Answer
(a) The maximum compression of the spring is 0.60 meters.
(b) The maximum value of $v_0$ is 1.50 m/s.
Work Step by Step
(a) The potential energy in the spring will be equal to the block's initial kinetic energy.
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$x^2 = \frac{mv^2}{k}$
$x = \sqrt{\frac{mv^2}{k}}$
$x = \sqrt{\frac{(5.00~kg)(6.00~m/s)^2}{500~N/m}}$
$x = 0.60~m$
The maximum compression of the spring is 0.60 meters.
(b) The potential energy in the spring will be equal to the block's initial kinetic energy.
$\frac{1}{2}mv_0^2 = \frac{1}{2}kx^2$
$v_0^2 = \frac{kx^2}{m}$
$v_0 = \sqrt{\frac{kx^2}{m}}$
$v_0 = \sqrt{\frac{(500~N/m)(0.150~m)^2}{5.00~kg}}$
$v_0 = 1.50~m/s$
The maximum value of $v_0$ is 1.50 m/s.
