#### Answer

(a) The rock leaves the spring with a speed of 15 m/s
(b) The contestants will go hungry.

#### Work Step by Step

(a) The rock's kinetic energy when it leaves the spring will be equal to the energy $U_s$ stored in the spring initially. We can find the speed of the rock.
$KE = U_s$
$\frac{1}{2}mv^2=\frac{1}{2}kx^2$
$v^2=\frac{kx^2}{m}$
$v=\sqrt{\frac{kx^2}{m}}$
$v=\sqrt{\frac{(1000~N/m)(0.30~m)^2}{0.400~kg}}$
$v = 15~m/s$
The rock leaves the spring with a speed of 15 m/s.
(b) The potential energy of the rock at maximum height will be equal to the rock's kinetic energy when it leaves the spring. We can find the maximum height $h$ reached by the rock.
$PE = KE$
$mgh = \frac{1}{2}mv^2$
$h = \frac{v^2}{2g}$
$h = \frac{(15~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 11.5~m$
Since the rock's maximum height is less than the 15 meters required to reach the fruit, the contestants will go hungry.