Answer:
c) 359%
Explanation:
k = spring constant of the spring
x₁ = initial stretch of the spring = 7 cm = 0.07 m
x₂ = final stretch of the spring = 15 cm = 0.15 m
Percentage increase in energy is given as
[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{((0.5)k{x_{2}}^{2} - {(0.5)k x_{1}}^{2})(100)}{(0.5)k{x_{1}}^{2}}[/tex]
[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({x_{2}}^{2} - {x_{1}}^{2})(100)}{{x_{1}}^{2}}[/tex]
[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({0.15}^{2} - {0.07}^{2})(100)}{{0.07}^{2}}[/tex]
[tex]\frac{\Delta U \times 100}{U_{1}}[/tex] = 359%
