How come I can't use m(v^2/r)=kx for this problem?

[kabailey]

Relaxed spring is sitting on a horizontal surface. A block is attached at one of its ends is kicked with a horizontal velocity, v1, given to it. The block will move and stretch. Find the distance, x, the spring will stretch. X is in meters.




Energy 1/2 mv1^2 = 1/2 mv2^2 + 1/2 k x^2
Angular momentum mv1l0 = mv2(l0+x)



The problem was solved using kinetic energy and conservation of angular momentum formulas.

My question is, why doesn't m(v^2/r)=kx apply for this problem? Is it because m(v^2/r) is to only be used in uniform circular motion and the object in the problem moves in the shape of an ellipse?

 

[cepheid]

I have no idea what path this thing moves on, but if the string stretches, then it can't be a circle.

 

[kabailey]

I have no idea what path this thing moves on, but if the string stretches, then it can't be a circle.

understood, but does m(v^2/r) only apply to a circle?

 

[Steely Dan]

understood, but does m(v^2/r) only apply to a circle?

Yes.

 

[kabailey]

Yes.

thank you!