Answer:
Energy needed = 54.02 J
Explanation:
the Energy in an elastic spring from hookes law is given as
F= ke , therefore the energy (E) is
E = [tex]\frac{1}{2} Ke^{2}[/tex]
K = 19.5 N/cm
e = 1.39cm
E = [tex]\frac{1}{2}[/tex] x 19.5 x 1.39
E = 13.55 J
The energy to stretch the spring for 6.93cm is
E = [tex]\frac{1}{2}[/tex] x 19.5 x 6.93
E = 67.57 J
The more energy needed for the further stretch is
67.57 - 13.55
Energy needed = 54.02 J
