Chapter 6_solutions - Chapter 6 Conservation of Energy 2. A...

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1 Chapter 6 Conservation of Energy 2. A 20.0 N force is applied at an angle of 40.0 degrees above the horizontal to a 4.00 kg box. The box moves a horizontal distance of 4.00 meters. The work done by the 20.0 N force is: A. 75.0 J. B. 61.3 J. C. 50.1 J. D. 46.3 J. E. 40.5 J J 3 . 61 40 cos ) 4 )( 20 ( cos m N d F d F W 3. A 30.00 kg box slides up a 12.00 meters long incline at an angle of 30.00 degrees above the horizontal. The change in gravitational potential energy is: A. 2110 J. B. 1467 J. C. 1080 J. D. 1764 J. E. 760.0 J. We need the vertical height h, which is h = d sin = (12m) sin 30 o = 6m. Then E pot = m g h = (30 kg)(9.8 m s -2 )(6 m)=1,764 J 5. A horizontal force is applied to a 4.0 kg box. The box starts from rest moves a horizontal distance of 8.0 meters and obtains a velocity of 6.0 m/s. The system is frictionless. The horizontal force is: A. 3.0 N. B. 5.0 N. C. 7.0 N. D. 9.0 N. E. 11 N. This can be done in two ways: 1) Using kinematics and Newton s 2 nd law: First find the acceleration. Using v f 2 -v i 2 = 2 a x, we find a = v f 2 /2 x = (6 m/s) 2 /(2*8m) = 2.25 m/s 2 in this case. Then the force is given by F = m a = (4.0 kg)(2.25 m/s 2 ) = 9.0 N. 2) Using energy considerations. Since we have no friction in this case, the mechanical energy is conserved. Therefore the work done on the system is all converted into kinetic energy. This means W = E kin . The kinetic energy is E kin = ½ m v 2 = 72 J. The work is W = F d and therefore F = W/d = E kin /d = (72 J)/(8.0 m) = 9.0N 10. As a hiker descends a hill, the work done by gravity on the hiker is:
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Chapter 6_solutions - Chapter 6 Conservation of Energy 2. A...

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