PHY 101 Exam 4

PHY 101 Exam 4 - ! ! f = N ! ! F = ma The only force acting...

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x x 2 2 x F ma The only force acting on the crate is friction. f N (mg) ma a g (0.250)(9.8m /s ) 2.45m /s = = μ = μ = = μ = = ! ! nd A B First, we assume a direction of motion, which for this problem would be the y-direction, and let it be positive. Let's apply Newton's 2 Law to each block individually: Block A: T = m a Block B: m g T - B B B A B B A m a Because the blocks are connected, the acceleration must be the same. Notice that we have +T and -T in the 2 equations. If we add them together, T will be eliminated. m g a(m m ) m g (1.60 a (m m ) = = + = = + 2 2 kg)(9.8m / s ) 3.14m / s (3.40 1.60)kg = + f N = μ ! !
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The apparent weight is the Normal force. 2 2 2 2 F ma Everything is happening in the y-direction, so we can drop the vector notation. v N mg m Substitute 7mg for the Normal force. r v 7mg mg m Notice that the pilot's mass cancels out. r v (320m / s) r 8g 6 = ± - = - = = = ! ! 2
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This note was uploaded on 10/13/2011 for the course PHY 101 taught by Professor Ashkenkai during the Fall '08 term at FIU.

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PHY 101 Exam 4 - ! ! f = N ! ! F = ma The only force acting...

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