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Unformatted text preview: F N = 0, hence mg r v m + = 2 min We have gr v = 2 min A B h 2r F G F N Now we use the conservation of energy: B B A A PE KE PE KE + = + The block starts from rest at point A, so KE A = 0 , and we choose the ground level as the reference level, then B B A mgy mv mgy + = + 2 2 1 The mass cancels out and we have gr r g gr r g v gh B 5 . 2 ) 2 ( 2 1 ) 2 ( 2 1 2 min min = + = + = Therefore, r h 5 . 2 min = This is the minimum height from which the block needs to be released in order for it to stay on the track even at the top of the loop....
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