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I only need help in part b.I use (mgx)(.5kx^2)=.5mv^2 and my answter is wrong help me figure out where im wrong please.?
A 1.0-kg object is suspended from a vertical spring whose spring constant is 112 N/m.
(a) Find the amount by which the spring is stretched from its unstrained length.
.0875 m
(b) The object is pulled straight down by an additional distance of 0.18 m and released from rest. Find the speed with which the object passes through its original position on the way up.
1 Answer
- Andrew SmithLv 76 years ago
We can only help you find out WHERE you are wrong if you show us what you have done.
Merely presenting us with the question isn't any use.
however from your info you include mgx. And this should not be done.
If you include mgx as energy you also need to work from the rest length of the spring and this makes the problem more difficult.
If you work on NET force instead then you get
1/2 k x^2 = 1/2 m v^2
k x^2 = m v^2
v = sqrt( k x^2 / m)
= sqrt(112 * 0.18^2 / 1)
=1.9 m/s
If you were foolish enough to want to use gravity then you would need to do
mg(L+x) + 1/2 k(L+x)^2 = mg(L) + 1/2 k L^2 + 1/2 m v^2
Where L is the rest length (0.0875) and x is the extension from that (0.18)
As you can see we have turned an easy problem into a harder one.