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.(adsbygoogle = window.adsbygoogle || []).push({});

(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.

Thanks in advance

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# Homework Help: Hooke's law with a spring

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