Answer
(a) The maximum speed is 5.81 m/s, which occurs at the point of equilibrium.
(b) The maximum compression of spring 1 is 15.0 cm.
Work Step by Step
(a) The maximum speed occurs as the block reaches the point of equilibrium. At this point, all the potential energy in the springs has been converted into kinetic energy.
$K = U_1+U_2$
$\frac{1}{2}mv^2 = \frac{1}{2}k_1x^2+\frac{1}{2}k_2x^2$
$v^2 = \frac{k_1x^2+k_2x^2}{m}$
$v = \sqrt{\frac{k_1x^2+k_2x^2}{m}}$
$v = \sqrt{\frac{(2500~N/m)(0.15~m)^2+(2000)(0.15~m)^2}{3.00~kg}}$
$v = 5.81~m/s$
The maximum speed is 5.81 m/s, which occurs at the point of equilibrium.
(b) The maximum compression of spring 1 is 15.0 cm, at which point the potential energy in the two springs is equal to the initial potential energy in the springs.
