Question
A hoop of radius 0.50 m and a mass of 0.20 kg is released from rest and allowed to roll down an inclined plane. How fast is it moving after dropping a vertical distance of 3.0 m ?
Answers
Henry
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*3 = 58.8
V = 7.67 m/s.
V^2 = 0 + 19.6*3 = 58.8
V = 7.67 m/s.
James
It is 5.4 m/s