Answer
(a) $v = 2.21~m/s$
(b) The spring is compressed a distance of $20.9~cm$
(c) The block rises up to its original height of $0.50~m$
Work Step by Step
(a) We can use conservation of energy to find the speed when the block is at a height of $0.25~m$:
$\frac{1}{2}mv^2+mgh_2 = mgh_1$
$v^2 = 2g~(h_1-h_2)$
$v = \sqrt{2g~(h_1-h_2)}$
$v = \sqrt{(2)(9.80~m/s^2)~(0.50~m-0.25~m)}$
$v = 2.21~m/s$
(b) By conservation of energy, the elastic potential energy stored in the spring will be equal to the initial gravitational potential energy:
$\frac{1}{2}kx^2 = mgh$
$x^2 = \frac{2mgh}{k}$
$x = \sqrt{\frac{2mgh}{k}}$
$x = \sqrt{\frac{(2)(2.0~kg)(9.80~m/s^2)(0.50~m)}{450~N/m}}$
$x = 0.209~m = 20.9~cm$
The spring is compressed a distance of $20.9~cm$
(c) When the block goes back to the left, by conservation of energy, the block rises up to its original height of $0.50~m$
