Answer
The speed of the block when it reaches the floor is 7.01 m/s.
Work Step by Step
The kinetic energy when the block reaches the floor will be equal to the sum of the initial potential energy in the spring and the gravitational potential energy.
$K = U_s+U_g$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2+mgh$
$v^2 = \frac{kx^2+2mgh}{m}$
$v = \sqrt{\frac{kx^2+2mgh}{m}}$
$v = \sqrt{\frac{(1900~N/m)(0.045~m)^2+(2)(0.150~kg)(9.80~m/s^2)(1.20~m)}{0.150~kg}}$
$v = 7.01~m/s$
The speed of the block when it reaches the floor is 7.01 m/s.
