PE = m*V^2 = 0.164 J.
.0508*V^2 = 0.164
V^2 = 3.228
V = 1.80 m/s.
What linear speed must a 0.0508-kg hula hoop have if its total kinetic energy is to be 0.164 J? Assume the hoop rolls on the ground without slipping.
1 answer