Posted by Mary on .
Can you please clarify your response to my post Wednesday April 4th at 8:53pm and 8:54pm.
Thanks for all of your help!!!
pleasde repost it.
Yes, repost it. Go to it, copy, and paste it here. Thanks.
A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 3.0 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?
For Further Reading
Physics - bobpursley, Wednesday, April 4, 2007 at 9:07pm
The spring compression had 1/2 k x^2 of work done on it. Assuming no losses to friction, then the energy that went into it was mg(h+x). Calculate h
So what is the question?
I lead you to mg(h+x)= 1/2 kx^2
solve for h. If questions, you have to ask specific questions.
Please tell me where I went wrong.
mg (h+x) = 1/2 K x^2
h = 1/2 (450)(3.0)^2/0.30kg x 9.81
h = 687.054
Please disregard. I figured it out. Thanks!