Answer
(a)$1.6Kg$
(b) greater than
(c) $0.63m/s$
Work Step by Step
(a) We can find the required mass as follows:
$W=\frac{1}{2}Kx^2=\Delta K.E$
$\frac{1}{2}Kx^2=\frac{1}{2}mv_f^2-0$
This can be rearranged as:
$m=\frac{Kx^2}{v_f^2}$
We plug in the known values to obtain:
$m=\frac{(750)(0.041)^2}{(0.88)^2}$
$m=1.6Kg$
(b) We know that
$v_f\propto \frac{1}{\sqrt m}$. This shows that if the mass is doubled then the speed is decreased by a factor of $\sqrt 2$. Thus, the final speed will be greater than $0.44m/s.$
(c) We can find the required final speed as follows:
$v_f=\sqrt {\frac{Kx^2}{m}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{(750)(0.041)^2}{2\times 1.6}}$
$v_f=0.63m/s$
