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Posted by on Tuesday, September 27, 2011 at 7:08pm.

When a 3.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.07 cm.
(a) What is the force constant of the spring?
N/m

(b) If the 3.00-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?
cm

(c) How much work must an external agent do to stretch the same spring 9.10 cm from its unstretched position?
J

  • Physics - , Tuesday, September 27, 2011 at 7:22pm

    k=force/stretch= 3g/.0307 in N/m

    b. x=force/k=1.5/k I bet it is half of 3.07

    c. work= 1/2 k x^2 watch units, x has to be in m.

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