Soln to 5-17 and 5-103

Soln to 5-17 and 5-103 - 5-17 The given factors are M:= 35...

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W 2361 J = W F d = F 169.7 N = F M g μ tan φ ( 29 + 1 μ tan φ ( 29 - = F M g μ cos φ ( 29 sin φ ( 29 + cos φ ( 29 μ sin φ ( 29 - = F cos φ ( 29 μ sin φ ( 29 - ( 29 M g μ cos φ ( 29 sin φ ( 29 + ( 29 = Solving for F, we find F cos φ ( 29 M g sin φ ( 29 - μ M g cos φ ( 29 F sin φ ( 29 + ( 29 = Therefore, when we substitute N into this equation, we get F cos φ ( 29 M g sin φ ( 29 - μ N - 0 = gives n F x 0 = N M g cos φ ( 29 F sin φ ( 29 + = N M g cos φ ( 29 - F sin φ ( 29 - 0 = gives n F y 0 = d 13.9m = d 3.6 m sin φ ( 29 = h 3.6 m = μ 0.2 = φ 15 deg = M 35 kg = The given factors are: 5-17
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E f E i = Therefore K f U f + E f = gives us 1 2 K y 2 M g y ( 29 - E f - 0 = The solution to this quadratic equation is y M g M g ( ) 2 4 k 2 E f + + k = y 0.166 m = Part (b) When y = 0.05 m, what is the velocity of the ball? y 0.05 m = K 1 2 M v 2 = U M - g ⋅ ∆ y 1 2 k ⋅ ∆ y 2 + = and K + U still equals E
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This note was uploaded on 01/05/2012 for the course PHYS 1401 taught by Professor Jamesr.boyd during the Spring '09 term at Collins.

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Soln to 5-17 and 5-103 - 5-17 The given factors are M:= 35...

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