Answer
The coefficient of kinetic friction between the block and the tabletop is 0.41.
Work Step by Step
The work done by friction will be equal in magnitude to the initial potential energy in the spring.
$F_f~d = \frac{1}{2}kx^2$
$mg~\mu_k~d = \frac{1}{2}kx^2$
$\mu_k = \frac{kx^2}{2mg~d}$
$\mu_k = \frac{(100~N/m)(0.20~m)^2}{(2)(0.50~kg)(9.80~m/s^2)(1.00~m)}$
$\mu_k = 0.41$
The coefficient of kinetic friction between the block and the tabletop is 0.41.
