Answer:
the answer is B
Explanation:
we assume that all of these springs have tha same k , the spring constant
elastic energy for the initial system = Ep1= [tex]\frac{kd^{2} }{2}[/tex]
for the new changed system , the spring constant = k1+k2 = k+k = 2k , but the displacement is also reduced to 0.5 d
so Ep2 = [tex]\frac{2k(0.5d)^{2} }{2} = \frac{kd^2}{4}[/tex]
so the new total Ep stored in the new system is 1/2Ep
hope it helps
