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?
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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
DrRuss/bobpursley/drwls please help!!! - bobpursley, Friday, April 6, 2007 at 4:27pm
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.
DrRuss/bobpursley/drwls please help!!! - Mary, Friday, April 6, 2007 at 8:41pm
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
Thnaks for your help but I figured it out. I was suppose to convert it to m first then change the answer back to cm. Thanks again!
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