Sliding Block Attached to a Spring

[AltruistKnight]

Homework Statement

A block with mass 2.90kg is attached as shown to a spring with a force constant of 492.0N/m. The coefficient of kinetic friction between the block and the surface on which it slides is 0.170. The block is pulled 5.30cm to the right of its equilibrium position and then released from rest. What is the speed of the block as it passes by its equilibrium position?

Rough image:

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Homework Equations

PEspring=1/2kx^2
Fspring=kx
Ffriction=(mu)*Force Normal
Work=F*d

The Attempt at a Solution

First, I found the force of Friction by multiplying the coefficient of friction by the force normal of the block due to gravity ("ma=mg=2.9*9.81=28.449N). Then, I attempted to use W=F*d to find the work done by friction. I then decided to try finding Wspring=F*d=kx*d (Where "d"=x, thus it would be "x^2"), and from there assumed that the Wnet=Wspring-Wfriction. However, that doesn't appear to be the proper method. In particular, if someone could explain the relationship of Work to Kinetic Energy AS WELL AS Potential Energy, that would be very helpful.

 

[tiny-tim]

Hi AltruistKnight!

PEspring=1/2kx^2
…
First, I found the force of Friction by multiplying the coefficient of friction by the force normal of the block due to gravity ("ma=mg=2.9*9.81=28.449N). Then, I attempted to use W=F*d to find the work done by friction.

correct

I then decided to try finding Wspring=F*d=kx*d (Where "d"=x, thus it would be "x^2"),

no, W = F*d only works if F is constant

if it isn't, you use W = ∫ F dx, which in this case is -1/2 kx² (= -∆PE)

(btw, that was in your equations above … why didn't you use it? )

now carry on ​

In particular, if someone could explain the relationship of Work to Kinetic Energy AS WELL AS Potential Energy, that would be very helpful.

the work-energy theorem says work done = change in mechanical energy (ie change in KE + PE)

(and potential energy is defined as minus the work done by a conservative force)