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bearhug
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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.
Thanks in advance
(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