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Hooke's law with a spring 
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#1
Oct206, 05:46 PM

P: 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)^21/2(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. Thanks in advance 


#2
Oct206, 06:52 PM

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P: 41,310

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 springwhich is the weight of the hanging mass: w = mg. 


#3
Oct206, 07:03 PM

P: 165




#4
Oct206, 07:12 PM

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P: 41,310

Hooke's law with a spring



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