Answer
(a) $0.948m/s$
(b) $1.29Kg$
Work Step by Step
(a) We can find the required speed as
$v_f=\sqrt{\frac{K}{m}x_f^2}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{955}{1.70}(-0.0400)^2}$
$v_f=0.948m/s$
(b) We know that
$m=K \frac{x_f^2}{v_i^2}$
We plug in the known values to obtain:
$m=(955)\frac{(-0.0400)^2}{(1.09)^2}$
$m=1.29Kg$
