When a 4.40-kg object is hung vertically on a certain light spring that obeys Ho

Question

When a 4.40-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.20 cm. (a) If the 4.40-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?
1 cm

(b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?
2 J (a) If the 4.40-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?
1 cm

(b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?
2 J

Explanation / Answer

a)

F=mg =4.4*9.8 =43.12 N

K=F/x =43.12/0.022

K=1960 N/m

now

F=1.5*9.8 =14.7 N

X=F/K =14.7/1960

X=0.75 cm

b)

W=(1/2)KX^2=(1/2)*1960*0.04^2

W=1.568 J

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