Answer:
[tex]v_f = 20 m/s[/tex]
Explanation:
Since the hoop is rolling on the floor so its total kinetic energy is given as
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2[/tex]
now for pure rolling condition we will have
[tex]v = R\omega[/tex]
also we have
[tex]I = mR^2[/tex]
now we will have
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}[/tex]
[tex]KE = mv^2[/tex]
now by work energy theorem we can say
[tex]W = KE_f - KE_i[/tex]
[tex]842 J = mv_f^2 - mv_i^2[/tex]
[tex]842 = 3(v_f^2) - 3\times 11^2[/tex]
now solve for final speed
[tex]v_f = 20 m/s[/tex]
