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

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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 - 2 cos 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 a) k ( x - x 0 ) 2 2 = mgh = mgl sin θ l = k ( x - x 0 ) 2 2 mg sin θ b) k ( x 0 - x 1 ) 2 2 = mg ( x 0 - x 1 ) sin θ + mg ( x 2 - x 0 ) sin θ + k ( x 2 - x 0 ) 2 2 k ( x 0 - x 1 ) 2 2 = mgl b sin θ + k ( l b + x 1 - x 0 ) 2 2 k ( x 0 - x 1 ) 2 2 = mgl b sin θ + k ( x 1 - x 0 ) 2 2 + kl 2 b 2 + 2 kl b ( x 1 - x 0 ) 2 l b = 2( x 0 - x 1 ) - 2 m k g sin θ x 0 - x 1 > 0 c) k ( x 0 - x 1 ) 2 2 = mg ( x 0 - x 1 ) sin θ + μ k mg cos θ ( x 0 - x 1 ) μ k = k ( x 0 - x 1 ) 2 mg cos θ - tan θ
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2 Problem 8.44 a) mv 2 f 2 = mv 2 i 2 + mgh v 2 f = v 2 i + 2 gh b) mv 2 i 2 + mgh = F h sinθ + mv 2 f 2 F = sin θ h m 2 ( v 2 f - v 2 i ) + mgh · Problem 8.74 down hill F = 0, F resistance = mg sin θ up hill F = 0, F bicyclist = mg sin θ + F resistance F bicyclist
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