How is Mechanical Energy Conserved in a Spring-Block System?

[ajmCane22]

Homework Statement

A 1.10 kg block slides with a speed of 0.885 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of 500 N/m. The block comes to rest after compressing the spring 4.15 cm.

Homework Equations

Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of (a) 0 cm, (b) 1.00 cm, (c) 2.00 cm, (d) 3.00, (e) 4.00 cm.

The Attempt at a Solution

For the potential energy of the spring I used 1/2kx^2 to get 0.431 J, and for the kinetic energy of the block I used 1/2mv^2 to get 0.431 J. So wouldn't the total mechanical energy of the system be:
PEspring + KEblock = 0.431 + 0.431 = 0.862 J? I tried that but am getting the answer wrong.

Please help

 

[rock.freak667]

For compression, you might need to put the work done by the spring as negative. Since I think the convention is that the work done by a spring in extension is positive.

 

[ajmCane22]

So then the total work would be zero?

 

[rock.freak667]

So then the total work would be zero?

Actually they told you that when x = 4.15 cm, v=0, so at that point E_total=PE. So it would just be the 0.431 J. At any other point before the maximum compression of 4.15 cm, E_total= KE + PE. (It is this at any point, it is just at the max compression, KE=0)

 

[ajmCane22]

I was making it so much hard than it needed to be. I finally got it. Thank you!