What is the block's displacement when it first comes to rest again?

[PhysicsMan999]

Homework Statement

1. 
   A block with mass m = 1.46 kg is attached to a spring with spring constant k = 52.0 N/m and negligible mass. The coefficient of kinetic friction between the block and the table is µk = 0.349. The block is displaced from its equilibrium position to the left by 0.203 m and released from rest. Calculate the block's displacement when it first comes to rest again.

Homework Equations

Thermal energy=Ff * x
Us= 0.5kx^2

The Attempt at a Solution

Ff= MuN= (1.46 x 9.8)(0.349)=4.99 N
Us(initial)= 0.5(52)(-0.203^2)
Us(final)= 0.5(52)Xfinal
initial thermal and spring energy=final thermal and spring energy
(4.99)(-0.203)+26(0.203^2)=4.99x +26x^2
i tried to solve it, plugged it into a quadratic equation but am still getting the wrong answer.

Any ideas on what I am doing wrong? Thanks!

 

[Doc Al]

Thermal energy=Ff * x

Instead of calling it thermal energy, call it energy dissipated by friction. Note that x is the distance traveled by the mass.

Us= 0.5kx^2

OK. Here x is the displacement from equilibrium. Careful with notation!

initial thermal and spring energy=final thermal and spring energy
(4.99)(-0.203)+26(0.203^2)=4.99x +26x^2

Careful! You start and end with pure spring potential energy, the difference being the energy lost to friction.

 

[PhysicsMan999]

Okay, so now I got:
4.99 (0.203+x)=26x^2-26(0.203^2)
plugged everything in and still didn't get the correct answer.

 

[Doc Al]

Okay, so now I got:
4.99 (0.203+x)=26x^2-26(0.203^2)

The two terms on the right hand side are in the wrong order. You want initial U minus final U.

 

[PhysicsMan999]

Okay awesome! Thank you very much haha I had been going at this one for a good 45 minutes! I feel pretty confident I understand it now just needed a bit of guidance. Thanks again!