Physics Solution Manual for 1100 and 2101

# 6 where wnc is the nonconservative work done by air

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Unformatted text preview: at given, the net work W is the total area under the graph. For the purpose of calculation, it is convenient to divide this area into two pieces, a triangle (from s = 0 m to s = 10.0 m) and a rectangle (from s = 10.0 m to s = 20.0 m). Both pieces have the same width (10.0 m) and height (10.0 N), so the triangle has half the area of the rectangle. 2W . m done on the object from s = 0 m to s = 10.0 m SOLUTION Solving W = 1 mvf2 for the final speed of the object, we obtain vf = 2 The net work W is the sum of the work W0,10 (the triangular area) and the work W10,20 done on the object from s = 10.0 m to s = 20.0 m (the rectangular area). We calculate the work W10,20 by multiplying the width (10.0 m) and height (10.0 N) of the rectangle: W10,20 = ( Width ) ( Height ) . The triangle’s area, which is the work W0,10, is half this amount: W0,10 = 1 W10,20 . Therefore, the net work 2 W = W0,10 + W10,20 done during the entire interval from s = 0 m to s = 20.0 m is 332 WORK AND ENERGY W = 1 W10,20 + W10,20 = 3 W10,20 = 2 2 3 2 ( Width ) ( Height ) We can now calculate the final speed of the object from vf = vf = 2 2W : m ( 3 ) ( Width )( Height ) = 3 (10.0 m )(10.0 N ) = 7.07 m/s 2 m 6.00 kg 76. REASONING AND SOLUTION The work done by the retarding force is given by Equation 6.1: W = ( F cos θ )s . Since the force is a retarding force, it must point opposite to the direction of the displacement, so that θ = 180° . Thus, we have W = ( F cosθ )s = (3.0 × 103 N)(cos 180°)(850 m) = –2.6 × 106 J The work done by this force is negative , because the retarding force is directed opposite to the direction of the displacement of the truck. ______________________________________________________________________________ 77. SSM REASONING The only two forces that act on the gymnast are his weight and the force exerted on his hands by the high bar. The latter is the (non-conservative) reaction force to the force exerted on the bar by the gymnast, as predicted by Newton's third law. This force, however, does no work because it points perpendicular to the circular path of motion. Thus, Wnc = 0 J, and we can apply the principle of conservation of mechanical energy. SOLUTION The conservation principle gives 1 mv 2 + mgh 2 4f 244f 14 3 Ef 1 2 = mv 0 + mgh0 24 14 244 3 E0 Since the gymnast's speed is momentarily zero at the top of the swing, v0 = 0 m/s. If we take hf = 0 m at the bottom of the swing, then h0 = 2 r , where r is the radius of the circular path followed by the gymnast's waist. Making these substitutions in the above expression and solving for vf , we obtain vf = 2 gh0 = 2 g(2r) = 2(9.80 m/s2 )(2 × 1.1 m) = 6.6 m/s ______________________________________________________________________________ Chapter 6 Problems 333 78. REASONING a. Since there is no air friction, the only force that acts on the projectile is the conservative gravitational force (its weight). The initial and final speeds of the ball are known, so the conservation of mechanical energy can be used to find the maximum height that the projectile attains. b. When air resistance, a nonconservative force, is present, it does negative work on the projectile and slows it down. Consequently, the projectile does not rise as high as when there is no air resistance. The work-energy theorem, in the form of Equation 6.6, may be used to find the work done by air friction. Then, using the definition of work, Equation 6.1, the average force due to air resistance can be found. SOLUTION a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that 2 2 Ef E0 1 mv + mgh = 1 mv + mgh 24 14 f244f 14 02440 3 24 3 The mass m can be eliminated algebraically from this equation since it appears as a factor in every term. Solving for the final height hf gives 1 2 hf = ( v02 − vf2 ) + h 0 g Setting h0 = 0 m and vf = 0 m/s, the final height, in the absence of air resistance, is 2 hf = 2 vo − vf 2g = (18.0 m / s )2 − ( 0 m/s )2 ( 2 9.80 m / s 2 ) = 16.5 m b. The work-energy theorem is Wnc = ( 1 mv 2 f 2 2 − 1 mv0 2 ) + ( mgh f − mgh0 ) (6.6) where Wnc is the nonconservative work done by air resistance. According to Equation 6.1, the work can be written as Wnc = ( FR cos 180° ) s , where FR is the average force of air resistance. As the projectile moves upward, the force of air resistance is directed downward, so the angle between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the displacement is the difference between the final and initial heights, s = hf – h0 = 11.8 m. With these substitutions, the work-energy theorem becomes ( ) 2 − FR s = 1 m vf2 − vo + mg ( hf − h0 ) 2 334 WORK AND ENERGY Solving for FR gives FR = = 1m 2 ( vf2 − vo2 ) + mg ( hf − h0 ) −s 1 2 ( 0.750 kg ) ( 0 m/s )2 − (18.0 m/s )2 + ( 0.750 kg ) ( 9.80 m/s2 ) (11.8 m ) = − (11.8 m ) 2.9 N ______________________________________________________________________________ 79. SSM REASONING AND SOLUTION We will assume that the tug-of-war rope remains parallel to the ground, so that the force that moves team B is in the same direction as the...
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