# Homework Help: Block on Spring without Friction

1. Dec 1, 2004

### 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??

2. Dec 1, 2004

### 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 .