Answer
(a) a factor of 4
(b) $3.85J$
Work Step by Step
(a) We know that when the mass attached to the spring is doubled, the spring stretch length will also be doubled and as a result the potential energy increases by a factor of 4.
(b) We can find the required spring potential energy as follows:
$\frac{U_2}{U_1}=\frac{\frac{1}{2}Kx_2^2}{\frac{1}{2}x_1^2}$
$\implies \frac{U_2}{U_1}=\frac{(2x_1)^2}{x_1^2}=4$
Now $U_2=4U_1$
$U_2=4(0.962J)=3.85J$
