HW_13_Solutions_F05 - Physics 112 - HW #13 Solutions 13-19...

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- 1 - Physics 112 - HW #13 Solutions Fall 2005 13-19 [Oscillation Function] x(t) = (7.40 cm) cos [ (4.16 s -1 )t – 2.42] ; compare with x(t) = x max cos [ ω t + φ ] (a) T = 2 π ω = 2 π 4.16 s -1 = 1.51 s (b) ω = k m , so k = m ω 2 = (1.50 kg)(4.16 s -1 ) 2 = 26.0 N/m (c) v max = ω x max = (4.16 s -1 )(7.40 cm) = 30.8 cm/s (d) F max = k x max = (26.0 N/m)(7.40 cm) = 1.92 N (e) x(1 sec) = (7.40 cm) cos [ (4.16 s -1 )(1 sec) – 2.42] = -1.25 cm v(1 sec) = -(7.40 cm)(4.16 s -1 ) sin [ (4.16 s -1 )(1 sec) – 2.42] = -30.3 cm/s, so speed is 30.3 cm/s a(1 sec) = - ω 2 x(1 sec) = -(4.16 s -1 ) 2 (-1.25 cm) = 21.6 cm/s 2 F(1 sec) = m a(1 sec) = (1.50 kg)(21.6 cm/s 2 ) = 0.324 N 13.27 [Energy in SHM] (a) E tot = K + U = ½ mv 2 + ½ kx 2 = ½ [(0.150 kg)(0.300 m/s) 2 + (300 N/m)(0.0120 m) 2 ] = 2.84 x 10 -2 J (b) ½ k x m 2 = E tot , so x m = 2E tot k = 2( 2.84 x 10 -2 J ) 300 N/m = 1.38 cm (c) ½ m v m 2 = E tot , so v m = 2E tot m = 2( 2.84 x 10 -2 J ) 0.150 kg = 0.615 m/s 13.71 [SHM of Butcher’s Scale] (a) First we need to know the speed v 1 of the steak just before it hits the scale. Let mass of steak be M =2.20 kg and mass of pan be m = 0.200 kg. Let speed of steak just before it hits pan be v 1 . Energy is conserved as steak falls: ½ Mv 1 2 = Mgh v 1 = 2gh = 2(9.81 m/s 2 )(0.40 m) = 2.80 m/s
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- 2 - Next we need to know the speed v 2 of the steak and the pan just after the steak hits the pan. Momentum is conserved in the collision:
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HW_13_Solutions_F05 - Physics 112 - HW #13 Solutions 13-19...

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