Energy/Work Q: Find Speed of Block in Frictionless & 0.350 Coeff Surfaces

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The discussion revolves around a physics problem involving a block attached to a spring, focusing on the concepts of work, kinetic energy (KE), potential energy (PE), and friction. The user seeks clarification on the equations used to calculate the speed of the block as it passes through equilibrium under two scenarios: one with a frictionless surface and another with a coefficient of friction of 0.350. There is confusion regarding the application of work-energy principles, particularly how work done by friction relates to the total energy of the system. The user suggests that in a frictionless scenario, the total energy is conserved, equating the kinetic energy to the potential energy stored in the spring. The thread highlights the importance of understanding the relationship between work, energy, and friction in solving such physics problems.
jared bernstein
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Hey guys I am in intro physics in college and took physics in high school. I have a general question about work/energy. In high school we used work= change in total energy and total energy = PE +KE +Q (Internal) which =work. In class there was a problem that read
2.00 kg block is attached to a spring of force constant 500 N/m. The block is pulled 4.50 cm to the right of equilibrium and released from rest.

(a) Find the speed of the block as it passes through equilibrium if the horizontal surface is frictionless.

(b) Find the speed of the block as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is 0.350

What I don't understand is that my teacher said KE + PE +PEs =Wf (which is work of friction) but in high school (the equation above the friction or Q is on the other side and she says that the friction is equal to W.

Can anyone explain!

Or simply do the problem out for me I am getting mixed messages!

thanks

first post btw
 
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At question b) is it right equation?
Maybe we can ignore PE for horizontal spring. We can write: KE+PEs+Wf= konstan.

@ is it your theory right? I Think: work= the change of KE or work= the change of PE. Sorry if I am wrong.
 
At question a). The problem said frictionless. I think, in the equilibrium, the KE =total energy. And at the initial position (pullet 4.50cm) the PE (from the spring)= total energy. So, we get: 1/2 m v² = 1/2 k x².
 
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