Hw07_Solutions

Hw07_Solutions - 1 Problem 7.69 a) W = mgh mv 2 b) mgh = 2...

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1 Problem 7.69 a) W = mgh b) mgh = mv 2 2 v = 2 gh c) mg ( h + x ) = kx 2 2 kx 2 - 2 mgx - 2 mgh = 0 x = 2 mg + p 4 m 2 g 2 + 8 mghk 2 k We don’t take the other root because it’s negative. Problem 7.72 A = h R 0 F ( x ) dx = mv 2 2 h R 0 Ax 3 dx = Ah 4 4 v = h 2 r A 2 m Problem 8.27 kx - mg 5 mg the maximum acceleration is when the maximum displacement kx 6 mg or the limiting case kx = 6 mg energy conservation mg ( h + x ) = mv 2 2 x = 6 mg k 36 m 2 g 2 k 2 k 2 = mg ± h + 6 mg k 12 m 2 g 2 k = mgh k = 12 mg h Problem 8.28 a) N=0 mg cos θ = mv 2 R v 2 = Rg cos θ energy conservation mv 2 2 = mgR (1 - cos θ ) v 2 = 2 gR (1 - cos θ ) cos θ = 2 - 2cos θ cos θ = 2 3 θ = 48 . 2 b) greater angle, because the angle depends only on the speed v 2 = Rg cos θ If friction is present, then it does some work. To get the same speed the object should descend more, which corresponds to the greater angle. Problem 8.42
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Hw07_Solutions - 1 Problem 7.69 a) W = mgh mv 2 b) mgh = 2...

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