When a 4.00 kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.50 cm. (a)If the object is replaced with one of mass 1.50 kg, how far will the spring stretch?(adsbygoogle = window.adsbygoogle || []).push({});

b) How much work must an external agent do (i.e. a force coming from the 'environment'), to stretch the same spring 4.00 cm from its unstretched position?

For a) I used the equation F= -kx to solve for k.

F=ma so ma=-kx since the object is vertical.

(4.00)(-9.8)= -k(.025m)

k= 1568 N/m

then I plugged it back in to solve for x

ma= -kx

(1.5)(-9.8)= -1568x

x= 0.00938 m

Is there something I'm doing wrong?

b) W= 1/2k(xi)^2 - 1/2k(xf)^2

W= 0-1/2(1568)(.04m)^2

= -1.25 J

The homework says this is wrong but I can't figure out what else to do?

Any help would be appreciated.

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# Homework Help: Hooke's law with spring

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