Answer
(a) v = 0.480 m/s
(b) The maximum speed is 0.566 m/s.
Work Step by Step
(a) $K+U_s = U_g$
$K = U_g - U_s$
$\frac{1}{2}mv^2= mgy - \frac{1}{2}ky^2$
$v^2= \frac{2mgy - ky^2}{m}$
$v= \sqrt{\frac{2mgy - ky^2}{m}}$
$v= \sqrt{\frac{(2)(3.00~kg)(9.80~m/s^2)(0.0500~m) - (900~N/m)(0.0500~m)^2}{3.00~kg}}$
$v = 0.480~m/s$
(b) The maximum speed occurs when the upward force of the spring is equal to the weight of the fish.
$ky= mg$
$y = \frac{mg}{k} = \frac{(3.00~kg)(9.80~m/s^2)}{900~N/m}$
$y = 0.03267~m$
$K+U_s = U_g$
$K = U_g - U_s$
$\frac{1}{2}mv^2= mgy - \frac{1}{2}ky^2$
$v^2= \frac{2mgy - ky^2}{m}$
$v= \sqrt{\frac{2mgy - ky^2}{m}}$
$v= \sqrt{\frac{(2)(3.00~kg)(9.80~m/s^2)(0.03267~m) - (900~N/m)(0.03267~m)^2}{3.00~kg}}$
$v = 0.566~m/s$
The maximum speed is 0.566 m/s.
