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(Solved): Part A IP A vertical spring stores 0.930 J in spring potential energy when a 3.5-kg mass is suspende ...



Part A IP A vertical spring stores 0.930 J in spring potential energy when a

3.5-kg

mass is suspended from it. By what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?

(U_(2))/(U_(1))=(1)/(2)

(U_(2))/(U_(1))=(1)/(\sqrt(2))

(U_(2))/(U_(1))=\sqrt(2)

(U_(2)^(**))/(U_(1))=2

(U_(i))/(U_(1))=4

(U_(2))/(U_(1))=8


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