Answer
The cylinder travels $~1.53~meters~$ along the incline.
Work Step by Step
We can find an expression for the total kinetic energy of the hollow cylinder at the bottom of the incline:
$KE = KE_{tr}+KE_{rot}$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}(MR^2)~(\frac{v}{R})^2$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}Mv^2$
$KE = Mv^2$
We can use conservation of energy to find the vertical height where the hollow cylinder stops moving:
$U_g = KE$
$Mgh = Mv^2$
$h = \frac{v^2}{g}$
$h = \frac{(3.00~m/s)^2}{9.80~m/s^2}$
$h = 0.9184~m$
We can find the distance the cylinder rolls along the incline:
$sin~\theta = \frac{h}{d}$
$d = \frac{h}{sin~\theta}$
$d = \frac{0.9184~m}{sin~37.0^{\circ}}$
$d = 1.53~m$
The cylinder travels $~1.53~meters~$ along the incline.
