Answer
(a) $3200\frac{N}{m}$
(b) $0.40\frac{m}{s}$
Work Step by Step
According to law of conservation of energy
$K.E_i+U_i=K.E_f+U_f$
$\implies \frac{1}{2}mv_i^2+0=0+\frac{1}{2}kx_{max}^2$
This can be rearranged as:
$K=\frac{mv_i^2}{x_{max}^2}$
We plug in the known values to obtain:
$K=\frac{(2.9)(1.6)}{(0.048)^2}$
$K=3200\frac{N}{m}$
(b) We can find the required initial speed as
$v_i=\sqrt{\frac{kx_{max}^2}{m}}$
We plug in the known values to obtain:
$v_i=\sqrt{\frac{(3200)(0.012)^2}{2.9}}$
$v_i=0.40\frac{m}{s}$
