Physics Solution Manual for 1100 and 2101

40 m above the ground solving for the final speed at

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Unformatted text preview: neglected, the only force that acts on the golf ball is the conservative gravitational force (its weight). Since the maximum height of the trajectory and the initial speed of the ball are known, the conservation of mechanical energy can be used to find the kinetic energy of the ball at the top of the highest point. The conservation of mechanical energy can also be used to find the speed of the ball when it is 8.0 m below its highest point. SOLUTION a. The conservation of mechanical energy, Equation 6.9b, states that 2 2 1 mv + mgh = 1 mv + mgh f f 0 0 2 2 123 KEf Solving this equation for the final kinetic energy, KEf, yields 308 WORK AND ENERGY KE f = 1 mv0 + mg ( h0 − hf 2 2 = 1 2 ) ( 0.0470 kg )( 52.0 m / s )2 + ( 0.0470 kg ) ( 9.80 m / s 2 ) ( 0 m − 24.6 m ) = 52.2 J b. The conservation of mechanical energy, Equation 6.9b, states that 1 mv 2 + mgh 24 14 f 244 f 3 = Ef 1 mv 2 + mgh 24 14 0244 0 3 Ef The mass m can be eliminated algebraically from this equation, since it appears as a factor in every term. Solving for vf and noting that the final height is hf = 24.6 m – 8.0 m = 16.6 m, we have that vf = v0 + 2 g ( h0 − hf ) = 2 ( 52.0 m / s )2 + 2 ( 9.80 m / s 2 ) ( 0 m − 16.6 m ) = 48.8 m / s ______________________________________________________________________________ 42. REASONING If air resistance is ignored, the only nonconservative force that acts on the person is the tension in the rope. However, since the tension always points perpendicular to the circular path of the motion, it does no work, and the principle of conservation of mechanical energy applies. SOLUTION Let the initial position of the person be at the top of the cliff, a distance h0 above the water. Then, applying the conservation principle to path 2, we have 1 mvf2 + mghf 24 14 244 3 Ef ( at water ) = 1 2 mv0 + mgh0 24 14 244 3 (6.9b) E0 ( at release point ) Solving this relation for the speed at the release point, we have v0 = vf2 + 2 g ( hf − h0 ) But vf = 13.0 m/s, since the speed of entry is the same for either path. Therefore, 2 v0 = vf2 + 2 g ( hf − h0 ) = (13.0 m/s ) + 2 ( 9.80 m/s 2 ) ( −5.20 m ) = 8.2 m/s _____________________________________________________________________________________________ Chapter 6 Problems 43. REASONING AND SOLUTION Since friction and air resistance are negligible, mechanical energy is conserved. Thus, the ball will have the same speed at the bottom of its swing whether it is moving toward or away from the crane. As the drawing shows, when the cable swings to an angle of 20.0°, the ball will rise a distance of 309 20.0º L L − L cos 20.0º L cos 20.0º h0 = 0 m h = L (1 – cos 20.0°) where L is the length of the cable. Applying the conservation of energy with h0 = 0 m gives 1 2 2 mv0 = 1 mvf2 + mgh 2 Solving for vf gives 2 vf = v0 − 2 gL (1 − cos 20.0° ) = ( 5.00 m/s ) 2 − 2 ( 9.80 m/s 2 ) (12.0 m )(1 − cos 20.0° ) = 3.29 m/s ______________________________________________________________________________ 44. REASONING To find the maximum height H above the end of the track we will analyze the projectile motion of the skateboarder after she leaves the track. For this analysis we will use the principle of conservation of mechanical energy, which applies because friction and air resistance are being ignored. In applying this principle to the projectile motion, however, we will need to know the speed of the skateboarder when she leaves the track. Therefore, we will begin by determining this speed, also using the conservation principle in the process. Our approach, then, uses the conservation principle twice. SOLUTION Applying the conservation of mechanical energy in the form of Equation 6.9b, we have 2 1 mvf2 + mghf = 1 mv0 + mgh0 2 2 14 244 4 3 14 244 4 3 Final mechanical energy at end of track Initial mechanical energy on flat part of track We designate the flat portion of the track as having a height h0 = 0 m and note from the drawing that its end is at a height of hf = 0.40 m above the ground. Solving for the final speed at the end of the track gives 2 vf = v0 + 2 g ( h0 − hf ) = ( 5.4 m/s ) 2 + 2 ( 9.80 m/s 2 ) ( 0 m ) − ( 0.40 m ) = 4.6 m/s 310 WORK AND ENERGY This speed now becomes the initial speed v0 = 4.6 m/s for the next application of the conservation principle. At the maximum height of her trajectory she is traveling horizontally with a speed vf that equals the horizontal component of her launch velocity. Thus, for the next application of the conservation principle vf = (4.6 m/s) cos 48º. Applying the conservation of mechanical energy again, we have 2 1 2 mvf + mghf 14 244 4 3 Final mechanical energy at maximum height of trajectory = 2 1 2 mv0 + mgh0 14 244 4 3 Initial mechanical energy upon leaving the track Recognizing that h0 = 0.40 m and hf = 0.40 m + H and solving for H give 1 2 H= 45. 2 mvf2 + mg ( 0.40 m ) + H = 1 mv0 + mg ( 0.40 m ) 2 2 v0 − vf2 2g = ( 4.6 m/s ) 2 − ( 4.6 m/s ) cos 48° 2 ( 9.80 m/s 2 ) 2 = 0.60 m SSM REASONING Friction and air resist...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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