Asked by Andtrew
If a yo yo has a speed of 3 m/s at the top of the circle, whose radius is 80 cm, What is its speed at the bottom
Answers
Answered by
Scott
the kinetic energy at the bottom is the kinetic energy at the top plus the gravitational potential across the circle
.5 m (Vb)^2 = [.5 m (Vt)^2] + [m g h]
dividing by .5 m
(Vb)^2 = (Vt)^2 + 2 g h
v^2 = 3^2 + [2 * 9.8 * (2 * .8)]
.5 m (Vb)^2 = [.5 m (Vt)^2] + [m g h]
dividing by .5 m
(Vb)^2 = (Vt)^2 + 2 g h
v^2 = 3^2 + [2 * 9.8 * (2 * .8)]
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