Answer
(a) The speed when the block loses contact with the spring is 0.747 m/s.
(b) The maximum speed of the block is 0.930 m/s.
Work Step by Step
(a) $K = U_s + W_f$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mg~\mu_k~x$
$v^2 = \frac{kx^2-2mg~\mu_k~x}{m}$
$v = \sqrt{\frac{kx^2-2mg~\mu_k~x}{m}}$
$v = \sqrt{\frac{(45.0~N/m)(0.280~m)^2-(2)(1.60~kg)(9.80~m/s^2)(0.300)(0.280~m)}{1.60~kg}}$
$v = 0.747~m/s$
The speed when the block loses contact with the spring is 0.747 m/s.
(b) The maximum speed will occur at the point where the force from the spring is equal in magnitude to the force of friction.
$kx = mg~\mu_k$
$x = \frac{mg~\mu_k}{k}$
$x = \frac{(1.60~kg)(9.80~m/s^2)(0.300)}{45.0~N/m}$
$x = 0.1045~m$
We can use this value of $x$ to find the maximum speed of the block.
$\frac{1}{2}mv^2 = \frac{1}{2}mv_{max}^2 +\frac{1}{2}kx^2 - mg~\mu_k~x$
$\frac{1}{2}mv_{max}^2 = \frac{1}{2}mv^2 -\frac{1}{2}kx^2 + mg~\mu_k~x$
$v_{max}^2 = \frac{mv^2 -kx^2 + 2mg~\mu_k~x}{m}$
$v_{max} = \sqrt{\frac{(1.60~kg)(0.747~m/s)^2 -(45.0~N/m)(0.1045~m)^2 + (2)(1.60~kg)(9.80~m/s^2)(0.300)(0.1045~m)}{1.60~kg}}$
$v_{max} = 0.930~m/s$
