ch6 problem40

Physics: Principles with Applications

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The block slides without friction, there is no nonconservative force acting on the block, thus its total mechanical energy is conserved. We should first determine what the minimum speed the block must have at point B so that it does not fall off the track. Once we find that, we can use the conservation of energy to find the height at point A. To determine the minimum speed at point B, let recall what we learned in last chapter. First draw a free-body diagram for the block. The net radial force is, + = + = mg F F F F N G N R From Newton’s 2 nd law, = = r v m ma F R R 2 Therefore, mg F r v m N + = 2 When the block falls off the track, it loses contact with the track, thus
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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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ch6 problem40 - F N = 0 hence mg r v m = 2 min We have gr v...

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