Hooke's law with a spring

1. bearhug

79
When a 4.00 kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5 cm.
(a) If the object is replaced with one of mass 1.45 kg, how far will the spring stretch?

F=ma
-ma=-kx
-(4)(9.8)=-k(0.025)
k=1568

-(1.5)(9.8)= -1568x
x=0.00938m
I keep being told that this is wrong and yet I have not been able to figure out where I went wrong. Can someone point out what the problem is.

b) how much work must an external agent do (i.e. a force coming from the 'environment'), to stretch the same spring 4.9 cm from its unstretched position?
Here I used k=1568 and W=1/2kxi^2- 1/2kxf^2
=1/2(1568)(0)^2-1/2(1568)(0.049)^2
W=1.25J or -1.25 J
Once again I'm being told that this is wrong and would like for someone to point out the problem.

Staff: Mentor

I don't see any fundamental problem here (except for your misuse of "F = ma"--see below). But why did you use 1.5 instead of the given 1.45 kg? (You can also solve this using ratios.)

One thing to point out: You are not using Newton's 2nd law here (F = ma), but you are using Hooke's law (F = kx). In this case, the F is the force that is stretching the spring--which is the weight of the hanging mass: w = mg.

I only see two problems: (1) Your signs are off. Note that change in spring energy (or anything) is always final - initial. The work done is positive. (2) Check your arithmetic.

Last edited: Oct 2, 2006
3. BishopUser

165
technicaly hooke's law is F= -kx and the work done by a spring is infact i-f.

Last edited: Oct 2, 2006

Staff: Mentor

True, but here we are finding the work done by an external agent in stretching the spring--which equals the change in the spring's energy--which is "final energy - initial energy". (The minus sign in Hooke's law tells you the direction of the spring's restoring force, which is opposite to the force applied by the external agent.)