Answer:
The hollow cylinder rolled up the inclined plane by 1.91 m
Explanation:
From the principle of conservation of mechanical energy, total kinetic energy = total potential energy
[tex]M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh[/tex]
The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.
[tex]\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh[/tex]
moment of inertia, I, of a hollow cylinder = ¹/₂mr²
substitute for I in the equation above;
[tex]\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\[/tex]
[tex]h = \frac{3}{4g}(v_1^2 -v_f^2)[/tex]
given;
v₁ = 5.0 m/s
vf = 0
g = 9.8 m/s²
[tex]h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m[/tex]
Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m
