Block on Spring without Friction

[Mivz18]

I'm having a bit of trouble with the second part of this problem:

A spring is stretched a distance of Dx = 40 cm beyond its relaxed length. Attached to the end of the spring is an block of mass m = 12 kg, which rests on a horizontal frictionless surface. A force of magnitude 35 N is required to hold the block at this position. The force is then removed.

a) When the spring again returns to its unstretched length, what is the speed of the attached object?

This I calculated to be 1.08 m/s by KEf - KEi = Wspring.

b) When the spring has returned only halfway (20 cm), what is the speed of the attached object?

This is where my trouble appears. I simply change my number 0.4 m to 0.2 m and get 0.7638 m/s . However, the online program I'm using says that is incorrect. Isn't that what you should do is change that number? Here is my work so far for this:

KEf - KEi = Wspring
1/2 * m*v^2 - 1/2 * m*v^2 = 1/2 * k*x^2
(1/2)(12 kg)(Vfinal ^2) - 0 = (1/2)k(0.20 m)^2
k = (spring force)/x
Since the spring force is equal and opposite to the force applied, it is 35 N.
Therefore, k = 35/0.20 , which therefore = 175 .
(1/2)(12 kg)(Vfinal ^2) = (1/2)(175)(0.20 m)^2
Then I solve for Vfinal and get my answer. However, it is wrong. Am I doing something wrong??

 

[Mivz18]

Nevermind, lol, I once again found my error. I realized that part A can be used in the determination of part B. Therefore, the Initial kinetic energy would contain the velocity I'm looking for while the Final Kinetic energy would contain the final velocity of 1.08. Therefore, the k constant would remain 87.5 and x would change to 0.20 m like I had originally thought. Thus, you would obtain 0.935 m/s .

 

[wais]

It looks like you are on the right track and your calculations are correct. However, there may be a mistake in the online program or in the way the question is being interpreted. It is possible that the question is asking for the speed of the object when the spring has returned to its unstretched length, rather than halfway. In that case, the answer would be the same as part a) and your calculation of 1.08 m/s would be correct. It's always a good idea to double check the wording of the question and make sure you are interpreting it correctly. If you are still unsure, you can always try reaching out to your teacher or instructor for clarification.