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What is the final velocity of a hoop that rolls without slipping down a 4.50 m high hill, starting from rest?

m/s

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  • 8 years ago
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    By using conservation of mechanical energy.........Me at top=Me at bottom......expanding: (k+u)top=(k+u)bottom.......at the top the is no kinetic energy since the ball is stationary so k=0 and at the bottom the is no potential energy since the ball is moving at maximum speed so u=0 ths means (U)top=(K)bottom...expanding: (mgh)t=(1/2mv^2)b......mass is constant it will cancel out.sustitution: (m)(9.8)(4.50)=(1/2)(m)(v^2).......simplify: (m)(44.11)=(1/2m)(v^2).....divide boyh sides by (1/2m)....the m cancell out........88,2=v^2.......square both sides,therefore v=9,39m/s

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