06prob_11

# 06prob_11 - Assigment#6 Page 1 of 3 UC Davis Physics 9A-C...

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Unformatted text preview: Assigment #6, 5/6/11 Page 1 of 3 UC Davis Physics 9A-C Assignment #6 Cole [7.9]  a)  The normal force is perpendicular to the displacement, so it does no work.  The work  done by gravity is  mgh  =  mgR  = 0. 98 J.  b)  When nonconservative forces do work, they change  the energy of the system:   W f  = E - Eø     õ  W f  =  1 2 mv 2- mgh  õ   v = 2 gh + W f m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟   =   v = 2 9.8m/s 2 ( ) 0.5 m ( ) + − 0.22 J 0.2 kg ⎛ ⎝ ⎜ ⎞ ⎠ ⎟  = 2.8 m/s. [7.30]  The magnitude of the friction force on the book is    f  =   μ k mg  = (0.25)(1.5 kg)(9.80 /s 2 ) = 3.68 N.  a) The work done during each part of the motion  is the same, and the total work done is  W  = -2(3.68 N)(8.0 m) = -59 J (round to two places). b) The magnitude of the displacement is Î r  =   2 (8 m) so the work done by friction is . W f = − 2 8.0 m ( ) 3.68 N ( )  = -42 J  c) The work is the same both coming and going, and the total  work done is the same as in part (a), -59 J.  d) The work required to go from one point to another  is path dependent, and the work required for a round trip is not zero, so friction is not a  conservative force. [7.32]  The force is  F x = − dU dx = C 6 d dx 1 x 6 = − 6 C 6 x 7 .  One hydrogen atom particle is at the origin  and the other at  x.   This force is on the particle at  x  and because it is negative, it acts toward the  atom at the origin, so it is an attractive force. [7.37]   a)   F r = − ∂ U dr = 12 a r 13 − 6 b r 7  b) Setting  F r  = 0 and solving for  r  gives  r min  = (2 a/b ) 1/6 . This is  the minimum of potential energy, so the equilibrium is stable. c) U ( r min ) = a r min 12 − b r min 6 = a ((2 a / b ) 1/6 ) 12 − b ((2 a / b ) 1/6 ) 6 = ab 2 4 a 2 − b 2 2 a = − b 2 4 a .   To separate the particles means to give enough energy to get them to zero energy, and requires  ÎE binding = 0 -  E o  = b 2 /4a . d) The expressions for  Eø  and  r min  in terms of  a  and  b  are   E = b 2 4 a r min 6 = 2 a b Multiplying the first by the second and solving for  b  gives  b = 2 E r min 6 , and substituting this into  the first and solving for  a  gives  a = E r min 12 .  Using the given numbers,....
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06prob_11 - Assigment#6 Page 1 of 3 UC Davis Physics 9A-C...

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