be replaced with d, which represents distance. The initial equation presented is deemed incorrect due to this misunderstanding. Additionally, the importance of considering the energy stored in the spring both initially and after the mass moves is emphasized, pointing out the need to account for changes in energy. The conversation ultimately seeks to clarify the correct application of the work-energy theorem in this context.
David112234
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Homework Statement


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Homework Equations


Work energy theorem

The Attempt at a Solution


.5mv2 - .5mv2 = (kx - μmg)d
final velocity is 0

½mv2 =(kx - μmg)d
solve for v2

((kx - μmg)d2)m
When this was wrong I tried integrating Fx but it was still wrong
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I miss a question in this problem statement ?
I miss what ##k## means (but I can guess)
I also miss what ##x## means, which is very important. Note that there is a ##d## already.
 
BvU said:
I miss a question in this problem statement ?
I miss what ##k## means (but I can guess)
I also miss what ##x## means, which is very important. Note that there is a ##d## already.
K is the spring constant
X is supposed to be d, I am just used to using x for position, in answers 3-4 i changed it to d
 
David112234 said:
K is the spring constant
X is supposed to be d, I am just used to using x for position, in answers 3-4 i changed it to d
If x is d, your very first equation is wrong.
Remeber that there is energy in the spring both initially and after the mass has moved distance d. What is the change in that energy?
 
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