HW_08_Solutions_F05_2

HW_08_Solutions_F05_2 - Physics 112 HW#8 Solutions...

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Physics 112 – HW #8 Solutions Fall 2005 1 7-46 [Loop-the-Loop] (a) Forces acting on the car when it is at point B: Σ F = m a : ( is +) N - mg = m v B 2 R N = mg + m v B 2 R Car doesn't fall from track N 0 v B 2 gR Conservation of Energy: K A + U gA = K B + U gB 0 + mgh = 1 2 mv B 2 + mg(2R) (Take y = 0 at ground level.) 1 2 m(gR) + mg(2R) = 5 2 mgR h 5 2 R (b) Forces acting on the car when it is at point C: Cons. of Energy: K A + U gA = K C + U gC 0 + mg(3.50 R) = 1 2 mv C 2 + mgR Solve for v C = 5gR = 5(9.8 m/s 2 )(20.0 m) = 31.3 m/s . a rad = v C 2 R = 5gR R = 5g ( ) Since the car's weight is the only vertical force acting on it: a tan = g ( ) mg m N mg m N a rad a tan a
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Physics 112 – HW #8 Solutions Fall 2005 2 7-72 [Fish on Spring] First, consider the fish hanging from the spring in equilibrium. In this case, the fish's acceleration is a = 0, so by Newton's 2nd Law the fish's weight, mg ( ), is balanced by the contact (or tension) force acting on the fish due to the spring, F s = kd ( ): Σ F = ma kd - mg = 0 k = mg d Why not use Cons. of Energy here? In lowering the fish gently down to its equilibrium position, the force due to the hand must do (-) work on the fish. So K + U g + U s isn't conserved, but K + U g + U s = W hand (< 0). Next, consider what happens when the fish is released from rest after being attached to the end of the unstretched spring. Now the fish descends freely past the equilibrium position (d) until it comes to rest momentarily before turning around and returning back up.
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HW_08_Solutions_F05_2 - Physics 112 HW#8 Solutions...

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